case study questions on electricity class 10

Case Study Questions Class 10 Science Chapter 12 Electricity

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Case study Questions Class 10 Science Chapter 12 are very important to solve for your exam. Class 10 Science Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 10 Science Chapter 12 Electricity

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In CBSE Class 10 Science Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Electricity Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

The heating effect of current is obtained by the transformation of electrical energy into heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser, etc. all are based on the heating effect of current.

(i) What are the properties of heating elements? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point.

Answer: (b) Low resistance, high melting point

(ii) What are the properties of an electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

The relationship between potential difference and the current was first established by George Simon Ohm. This relationship is known as Ohm’s law. According to this law, the current passed through a conductor is proportional to the potential difference applied between its ends provided the temperature remains constant i.e. I ∝ V or V = IR where R is the constant for the conductor and it is known as the resistance of the conductor. Although Ohm’s law has been found valid over a large class of materials, there are some materials that do not hold Ohm’s law.

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

Case Study 3

3.1) The current passing through an electric kettle has been doubled. The heat produced will become : (a) half (b) double (c) four times (d) one fourth

Answer(c) four times

3.2) The heat produced in a wire of resistance ‘a’ when a current ‘b’ flows through it in time ‘c’ is given by : (a) a 2 bc (b) abc 2 (c) ab 2 c (d) abc

Answer(c) ab2c

3.3) What are the properties of heating element ? (a) high resistance, high melting point (b) low resistance, high melting point (c) low resistance, high melting point (d) low resistance, low melting point

Answer (a) high resistance, high melting point

3.4) Calculate the heat produced when 96,000 coulombs of charge is transferred in one hour through a potential difference of 50 volts. (a) 4788 J (b) 4788 kJ (c) 478 kJ (d) 478 J

Answer (b) 4788 kJ

3.5) Which of the following characteristic is not suitable for a fuse wire ? (a) thin and short (b) low melting point (c) thick and short (d) high resistance

Answer (c) thick and short

Case Study 4

Substance through which charges cannot pass is called insulators. Glass, pure water, and all gases are insulators. Insulators are also called dielectrics. In insulators, the electrons are strongly bound to their atoms and cannot get themselves freed. Thus, free electrons are absent in insulators. Insulators can easily be charged by friction. This is due to the reason that when an electric charge is given to an insulator, it is unable to move freely and remains localized. But this does not mean that conductors cannot be charged. A metal rod can be charged by rubbing it with silk if it is held in a handle of glass or amber

4.1) Calculate the current in a wire if a 1500 C charge is passed through it in 5 minutes. (a) 2 A (b) 5 A (c) 3 A (d) 4 A

Answer (b) 5 A

4.2) Electrons and conventional current flows in : (a) The same direction (b) The opposite direction (c) Any direction (d) Can’t say

Answer (b) The opposite direction

4.3) If the current passing through a lamp is 5 A, what charge passes in 10 second ? (a) 0.5 C (b) 3 C (c) 5 C (d) 50 C

Answer (d) 50 C

4.4) One-coulomb charge is equivalent to the charge contained in : (a) 6.2 × 10 19 electrons (b) 2.6 × 10 18 electrons (c) 2.65 × 10 19 electrons (d) 6.25 × 10 18 electrons

Answer (d) 6.25 × 1018 electrons

4.5) When an electric lamp is connected to 12 V battery, it draws a current of 0.5 A. The power of the lamp is : (a) 0.5 W (b) 6 W (c) 12 W (d) 24 W

Answer (b) 6 W

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Case Study Questions for Class 10 Science Chapter 12 Electricity

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Question 1:

Read the following and answer the questions from (i) to (v) given below: The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure).

case study questions on electricity class 10

Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved.

(i) The SI unit of electric energy per unit time is (a) joule (b) joule-second (c) watt (d) watt-second

(ii) Kilowatt-hour is equal to (a) 3.6 ×10 4 J (b) 3.6 ×10 6 J (c) 36 ×10 6 J (d) 36 ×10 4 J

(iii) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is (a) doubled (b) half (c) remains same (d) four times

(iv) The power of a lamp is 60 W. The energy consumed in 1 minute is (a) 360 J (b) 36 J (c) 3600 J (d) 3.6 J

(v) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 Ω for 30 minutes. (a) 40 kJ (b) 60 kJ (c) 10 kJ (d) 90 kJ

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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

Here we are providing you with Class 10 Science Chapter 12 Electricity Case Study Questions, by practicing these Case Study and Passage Based Questions will help you in your Class 10th Board Exam.

Case Study Chapter 12 Electricity

Here, we have provided case-based/passage-based questions for Class 10 Science Chapter 12 Electricity

Case Study/Passage Based Questions

Question 1:

Answer: (b) Low resistance, high melting point

Answer: (c) High resistance, low melting point

(iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(a) doubled	(b) halved
(c) four times	(d) one fourth times

Answer: (a) doubled

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(a) 4 times	(b) 2 times
(c) 6 times	(d) 8 times

Answer: (b) 2 times

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 ohm, the amount of heat produced is

(a) 250 J	(b) 5000J
(c) 750J	(d) 1000J

Answer: (c) 750J

Question 2:

2.1) Name the law which is illustrated by the VI graph. (a) Lenz law (b) Faraday’s law (c) Ohm’s law (d) Newton’s law

Answer(c) Ohm’s law

2.2) By increasing the voltage across a conductor, the (a) current will decrease (b) current will increase (c) resistance will increase (d) resistance will decrease

Answer(b) current will increase

2.3) When a battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is (a) 9 Ohm (b) 0.9 Ohm (c) 90 Ohm (d) 900 Ohm

Answer(c) 90 Ohm

2.4) If both the potential difference and resistance in a circuit are doubled then : (a) current remains same (b) current becomes double (c) current becomes zero (d) current becomes half

Answer(a) current remains same

2.5) Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become : (a) double (b) half (c) one fourth (d) 4 time

Answer(b) half

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Electricity Case Study Based Questions Class 10

Students who are studying in CBSE class 10 board, need to get the knowledge about the Electricity Case Study Based Questions. Case based questions are generally based on the seen passages from the chapter Electricity. Through solving the case based questions, students can understand each and every concept.

With the help of Electricity Case Study Based Questions, students don’t need to memorise each answer. As answers for these case studies are already available in the given passage. Questions are asked through MCQs so student’s won’t take time to mark the answers. These multiple choice questions can help students to score the weightage of Electricity.

Electricity Case Study Based Questions with Solutions

Selfstudys provides case studies for the Class 10 Science chapter Electricity with solutions. The Solutions can be helpful for students to refer to if there is a doubt in any of the case studies problems. The solutions from the Selfstudys website are easily accessible and free of cost to download. This accessibility can help students to download case studies from anywhere with the help of the Internet.

Electricity Case Study Based Questions with solutions are in the form of PDF. Portable Document Format (PDF) can be downloaded through any of the devices: smart phone, laptop. Through this accessibility, students don't need to carry those case based questions everywhere.

Features of Electricity Case Study Based Questions

Before solving questions, students should understand the basic details of Electricity. Here are the features of case based questions on Electricity are:

These case based questions start with short or long passages. In these passages some concepts included in the chapter can be explained.
After reading the passage, students need to answer the given questions. These questions are asked in the Multiple Choice Questions (MCQ).
These case based questions are a type of open book test. These case based questions can help students to score well in the particular subject.
These Electricity Case Study Based Questions can also be asked in the form of CBSE Assertion and Reason .

Benefits of Solving Electricity Case Study Based Questions

According to the CBSE board, some part of the questions are asked in the board exam question papers according to the case studies. As some benefits of solving Electricity Case Study Based Questions can be obtained by the students. Those benefits are:

Through solving case studies students will be able to understand every concept included in the chapter Electricity
Passages included in the case study are seen passages, so students don’t need to struggle for getting answers. As these questions and answers can be discussed by their concerned teacher.
Through these students can develop their observation skills. This skill can help students to study further concepts clearly.
Case studies covers all the concepts which are included in the Electricity

How to Download Electricity Case Based Questions?

Students studying in CBSE class 10 board, need to solve questions based on case study. It is necessary for students to know the basic idea of Electricity Case Study Based Questions. Students can obtain the basic idea of case based questions through Selfstudys website. Easy steps to download it are:

Open Selfstudys website.

Electricity Case Study, Electricity Case Based Questions

Bring the arrow towards CBSE which is visible in the navigation bar.

A pop-up menu will appear, Select case study from the list.

New page will appear, select 10 from the list of classes.

Select Science from the subject list.

And in the new page, you can access the Electricity Case Study Based Questions.

Tips to solve Electricity Case Study Questions-

Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science.

Generally, the case based questions are in the form of Multiple Choice Questions (MCQs).
Students should start solving the case based questions through reading the given passage.
Identify the questions and give the answers according to the case given.
Read the passage again, so that you can easily answer the complex questions.
Answer according to the options given below the questions provided in the Electricity Case Study Based Questions.

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Case Study Chapter 12 Electricity

Please refer to Chapter 12 Electricity Case Study Questions with answers provided below. We have provided Case Study Questions for Class 10 Science for all chapters as per CBSE, NCERT and KVS examination guidelines. These case based questions are expected to come in your exams this year. Please practise these case study based Class 10 Science Questions and answers to get more marks in examinations.

Case Study Questions Chapter 12 Electricity

Case/Passage – 1

Two tungston lamps with resistances R1 and R2 respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: R 1 > R 2 .

Question: Which lamp will glow more brightly when they are connected in parallel? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: Which lamp will glow more brightly when they are connected in series? (a) Bulb having lower resistance (b) Bulb having higher resistance (c) Both the bulbs (d) None of the two bulbs

Question: If the lamp of resistance R 2 now burns out and the lamp of resistance R1 alone is plugged in, will the illumination increase or decrease? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) None

Question: If the lamp of resistance R 1 now burns out, how will the illumination produced change? (a) Net illumination will increase (b) Net illumination will decrease (c) Net illumination will remain same (d) Net illumination will reduced to zero

Question: Would physically bending a supply wire cause any change in the illumination? (a) Illumination will remain same (b) Illumination will increase (c) Illumination will decrease (d) It is not possible to predict from the given datas

Case/Passage – 2

The rate at which electric energy is dissipated or consumed in an electric circuit. This is termed as electric power, P = IV, According to Ohm’s law V = IR We can express the power dissipated in the alternative forms P =I 2 R=V 2 /R

If 100W – 220V is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts. The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy. The total energy used in a circuit = power of the electric circuit × time.

Question: Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) V 2 /R

Question: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in sereis and then in parallel in an electric circuit. The ratio of heat produced in series and in parallel combinations would be– (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Question: In an electrical circuit, two resistors of 2Ω and 4Ω respectively are connected in series to a 6V battery. The heat dissipated by the 4Ω resistor in 5s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J

Question: In an electrical circuit three incandescent bulbs. A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A (d) Brightness of bulb C will be less than that of B

Question: An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be– (a) 100 W (b) 75 W (c) 50 W (d) 25 W

Case/Passage – 3

Answer the following questions based on the given circuit.

Question: The equivalent resistance between points A and B is (a) 7Ω (b) 6Ω (c) 13Ω (d) 5Ω

Question: The potential drop across the 3Ω resistor is (a) 1 V (b) 1.5 V (c) 2 V (d) 3 V

Question: The current flowing through in the given circuit is (a) 0.5 A (b) 1.5 A (c) 6 A (d) 3 A

Case/Passage – 4

Answer the following questions based on the given circuit.

Question: The current through each resistor is (a) 1 A (b) 2.3 A (c) 0.5 A (d) 0.75 A

Question: The equivalent resistance between points A and B, is (a) 12 Ω (b) 36 Ω (c) 32 Ω (d) 24 Ω

Question: The potential drop across the 12Ω resistor is (a) 12 V (b) 6 V (c) 8 V (d) 0.5 V

Case/Passage – 5

Question: The equivalent resistance between points A and B (a) 6.2 Ω (b) 5.1 Ω (c) 13.33 Ω (d) 1.33 Ω

Question: The current through the 4.0 ohm resistor is (a) 5.6 A (b) 0.98 A (c) 0.35 A (d) 0.68 A

Question: The current through the battery is (a) 2.33 A (b) 3.12 A (c) 4.16 A (d) 5.19 A

Case/Passage – 6

Question: The total resistance of the circuit is (a) 2 Ω (b) 4 Ω (c) 1.5 Ω (d) 0.5 Ω

Question: The current flowing through 6Ω resistor is (a) 0.50 A (b) 0.75 A (c) 0.80 A (d) 0.25

Question: The current flowing through 0.5Ω resistor is (a) 1 A (b) 1.5 A (c) 3 A (d) 2.5 A

Case/Passage – 7

Ohm’s law gives the relationship between current flowing through a conductor with potential difference across it provided the physical conditions and temperature remains constant. The electric current flowing in a circuit can be measured by an ammeter. Potential difference is measured by voltmeter connected in parallel to the battery or cell. Resistances can reduce current in the circuit. A variable resistor or rheostat is used to vary the current in the circuit.

Question. Which type of conductor is represented by the graph given alongside?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What is the slope of graph in (i) equal to? (a) V (b) I (c) R (d) VI

Question. Which of the following is the factor on which resistance of a conductor does not depend? (a) Length (b) Area (c) Temperature (d) Pressur

Question. What type of conductor is represented by the following graph?

(a) Non-ohmic conductor like thermistor (b) Non-ohmic conductor like metal filament (c) Ohmic conductor like copper (d) None of these

Question. What type of conductors are represented by the following graph?

Study this table related to material and their resistivity and answer the questions that follow.

Question. Which of the following is used in transmission wires? (a) Cr (b) Al (c) Zn (d) Fe

Question. Which is the best conducting metal? (a) Cu (b) Ag (c) Au (d) Hg

Question. Which of the following is used as a filament in electric bulbs? (a) Nichrome (b) Tungsten (c) Manganese (d) Silver

Question. What is the range of resistivity in metals, good conductors of electricity? (a) 10–8 to 10–6 Wm (b) 10–6 to 10–4 Wm (c) 1010 to 1014 Wm (d) 1012 to 1014 Wm

Question. Which property of the alloy makes it useful in heating devices like electric iron, toasters, immersion rods, etc.? (a) Higher resistivity (b) Do not oxidise at low temperature (c) Do not reduce at high temperature (d) Oxidise at high temperature

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Case Study Questions Class...

Case Study Questions Class 10 Science

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Download Case study questions for CBSE class 10 Science in PDF format from the myCBSEguide App . We have the new pattern case study-based questions for free download. Class 10 Science case study questions

This article will guide you through:

What are case study questions?

Sample Papers with Case Study questions
Class 10 Science Case Study question examples
How to get case-based questions for free?
How to attempt the case-based questions in Science?

Questions based on case studies are some real-life examples. The questions are asked based on a given paragraph i.e. Case Study. Usually, 4-5 questions are asked on the basis of the given passage. In most cases, these are either MCQs or assertion & reason type questions. Let’s take an example to understand. There is one paragraph on how nitrogen is generated in the atmosphere. On the basis of this paragraph, the board asks a few objective-type questions. In other words, it is very similar to the unseen passages given in language papers. But the real cases may be different. So, read this article till the end to understand it thoroughly.

What is CBE?

CBSE stands for competency-based education. The case study questions are part of this CBE. The purpose of CBE is to demonstrate the learning outcomes and attain proficiency in particular competencies.

Questions on Real-life Situations

As discussed the case study questions are based on real-life situations. Especially for grade 10 science, it is very essential to have the practical knowledge to solve such questions. Here on the myCBSEguide app, we have given many such case study paragraphs that are directly related to real-life implications of the knowledge.

Sample Papers with Case Study Questions

Class 10 Science Sample Papers with case study questions are available in the myCBSEguide App . There are 4 such questions (Q.No.17 to 20) in the CBSE model question paper. If you analyze the format, you will find that the MCQs are very easy to answer. So, we suggest you, read the given paragraph carefully and then start answering the questions. In some cases, you will find that the question is not asked directly from the passage but is based on the concept that is discussed there. That’s why it is very much important to understand the background of the case study paragraph.

CBSE Case Study Sample Papers

You can download CBSE case study sample papers from the myCBSEguide App or Student Dashboard. Here is the direct link to access it.

Case Study Question Bank

As we mentioned that case study questions are coming in your exams for the last few years. You can get them in all previous year question papers issued by CBSE for class 1o Science. Here is the direct link to get them too.

Class 10 Science Case Study Question Examples

As you have already gone through the four questions provided in the CBSE model question paper , we are proving you with other examples of the case-based questions in the CBSE class 10 Science. If you wish to get similar questions, you can download the myCBSEguide App and access the Sample question papers with case study-type questions.

Case-based Question -1

Read the following and answer any four questions: Salt of a strong acid and strong base is neutral with a pH value of 7. NaCl common salt is formed by a combination of hydrochloride and sodium hydroxide solution. This is the salt that is used in food. Some salt is called rock salt bed of rack salt was formed when seas of bygone ages dried up. The common salt thus obtained is an important raw material for various materials of daily use, such as sodium hydroxide, baking soda, washing soda, and bleaching powder.

Phosphoric acid
Carbonic acid
Hydrochloric acid
Sulphuric acid
Blue vitriol
Washing soda
Baking soda
Bleaching powder

Case-based Question -2

V 1 + V 2 + V 3
V 1 – V 2 +V 2
None of these
same at every point of the circuit
different at every point of the circuit
can not be determined
20 3 Ω 203Ω
15 2 Ω 152Ω

Case-based Question -3

pure strips
impure copper
refined copper
none of these
insoluble impurities
soluble impurities
impure metal
bottom of cathode
bottom of anode

How to Attempt the Case-Based Questions in Science?

Before answering this question, let’s read the text given in question number 17 of the CBSE Model Question Paper.

All living cells require energy for various activities. This energy is available by the breakdown of simple carbohydrates either using oxygen or without using oxygen.

See, there are only two sentences and CBSE is asking you 5 questions based on these two sentences. Now let’s check the first questions given there.

Energy in the case of higher plants and animals is obtained by a) Breathing b) Tissue respiration c) Organ respiration d) Digestion of food

Now let us know if you can relate the question to the paragraph directly. The two sentences are about energy and how it is obtained. But neither the question nor the options have any similar text in the paragraph.

So the conclusion is, in most cases, you will not get direct answers from the passage. You will get only an idea about the concept. If you know it, you can answer it but reading the paragraph even 100 times is not going to help you.

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Electricity Class 10 Important Questions with Answers Science Chapter 12

July 20, 2022 by Bhagya

We have given these Important Questions for Class 10 Science Chapter 12 Electricity to solve different types of questions in the exam. Previous Year Questions & Important Questions of Electricity Class 10 Science Chapter 12 will help the students to score good marks in the board examination.

Important Questions of Electricity Class 10 Science Chapter 12

Question 1. A current of 10 A flows through a conductor for two minutes. (i) Calculate the amount of charge passed through any area of cross section of the conductor. (ii) If the charge of an electron is 1.6 × 10 -19 C, then calculate the total number of electrons flowing. (Board Term I, 2013) Answer: Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s (i) Amount of charge Q passed through any area of cross-section is given by I = $\frac { Q }{ t }$ or Q = I × t ∴ Q = (10 × 120) A s = 1200 C

(ii) Since, Q = ne where n is the total number of electrons flowing and e is the charge on one electron ∴ 1200 = n × 1.6 × 10 -19 or n = $\frac { 1200 }{ 1.6×10^{-19} }$ = 7.5 × 10 21

Question 2. Define electric current. (1/5, Board Term 1,2017) Answer: Electric current is the amount of charge flowing through a particular area in unit time.

Question 3. Define one ampere. (1/5, Board Term 1,2015) Answer: One ampere is constituted by the flow of one coulomb of charge per second. 1 A = 1 C s -1

Question 4. Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so. (Board Term I, 2013) Answer: A cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor.

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 1

Question 7. State Ohms law. (AI 2019) Answer: It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically, V ∝ I V = RI where R is resistance of the conductor.

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 3

Current, I(A)	Voltage, V(V)
0	0
1	2
2	4
3	6
4	8

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 7

Question 11. State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor. (Board Term I, 2016) Answer: Refer to answer 7 and 8.

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 9

Question 13. Assertion (A) : The metals and alloys are good conductors of electricity. Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity. (a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A). (b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true. (2020) Answer: (c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity.

Question 14. A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l’ is (2020) (a) $\frac { A }{ 2 }$ (b) $\frac { 3A }{ 2 }$ (c) 2A (d) 3A Answer: (c) : The resistance of a conductor of length!, and area of cross section, A is R = ρ$\frac { l }{ A }$ where ρ is the resistivity of the material. Now for the conductor of length 21, area of cross-section A’ and resistivity ρ. R’ = ρ$\frac { l’ }{ A’ }$ = ρ$\frac { 2l }{ A’ }$ But given, R = R’ ⇒ ρ$\frac { l }{ A }$ = ρ$\frac { 2l }{ A }$ or A’ = 2A

Question 15. Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater. Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals. (a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A). (b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A). (c) (A) is true, but (R) is false. (d) (A) is false, but (R) is true. (2020) Answer: (a)

Question 16. How is the resistivity of alloys compared with those of pure metals from which they may have been formed? (Board Term I, 2017) Answer: The resistivity of an alloy is generally higher than that of its constituent metals.

Question 17. (i) List three factors on which the resistance of a conductor depends. (ii) Write the SI unit of resistivity. (Board Term 1, 2015) Answer: (i) Resistance of a conductor depends upon the following factors: (1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R). R ∝ I

(2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance. R ∝ $\frac { 1 }{ A }$

(3) Nature of conductor. (ii) SI unit of resistivity is Ω m.

Question 18. Calculate the resistance of a metal wire of length 2m and area of cross section 1.55 × 10 6 m², if the resistivity of the metal be 2.8 × 10 -8 Ωm. (Board Term I, 2013) Answer: For the given metal wire, length, l = 2 m area of cross-section, A = 1.55 × 10 -6 m² resistivity of the metal, p = 2.8 × 10 -8 Ω m Since, resistance, R = ρ$\frac { l }{ A }$ So R = ($\frac { 2.8×10^{-8}×2 }{ 1.55×10^{-6} }$)Ω = $\frac { 5.6 }{ 1.55 }$ × 10 -2 Ω = 3.6 × 10 -2 Ω or R = 0.036Ω

Question 19. (a) List the factors on which the resistance of a conductor in the shape of a wire depends. (b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason. (c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018) Answer: (a) Refer to answer 17 (i). (b) Metal have very low resistivity and hence they are good conductors of electricity. Whereas glass has very high resistivity so glass is a bad conductor of electricity. (c) Alloys are commonly used in electrical heating devices due to the following reasons (i) Alloys have higher resistivity than metals (ii) Alloys do not get oxidised or burn readily.

Question 20. Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms. (Board Term I, 2017) Answer: We are given, the length of wire, l = 1 m, radius of wire, r = 0.01 cm = 1 × 10 -4 m and resistance, R = 20Ω As we know, R = ρ$\frac { l }{ A }$, where ρ is resistivity of the material of the wire. ∴ 20Ω.= ρ$\frac{l}{\pi r^{2}}$ = ρ$\frac{1 \mathrm{~m}}{3.14 \times\left(10^{-4}\right)^{2} \mathrm{~m}^{2}}$ ∴ ρ = 6.28 × 10 -7 Ω m

Question 21. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10 -8 Ω m. Calculate the length of this wire to make it resistance 100 Ω. How much does the resistance change if the diameter is doubled without changing its length? (Board Term I, 2015) Answer: Given; resistivity of copper = 1.6 × 10 -8 Ω m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 Ω Radius of wire, r = $\frac {d }{ 2}$ = $\frac {0.5 }{ 2 }$ mm = 0.25 mm = 2.5 × 10 -4 m Area of cross-section of wire, A = nr² ∴ A = 3.14 × (2.5 × 10 -4 )² = 1.9625 × 10 -7 m² = 1.9 × 10 -7 m² As, R = ρ$\frac { l }{ A }$ ∴ 100 Ω = $\frac{1.6 \times 10^{-8} \Omega \mathrm{m} \times l}{1.9 \times 10^{-7} \mathrm{~m}^{2}}$ l = 1200 m If diameter is doubled (d’ = 2d), then the area of cross-section of wire will become A’ = πr² = π($\frac { d’ }{ 2 }$)² = π($\frac { 2d }{ 2 }$)² = 4A Now R ∝ $\frac { 1 }{ A }$, so the resistance will decrease by four times or new resistance will be R’ = $\frac { R }{ 4 }$ = $\frac { 100 }{ 4 }$ = 25Ω

Question 22. The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10 -8 ohm meter, find the length of the wire. (Board Term I, 2014) Answer: Here, r = 0.01 cm = 10 -4 m, ρ = 50 × 10 -8 Ω m and R = 10 Ω As, R = ρ$\frac { l }{ A }$ or l = $\frac{R A}{\rho}=\frac{R}{\rho}\left(\pi r^{2}\right)$ so l = $\frac{10}{50 \times 10^{-8}} 3.14 \times\left(10^{-4}\right)^{2}$ = 0.628 m = 62.8 cm

Question 23. A wire has a resistance of 16 Ω. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance? (Board Term I, 2013) Answer: When wire is melted, its volume remains same, so, V’ = V or A’l’ = Al Here, l’ = $\frac { l }{ 2 }$ Therefore, A’ = 2 A Resistance, R = ρ$\frac { l }{ A }$ = 16 Ω Now, R’ = $\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{(l / 2)}{2 A}=\frac{1}{4} \rho \frac{l}{A}$ So, R’ = $\frac { R }{ 4 }$ = $\frac { 16 }{ 4 }$ = 4 Ω Percentage change in resistance, = $\left(\frac{R-R^{\prime}}{R}\right) \times 100=\left(\frac{16-4}{16}\right)$ × 100 = 75%

Question 24. If the radius of a current carrying conductor is halved, how does current through it change? (2/5 Board Term I, 2014) Answer: If the radius of conductor is halved, the area of cross-section reduced to ($\frac { 1 }{ 4 }$) of its previous value. Since, R ∝ $\frac { 1 }{ A }$, resistance will become four times From Ohm’s law, V = IR For given V, I ∝ $\frac { 1 }{ R }$ So, current will reduce to one-fourth of its previous value.

Question 25. Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10 -8 Ω m. (Board Term I, 2014) Answer: Resistance is the property of a conductor to resist the flow of charges through it. Factors affecting resistance of a conductor: Refer to answer 17(i) Rheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit. We are given, length of wire, l = 50 cm = 50 × 10 -2 m cross-sectional area, A = 0.01 mm² = 0.01 × 10 -6 m² and resistivity, ρ = 5 x 10 -8 Ω m. As, resistance, R = ρ$\frac { l }{ A }$ ∴ R = $\left(\frac{5 \times 10^{-8} \times 50 \times 10^{-2}}{0.01 \times 10^{-6}}\right)$ Ω = 2.5 Ω

Question 26. If a person has five resistors each of value $\frac { 1 }{ 5 }$ Ω, then the maximum resistance he can obtain by connecting them is (a) 1 Ω (b) 5 Ω (c) 10 Ω (d) 25 Ω (2020) Answer: (a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus, R s = $\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$ 1 Ω

Question 27. The maximum resistance which can be made using four resistors each of 2 Ω is (a) 2 Ω (b) 4 Ω (c) 8 Ω (d) 16 Ω (2020) Answer: (c) : A group of resistors can produce maximum resistance when they all are connected in series. ∴ R s = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω

Question 28. The maximum resistance which can be made using four resistors each of resistance $\frac { 1 }{ 2 }$ Ω is (a) 2 Ω (b) 1 Ω (c) 2.5 Ω (d) 8 Ω (2020) Answer: (a) The maximum resistance can be produced from a group of resistors by connecting them in series. Thus, R s = $\frac { 1 }{ 2 }$ Ω + H $\frac { 1 }{ 2 }$ Ω + $\frac { 1 }{ 2 }$ Ω + $\frac { 1 }{ 2 }$ Ω = 2 Ω

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 10

Question 30. List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013) Answer: (a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.

(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 11

Question 34. (a) A 6 Ω resistance wire is doubled on itself. Calculate the new resistance of the wire. (b) Three 2 Ω resistors A, B and C are connected in such a way that the total resistance of the combination is 3 Ω. Show the arrangement of the three resistors and justify your answer. (2020) Answer: (a) Given resistance of wire, R = 6 Ω Let l be the length of the wire and A be its area of cross-section. Then R = $\frac { ρl }{ A }$ = 6 Ω Now when the length is doubled, l’ = 2l and A’ = $\frac { A }{ 2 }$ ∴ R’ = $\frac{\rho(2 l)}{A / 2}=\frac{4 \rho l}{A}$ = 4 × 6 Ω = 24 Ω

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 16

(ii) Since, potential difference across each resistor connected in parallel is same. So, V 1 = V 2 = V 3 = 6 V Applying Ohm’s law, V 1 = I 1 R 1 or I 1 = $\frac { V_1 }{ R_1 }$ or I 3 = $\frac { 6 }{ 3 }$ A = 2A Similarly, I 2 = $\frac { 6A }{ 4 }$ = 1.5 A and I 3 = $\frac { 6 }{ 6 }$ A = 1 A

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 24

Precautions: (i) All the connections are neat and tight. (ii) Ammeter is connected with the proper polarity, i.e., positive terminal of the ammeter should go to positive terminal and negative terminal of ammeter to the negative terminal of the battery or cell used.

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 26

(ii) Total current drawn from battery, I = I 1 + I 2 + I 3 = 0.2 + 0.6 + 1.2 = 2 A (iii) Equivalent resistance of the circuit, R eq can be obtained by Ohm’s law V= I R eq So, 6 V = 2 A × R eq or, R eq = $\frac { 6 }{ 2 }$ = 3 Ω Aliter, $\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$ $\frac{1}{30}+\frac{1}{10}+\frac{1}{5}=\frac{1+3+6}{30}=\frac{10}{30}=\frac{1}{3}$ or R eq = 3 Ω

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 27

Question 44. The resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become (a) two times (b) half (c) one-fourth (d) four times (2020) Answer: (a) : We know, H = I²Rt = $\frac { V^2 }{ 4 }$. t Now when, R’ = $\frac { R }{ 24 }$, V’ = V and t’ = t H’ = $\frac{V^{\prime 2} t^{\prime}}{R^{\prime}}=\frac{V^{2} t}{R / 2}=\frac{2 V^{2} t}{R}$ = 2H

Question 45. (a) Write the mathematical expression for Joules law of heating. (b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V. (2020) Answer: (a) The Joule’s law of healing implies that heat produced in a resistor is (i) directly proportional to the square of current lor a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor. i.e., H = I² Rt (b) Given, charge q = 96000 C, time t = 2 h = 7200 s and potential difference V = 40 V We know, H = I²Rt = $\frac{Q^{2}}{t^{2}} \times \frac{V}{Q}$ × t × t = VQ = 40 × 96000 = 3.84 × 10 6 J = 3.84 MJ

Question 46. Write Joules law of heating. (1/3, 2018) Answer: Refer to answer 45(a).

Question 47. Explain the use of an electric fuse. What type of material is used for fuse wire and why? (Board Term I, 2016) Answer: Electric fuse protects circuits and appliances by stopping the flow of any unduly high electric current. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.

Question 48. (a) Why is tungsten used for making bulb filaments of incandescent lamps? (b) Name any two electric devices based on heating effect of electric current. (2/5, Board Term I, 2015) Answer: (a) (i) Tungsten is a strong metal and has high melting point (3380°C). (ii) It emits light at high temperatures (about 2500°C). (b) Electric laundry iron and electric heater are based on heating effect of electric current.

Question 49. A fuse wire melts at 5 A. If it is desired that the fuse wire of same material melt at 10 A, then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reasons for your answer. (3/5, Board Term I, 2014) Answer: Let the resistance of the wire be R, heat produced in the fuse at 5 A in Is is H=(5)²R ( H – I²Rt) 50. fuse melts at (5)²R joules of heat. Let, the resistance of new wire is R’ So, heat produced in 1 second = (10)²R’ To prevent it from melting (5)²R = (10)²R’ or R’ = $\frac { R }{ 4 }$ As R ∝ $\frac { 1 }{ A }$ ∴ cross-sectional area of new fuse wire is four times the first fuse. Now, A = πr², so new radius is twice the previous one. So, at 10 A, the new fuse wire of same material and length has larger radius than the earlier one.

Question 50. What is heating effect of current? List two electrical appliances which work on this effect. (2/5, Board Term I, 2013) Answer: If only resislors are connected to the battery, the source energy continually gets dissipated entirely in the form of heal. This is known as healing effect of current, ’file amount of heat (77) produced in time t is given by Joule’s law of heating. H = I²Rt Where, 7 is current flowing through resistor R. The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some common devices based on heating effect of current.

Question 51. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be (a) 0.4 A (b) 0.6 A (c) 0.8 A (d) 1A (2020) Answer: (d) : Given power of first bulb, P 1 = 100 W and second bulb P 2 = 40 W Current through 100 W bulb, I 1 = 1 A Current through 40 W bulb, I 2 = ? Since both the bulbs are connected in series, the electric current passing through both the bulbs are same i.e., I 2 = 1 A.

Question 52. Write the relation between resistance (R) of filament of a bulb, its power (P) and a constant voltage V applied across it. (Board Term I, 2017) Answer: P = $\frac { V^2}{ R }$

Question 53. Power of a lamp is 60 W. Find the energy in joules consumed by it in Is. (Board Term I, 2016) Answer: Here, power of lamp, P = 60 W time, t = 1 s So, energy consumed = Power × time = (60 × 1)J = 60 J

Question 54. Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V. (2/3, 2018, Board Term I, 2014) Answer: Since both the bulbs are connected in parallel and to a 220 V supply, the voltage across each bulb is 220 V. Then Current drawn by 100 W bulb, I 1 = $\frac { power rating}{ voltage applied }$ = $\frac { 100 W}{ 220 V }$ = 0.454 A Current drawn by 60 W bulb, I 2 = $\frac { 60 W}{ 220 V }$ = 0.273 A Total current drawn from the supply line, I = I 1 + I 2 = 0.454 A + 0.273 A = 0.727 A = 0.73 A

Question 55. How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts. (Board Term I, 2016) Answer: Here, V = 220 V, R = 55 Ω By Ohm’s law V = IR ∴ 220 = 7 × 55 or I = 4A Wattage of electric iron = Power = $\frac{V^{2}}{R}=\frac{(220)^{2}}{55}$ = 880 W

Question 56. An electric iron has a rating of 750 W; 200 V. Calculate: (i) the current required. (ii) the resistance of its heating element. (iii) energy consumed by the iron in 2 hours. [Board Term 1, 2015] Answer: Here, P = 750 W, V = 200 V (i) As P = V7 I = P/V= (750/200) A = 3.75A (ii) By Ohm’s law V = IR or R = V/I ∴ R = $\frac { 200}{ 3.75 }$ Ω = 53.3 Ω (iii) Energy consumed by the iron in 2 hours = P × t = 750 W × 2h = 1.5 kWh or E = (750 × 2 × 3600) J = 5.4 × 10 6 J

Question 57. An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb. (1/3, Board Term I, 2015) Answer: Here, V= 220 V,/= 2.5 A Power of the bulb P = VI = 220 × 2.5 W = 550 W

Question 58. (a) Define power and state its SI unit. (b) A torch bulb is rated 5 V and 500 mA. Calculate its (i) power (ii) resistance (iii) energy consumed when it is lighted for 2 $\frac { 1 }{ 2 }$ hours. Answer: (a) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit. P = VI = I²R = V²/R The SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of IV. 1 W = 1 volt × 1 ampere = 1 V A

(b) Given, V = 5 V and I = 500 mA = 0.5 A (i) Power, P = V × 7 = 5 × 0.5 = 2.5 W (ii) As, P = $\frac{V^{2}}{R} \Rightarrow R=\frac{V^{2}}{P}=\frac{25}{2.5}$ = 10 Ω (iii) Given, time t = 2.5 hrs = 9000 s ∴ The energy consumed, E = P × t = 2.5 × 9000 = 2.25 × 10 4 J = 6.25 Watt hour

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 29

(ii) Potential difference across conductor $\frac { Total voltage }{ Total resistance }$ × resistance of conductor = $\frac { 6}{ 24 }$ × 4 = 1 V

(d) Power of the lamp = (current)² × resistance of lamp = (0.25)² × 20 = 1.25 W

Electricity Class 10 Important Questions with Answers Science Chapter 12 Img 31

Question 62. A bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. (AI 2019) Answer: In first case, P = 40 W, V = 220 V Current drawn l = $\frac { P}{ V }$ = $\frac { 40}{ 220 }$ = 0.18 A Also, resistance of bulb, R = $\frac{V^{2}}{P}=\frac{(220)^{2}}{40}$ = 1210 Ω In second case, P = 25 W, V = 220 V Current drawn, I = $\frac { P}{ V }$ = $\frac { 25}{ 220 }$ = 0.11 A Also, resistance of the bulb, R = $\frac { V^2}{ P }$ = $\frac { (220)^2}{ 25 }$ = 1936 Ω Hence, by replacing 40 W bulb to 25 W bulb, having same source of voltage the amount of current flows decreases while resistance increases.

Question 63. (a) How two resistors, with resistances R 1 Ω and R 1 Ω respectively are to be connected to a battery of emf V volts so that the electrical power consumed is minimum? (b) In a house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 3.60 per kWh. (Board Term I, 2017) Answer: (a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series. (b) Power of each bulb P 1 = 100 watt Total power of 3 bulbs, P 1 = 3 × 100 = 300 watt Energy consumed by bulbs in 1 day E 1 = P 1 × t = 300 watt × 5 hours. = 1500 Wh = 1.5 kWh Power of each fan = 50 watt Total power of 2 fans = 2 × 50 watt P 2 = 100 watt Energy consumed by fans in 1 day E 2 = P 2 × t = 100 watt × 10 hours = 1000 watt hour = 1 kWh Energy consumed by heater, E 3 = 1 kW × 1/2 h = 0.5 kWh Total energy consumed in one day E = E 1 + E 2 + E 3 = (1.5 + 1 + 0.5) kWh = 3 kWh Total energy consumed in a month of 31 days = E × 31 = (3 × 31) kWh = 93 kWh Cost of energy consumed = Rs (93 × 3.60) = Rs 334.80

Question 64. (a) An electric bulb is connected to a 220 V generator. If the current drawn by the bulb is 0.50 A, find its power. (b) An electric refrigerator rated 400 W operates 8 hours a day. Calculate the energy per day in kWh. (c) State the difference between kilowatt and kilowatt hour. (3/5, Board Term I, 2013) Answer: (a) Here, V = 220 V, I = 0.50 A Power of the bulb, P = VI = (220 × 0.5)W = 110 W (b) Energy consumed by electric refrigerator in a day = Power x time = 400 W × 8 h = 3200 Wh = 3.2 kWh (c) Kilowatt is unit of power and kilowatt hour is a unit of energy.

Question 65. (i) State one difference between kilowatt and kilowatt hour. Express 1 kWh in joules. (ii) A bulb is rated 5V; 500 mA. Calculate the rated power and resistance of the bulb when it glows. (Board Term I, 2013) Answer: (i) Refer to answer 64(c). 1 kWh = 1000 W × 1 h = 1000 W × 3600 s = 3600000 J = 3.6 × 10 6 J

(ii) Here, V = 5 V, I = 500 mA = 0.5 A Power rating of bulb is P = VI = ( 5 × 0.5)W = 2.5W Resistance of the bulb is R = V/I = (5/0.5) Ω = 10 Ω

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NCERT Exemplar Class 10 Science Solutions for Chapter 12 - Electricity

Ncert exemplar solutions class 10 science chapter 12 – free pdf download.

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity are the study materials necessary for you to understand the questions that can be asked from the Class 10 Science Electricity chapter. It is crucial for students to get acquainted with this chapter in order to score excellent marks in their CBSE Class 10 examination. This solution provides answers to the questions provided in NCERT Class 10 Exemplar book. This page has answers to 18 MCQs, 11 short answer questions and 7 long answer questions.

To help students grasp all the concepts clearly and in-depth, we are offering free NCERT Exemplar for Class 10 Science Chapter 12 here. These exemplars will enable students to learn the correct answers to all the questions given at the end of the chapter. These NCERT Exemplars are prepared by experts and can be used by students as an effective learning tool to improve their conceptual understanding.

Take a closer look at Class 10 Science Chapter 12 NCERT Exemplar below.

Download the PDF of NCERT Exemplar for Class 10 Science Chapter 12 – Electricity

NCERT Exemplar solutions class 10 science Chapter 12 part 01

Access Answers to NCERT Exemplar Class 10 Science Chapter 12 – Electricity

Multiple choice questions.

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

ncert solutions class 10 science chapter 12 electricity fig 1

(a) maximum in (i)

(b) maximum in (ii)

(d) the same in all the cases

The answer is (d) the same in all the cases

Explanation:

There are no changes in any of the circuits, hence current will be same in all the circuits.

2. In the following circuits (Figure 12.2), the heat produced in the resistor or combination of resistors connected to a 12 V battery will be

ncert solutions class 10 science chapter 12 electricity fig 2

(a) same in all the cases

(b) minimum in case (i)

(d) maximum in case(iii)

The answer is (c) maximum in case(ii)

Explanation

Here two transistors are in series. In figure (iii) total resistance will be less than individual resistances as they are connected parallel. Higher resistance produces more heat hence option c) is the right answer.

3. Electrical resistivity of a given metallic wire depends upon

(a) its length

(b) its thickness

(d) nature of the material

The answer is (d) nature of the material

4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

Answer is (a) 10 20

n = 16 /1.6 x 10 -19

n = 10 x 10 19

n = 10 20 electrons

The number of electrons flowing is 10 20 electrons

5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

ncert solutions class 10 science chapter 12 electricity fig 3

The answer is (b) (ii)

6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

The answer is (d) 1 Ω

Maximum resistance is obtained when resistors are connected in series.

7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?

Answer is (b) 1/25 Ω

Minimum resistance is obtained when resistors are connected parallel

1/R = 5 + 5 + 5 +5 +5= 25 Ω

8. The proper representation of the series combination of cells (Figure 12.4) obtaining maximum potential is

ncert solutions class 10 science chapter 12 electricity fig 5

The answer is (a) (i)

Here positive terminal of the next cell is adjacent to the negative terminal of the previous cell.

9. Which of the following represents voltage?

(a) $\begin{array}{l}\frac{Work done}{Current\times Time}\end{array} $

(b) Work done × Charge

(d) Work done × Charge × Time

10. A cylindrical conductor of length l and uniform area of crosssection A has resistance R. Another conductor of length 2l and resistance R of the same material has an area of cross-section

Answer is (c) 2A

When Length doubles

11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

ncert solutions class 10 science chapter 12 electricity fig 6

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(d) R2 > R3 > R1

The answer is (c) R3 > R2 > R1

Current flow is inversely proportional to resistance. Highest resistance will show less flow of current hence answer is c).

12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

Answer is (c) 300 %

Heat generated by a resistor is directly proportional to the square of the current. Hence, when the current becomes double, dissipation of heat will multiply by 2 =4. This means there will be an increase of 300%.

13. The resistivity does not change if

(a) the material is changed

(b) the temperature is changed

(d) both material and temperature are changed

Answer is (c) the shape of the resistor is changed

14. In an electrical circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a) The brightness of all the bulbs will be the same

(b) The brightness of bulb A will be the maximum

(d) The brightness of bulb C will be less than that of B

Answer is (c) Brightness of bulb B will be more than that of A

Bulbs are connected in parallel so the resistance of combination would be less than the arithmetic sum of the resistance of all the bulbs. So. there will be no negative effect on the flow of current. As a result, bulbs would glow according to their wattage.

15. In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

Answer is (c) 20 J

Equivalent resistance of the circuit is R = 4+2 = 6Ω

current, I= V/R = 6/6= 1A

the heat dissipated by 4-ohm resistor is, H = I 2 Rt = 20J

16. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

The answer is (d) 5 A

Or 1000 w = 220v x I

I = $\begin{array}{l}\frac{1000w}{220v}\end{array} $ = 4.54 A

17. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have

(a) same current flowing through them when connected in parallel

(b) same current flowing through them when connected in series

(d) different p

The answer is (b) same current flowing through them when connected in series

In series combination current does not get divided into branches because resistor receives a common current.

18. Unit of electric power may also be expressed as

(a) volt-ampere

(b) kilowatt-hour

(d) joule second

The answer is (a) volt-ampere

Volt-ampere (VA) is the unit used for the apparent power in an electrical circuit. A watt-second (also watt-second, symbol W s or W. s) is a derived unit of energy equivalent to the joule. The joule-second is the unit used for Planck’s constant.

Short Answer Questions

19. A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

ncert solutions class 10 science chapter 12 electricity fig 7

20. Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

ncert solutions class 10 science chapter 12 electricity fig 9

Current P= I 2 R

18W = I 2 x 2Ω

I 2 = 18W/ 2Ω

Maximum value of current passing through A is 3A.

Current through B = Current through C = 1/2 x Current through A

Current through B = Current through C = 1/2 x 3

Current through B = Current through C = 1.5 A

21. Should the resistance of an ammeter be low or high? Give reason.

Resistance of ammeter should be zero because ammeter should not affect the flow of current.

22. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

ncert solutions class 10 science chapter 12 electricity fig 10

Total resistance for parallel combination of 40 resistors can be calculated as follows:

Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 20 resistance will be same as potential difference across the other two resistors which are connected in parallel.

23. How does use of a fuse wire protect electrical appliances?

Fuse wire has great resistance than the main wiring. When there is significant increase in the electric current. Fuse wire melts to break the circuit. This prevents damage of electrical appliance.

24. What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Property of the conductor which resists the flow of electric current is called resistivity. Resistance for a particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow.

When length is doubled resistance becomes double and current flow reduces to half. This is the reason for the decrease in ammeter reading.

25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Commercial unit of electrical energy is kilowatt/hr

1 kw/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 10 6 J

26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

1) Let R be the resistance of the electric lamp. In series total resistance = 5 + R

1 = 10/5+R

2) V across Lamp + conductor = 10 V

V acoess Lamp = I × R = 1 * 5 = 5 Volt

27. Why is parallel arrangement used in domestic wiring?

Parallel arrangement used in domestic wiring because it provides the same potential difference across each electrical appliance.

28. B1 , B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

What happens to the glow of the other two bulbs when the bulb B1 gets fused?
What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused?
How much power is dissipated in the circuit when all the three bulbs glow together?

i) Potential difference does not get divided in parallel circuit. Hence glowing of other bulbs will not get affected when bulb one is fused.

ii) Ammeter A shows a reading of 3A. This means each of the Al. A2, and A3 show IA reading.

iii) R= V/I = 4.5V/3A= 1.5Ω

Now P= I 2 R

= (3A) 2 x 1.5 Ω

Long Answer Questions

29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.

(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.

(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

(Resistance of the bulbs in series will be three times the resistance of single bulb. Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly.

The bulbs in series combination will stop glowing as the circuit is broken and current is zero. However the bulbs in parallel combination shall continue to glow with the same brightness.

30. State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Ohm’s law states that at constant temperature potential difference (voltage) across an ideal conductor is proportional to the current through it.

ncert solutions class 10 science chapter 12 electricity fig 11

Verification of Ohm’s law

Set up a circuit as shown in Fig. consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)

First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given

Next, connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.

Repeat the above steps using three cells and then four cells in the circuit separately.

ncert solutions class 10 science chapter 12 electricity fig 12

31. What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Resistivity is an inherent property of a conductor which resists the flow of electric current. Resistivity of each material is unique. SI unit of resistance is Ωm.

Experiment to study the factors on which the resistance of conducting wire depends.

Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5 A range), a plug key and some connecting wires.

Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4.

Observation:

It is observed that resistance depend on material of conductor

Length of conductor determines resistance

Resistance depends on area of cross-section.

ncert solutions class 10 science chapter 12 electricity fig 13

Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.

Now repeat the above step with the 10 W bulb in the gap XY. Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate?

You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Collect three resistors R1, R2, R3 in series to make the circuit.
Use ammeter to see the changes observed in the current flow.
Remove R1 and take the reading of potential difference of R2 and R3
Remove R2 and take the reading of potential difference of R1 and R3

ncert solutions class 10 science chapter 12 electricity fig 14

Ammeter reading was the same in each case so it can be inferred that the current remains the same in the circuit. To cross-check one can place ammeter and different places and observe the current flow.

33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Take three resistors R 1 , R 2 and R 3 , connect them in parallel to make a circuit; as shown in the figure.

Use voltmeter to take reading of potential difference of three resistors in parallel combination.

Now, remove the resistor R1 and take the reacting of the potential difference of remaining resistors combination.

Then, remove the resistor R 2 , and take the reading of potential difference of remaining resistor.

ncert solutions class 10 science chapter 12 electricity fig 15

In each case Voltmeter reading was the same which shows that the same potential difference exists across three resistors connected in a parallel arrangement.

34. What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

According to Joules heating effect heat produces in a resistor is

Directly proportional to square of current for the given resistor.
Directly proportional to resistance for a given current,
Directly proportional to the time of current flowing through the resistor.

This can be expressed as

H is heating effect, I is electric current, R is resistance and t is time.

Experiment to demonstrate Joules law of heating

Take a water heating immersion rod and connect to a socket which is connected to the regulator. It Is important to recall that a regulator controls the amount of current flowing through a device.
Keep the pointer of regulator on the minimum and count the time taken by immersion rod to heat a certain amount of water.
Increase the pointer of the regulator to the next level. Count the time taken by immersion rod to heat the same amount of water.
Repeat above step for higher levels on regulator to count the time.

It is seen that with an increased amount of electric current, less time is required o heat the same amount of water. This shows Joule’s Law of Heating.

Application:

Electric toaster, oven, electric kettle and electric heater etc. work on the basis of leafing effect of current.

35. Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination

(b) Current flowing through 4 Ω resistor

(d) Power dissipated in 4 Ω resistor (e) Difference in ammeter readings, if any

ncert solutions class 10 science chapter 12 electricity fig 16

NCERT Exemplar Class 10 Chapter 12 Electricity

Sometimes we might have wondered about what constitutes electricity or how does it flow in an electric circuit, or what controls and regulates the current through an electric circuit? In Chapter 12 Electricity, students will find answers to these questions. They will also learn about other topics like the heating effect of electric current and its applications, the circuit diagram, Ohm’s law , resistors and conductors, electrical potential and potential difference.

Topics covered in Class 10 NCERT Exemplar Solutions for Science Chapter 12 Electricity

Introduction
The potential difference – Definition of volt and voltmeter
Ohm’s law – Ohm and resistance
Factors on which the resistance of the conductor depends – Resistivity
Resistors in series – Total/resultant/overall and voltage across each resistor
Resistors in parallel
The advantage of parallel combination over the series combination
Heating effect of an electric circuit – Joule’s law of the heating effect of electric current, electric fuse and electric power.

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Case Study Questions Class 10 Science Magnetic Effects of Electric Current

Case study questions class 10 science chapter 13 magnetic effects of electric current.

CBSE Class 10 Case Study Questions Science Magnetic Effects of Electric Current. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Magnetic Effects of Electric Current.

CBSE Case Study Questions Class 10 Science Magnetic Effects of Electric Current

Case Study – 1

Answer- It states that if we stretch thumb, forefinger or the index finger and the middle finger in such a way that they are mutually perpendicular to each other then the thumb gives the direction of the motion or the force acting on conductor, index finger gives the direction of magnetic field and the middle finger gives the direction of current.

State your observation in the galvanometer.

Explanation- Electromagnetic induction is a process by changing a magnetic field in a conductor, which induces a current in another conductor placed in nearby.

Case study: 4

If we stretched the thumb, forefinger and middle finger of our left hand so that they are mutually perpendicular to each other. If the forefinger gives the direction of magnetic field and middle finger gives the direction of electric current then the thumb gives the direction of motion or the force acting on the conductor.

The domestic electric circuit consist of red insulated cover called as live wire, wire with black insulation called as neutral wire and the wire with green insulation is called as Earth wire. We know that fuse is connected in series with the circuit to prevent the damaging of electrical appliances and circuit from overloading. Overloading occurs when live wire and the neutral wire comes in direct contact with each other. Because of which current through the circuit increases suddenly. Also, overloading may occurs because of connecting many appliances to a single socket. The Earth wire which is green in colour is connected to a metal plate deep in the earth near the house. This type of safety measure is used in appliances like electric press, toaster, table fan, refrigerator etc. The Earth wire is gives low resistance conducting path for the electric current. In this way it protects us from severe electric shock.

4) What is the main purpose of using fuse in electric circuit?

4) Because of Joule’s heating effect the heat produced causes the fuse to melt and to break the circuit. And thereby protect the circuit and electric appliances.

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Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023

Class 10th Science - Electricity Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10 Science Subject - Electricity, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Electricity case study questions with answer key.

Final Semester - June 2015

The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow of electrons in the current. The electric potential is defined as the amount of work done in bringing a unit positive test charge from infinity to a point in the electric field. The amount of work done in bringing a unit positive test charge from one point to another point in an electric field is defined as potential difference. $\begin{equation} V_{A B}=V_{B}-V_{A}=\frac{W_{B A}}{q} \end{equation}$ The SI unit of potential and potential difference is volt. (i) The 2 C of charge is flowing through a conductor in 100 rns, the current in the circuit is

(ii) Which of the following is true? (a) Current flows from positive terminal ofthe cell to the negative terminal of the cell outside the cell. (b) The negative charge moves from lower potential to higher potential. (c) The direction of flow of current in same as the direction of flow of positive charge. (d) All of these (iii) The potential difference between the two terminals of a battery, if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to other is

(iv) The number of electrons flowing per second in a conductor if 1A current is passing through it

(v) The voltage can be written as

The relationship between potential difference and current was first established by George Simon Ohm called Ohm's law. According to this law, the current through a metallic conductor is proportional to the potential difference applied between its ends, provided the temperature remain constant i.e. I $\begin{equation} \propto \end{equation}$ V or V = IR; where R is constant for the conductor and it is called resistance of the conductor. Although Ohm's law has been found valid over a large class of materials, there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. (i) If both the potential difference and the resistance in a circuit are doubled, then

(ii) For a conductor, the graph between V and I is there. Which one is the correct?

(iii) The slope of V - I graph (V on x-axis and I on y-axis) gives

(iv) When battery of 9 V is connected across a conductor and the current flows is 0.1 A, the resistance is

(v) By increasing the voltage across a conductor, the

The obstruction offered by a conductor in the path of flow of current is called resistance. The SI unit of resistance is ohm ( $\begin{equation} \Omega \end{equation}$ ). It has been found that the resistance of a conductor depends on the temperature of the conductor. As the temperature increases the resistance also increases. But the resistance of alloys like mangnin, constantan and nichrome is almost unaffected by temperature. The resistance of a conductor also depends on the length of conductor and the area of cross-section of the conductor. More be the length, more will be the resistance, more be the area of cross-section, lesser will be the resistance. (i) Which of the following is not will desired in material being used for making electrical wires?

< T	> T
= T

(iii) Two wires of same material one of length L and area of cross-section A, other is of length 2L and area A/2 . Which of the following is correct?

= R	= 4R
= 4R	= 2R

(iv) For the same conducting wire (a) resistance is higher in summer (b) resistance is higher in winter (c) resistance is same is summer or in winter (d) none of these (v) A wire of resistance 20 $\begin{equation} \Omega \end{equation}$ is cut into 5 equal pieces. The resistance of each part is

= V = V	+ V + V
+ V + V = 3V	+ V + V

(ii) When the three resistors each of resistance R ohm, connected in series, the equivalent resistance is

(iii) There is a wire oflength 20 cm and having resistance 20 $\begin{equation} \Omega \end{equation}$ cut into 4 equal pieces and then joined in series. The equivalent resistance is

= 2I = 3I	= 4I = 3I
= I = 3I	= 2I = I

Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. When the resistances are connected in series, the current in each resistance is same but the potential difference is different in each resistor. When the resistances are connected in parallel, the voltage drop across each resistance is same but the current is different in each resistor. (i) The household circuits are connected in

(v) Two resistances 10 $\begin{equation} \Omega \end{equation}$ and 3 $\begin{equation} \Omega \end{equation}$ are connected in parallel across a battery. If there is a current of 0.2 A in 10 .Q resistor, the voltage supplied by battery is

The heating effect of current is obtained by transformation of electrical energy in heat energy. Just as mechanical energy used to overcome friction is covered into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. The heat produced in a conductor, when a current flows through it is found to depend directly on (a) strength of current (b) resistance of the conductor (c) time for which the current flows. The mathematical expression is given by H = I 2 Rt. The electrical fuse, electrical heater, electric iron, electric geyser etc. all are based on the heating effect of current. (i) What are the properties of heating element? (a) High resistance, high melting point (b) Low resistance, high melting point (c) Low resistance, high melting point (d) Low resistance, low melting point. (ii) What are the properties of electric fuse? (a) Low resistance, low melting point (b) High resistance, high melting point. (c) High resistance, low melting point (d) Low resistance, high melting point (iii) When the current is doubled in a heating device and time is halved, the heat energy produced is

(iv) A fuse wire melts at 5 A. It is is desired that the fuse wire of same material melt at 10 A. The new radius of the wire is

(v) When a current of 0.5 A passes through a conductor for 5 min and the resistance of conductor is 10 $\begin{equation} \Omega \end{equation}$ , the amount of heat produced is

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes we use a bigger unit of electrical energy which is called kilowatt hour. 1 kilowatt-hour is equal to 3.6 x 106 joules of electrical energy. (i) The energy dissipated by the heater is E. When the time of operating the heater is doubled, the energy dissipated is

(ii) The power of a lamp is 60 W The energy consumed in 1 minute is

(iii) The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is $\begin{equation} ₹ \end{equation}$ 5 per kWh. Find the cost of running the refrigerator for one day?

32	16
8	4

(iv) Calculate the energy transformed by a 5 A current flowing through a resistor of 2 $\begin{equation} \Omega \end{equation}$ for 30 minutes?

(v) Which of the following is correct? (a) 1 watt hour = 3600 J (b) lkWh = 36x10 6 J (c) Energy (in kWh) = power (in W) x time (in hr) (d) $\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}$

(i) Total resistance of parallel combination is : (a) 2.4 Ω (b) 3 Ω (c) 6 Ω (d) 2 Ω (ii) Equivalent resistance of total circuit is : (a) 5 Ω (b) 9 Ω (c) 11 Ω (d) 13 Ω (iii) Total current in the circuit is : (a) 2 A (b) 4.5 A (c) 0.5 A (d) 10 A (iv) Current in 6 ohm resistance is (a) 0.3 A (b) 0.2 A (c) 4 A (d) 6 A (v) Potential across 3.6 ohm resistance will be : (a) 1.8 V (b) 2.6 V (c) 9 V (d) 4.5 V

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CBSE Class 10 Electricity Important Questions with Solutions

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Electricity Class 10 Questions and Answers

Chapter 12 of Class 10 Science is about ‘Electricity’. One of the most fundamental elements of our society is electricity. Since the beginning of the industrial revolution, electricity has contributed to the development of our civilization by powering numerous businesses and industries. Today, life would be in complete disarray if we were to lose this energy source. Therefore, it’s crucial for students to learn ideas about how electricity functions at the molecular level and explore its applications. Chapter 12 ‘Electricity’ will teach students about the fundamentals of electricity, the movement of current, and the operation of circuits as a whole.

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Throughout our question bank of Chapter 12 Class 10 Science Important Questions, students are given important problems to complete so they will be familiar with the questions which are expected in their final exams.

Students should not memorise the answer of science problems since it requires an in-depth understanding of the concepts involved. For each question in our Important Questions Class 10 Science Chapter 12, there is a step-by-step explanation prepared by our in-house Science subject experts. . This makes it easy for students to have complete faith and trust, to revise concepts covered in class.

The questions are compiled from the NCERT textbook, NCERT exemplar, and other reference books for providing the best and most authentic source of question bank. Furthermore, there are questions from previous year’s exams as well. In order to set the foundation for the chapter, it could be beneficial for students to refer to the Important Questions Class 10 Science Chapter 12.

There is a need for students to improve their understanding of the art of writing answers. They can practice representational diagrams by referring to them from NCERT textbook and from our NCERT chapter-wise solutions. Students can access our entire collection of Science Class 10 Chapter 12 Important Questions by registering on the Extramarks’ website. Besides, students can review other study materials on our Extramarks websites, such as NCERT solutions, revision notes, and the previous year’s question papers.

CBSE Class 10 Science Important Questions

CBSE Class 10 Science Important Questions are also available for the following chapters:



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Important Electricity Questions with Answers Class 10 Chapter 12

Given below is a set of questionnaires and their answers from our question bank of Class 10 Science Chapter 12 Important Questions .

For the best preparation, students are advised to go through this set of Important questions Class 10 Science Chapter 12. They can register on our website to get access to it.

Question 1: Which of the following does not represent electrical power in a circuit?

Answer : b) IR 2

Explanation:

Electrical power is represented by the expression P = VI . (Equation 1)

According to Ohm’s law,

Putting the value of V in ( Equation 1), we get

P = ( IR ) × I

Similarly, from Ohm’s law,

Putting the value of I in (Equation 1),

P = V × V / R = V 2 / R

It is thus clear that the equation IR 2 does not represent electrical power in a circuit. Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit.

Question 2: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be _____.

Answer : (d) 25 W

This expression demonstrates how much energy the electric bulb consumes.

P = VI = V 2 /R

The given formula can be used to calculate the light bulb’s resistance:

Putting the values, we get

R = (220)2/100 = 484 Ω

The resistance generally does not change when the voltage supply is decreased. Consequently, the amount of electricity used can be determined as follows:

P = (110) 2 V/484 Ω = 25 W

As a result, the electric bulb uses 25 W of power when it is operating at 110 V.

Question 3: What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

Answer: (d) 1 Ω

Explanation: Resistance is maximum when resistors are connected in series.

R= 1/ 5 + 1/ 5 + 1/ 5 + 1/ 5 + 1/ 5

Question 4: If the current ‘I’ through a resistor is increased by 100% (assuming that the temperature remains unchanged), the approximate increase in power dissipated will be

Answer: (c) 300 %

Explanation: The amount of heat produced by a resistor is inversely proportional to the square of the current. Therefore, the loss of heat will multiply by 2=4 when the current doubles. Accordingly, there will be a 300% increase.

Question 5: A piece of wire of resistance R is cut into five equal parts. These parts are then arranged in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____.

Answer: d) 25

The resistance is divided into five halves, each of which has a resistance of R/5.

Since we are aware that each component is linked to the others in parallel, we can compute the equivalent resistance as follows:

R’ = 5/ R + 5/ R + 5/ R + 5/ R + 5/ R

R’ = ( 5 + 5+ 5+ 5+ 5)/ R = 25/ R

R R’ = 25

The ratio of R/R′ is 25.

Question 6: The correct representation of the series combination of cells (Figure 12.4) obtaining maximum potential is

Answer: (a)

A cell’s positive terminal needs to be connected to the neighbouring cell’s negative terminal. The appropriate cell combination is represented by case I.

Question 7: Two pieces of conducting wire of the same material and of equal lengths and the equal diameters are first connected in series and then changed to parallel in a circuit across the same potential difference. The ratio of heat produced in both series and parallel combinations would be _____.

Answer 7: (c)

Let Rs and Rp represent the wires’ respective equivalent resistances when linked in series and parallel.

The ratio of heat generated in the circuit is provided by

H s/ H p = ( V 2 / R s) t/( V 2 / R p)/ t = R p/ R s

The equivalent resistance (Rs) of resistors connected in series is R + R = 2R

The equivalent resistance (Rp) of resistors connected in parallel is 1/ R + 1/ R = 2/R

Hence, the estimated ratio of the heat produced in series and parallel combinations would be

H s/Hp = 2R/( R/2) = 1/ 4

Thus, the ratio of heat produced is 1:4.

Question 8: What is the minimum resistance which can be made using five resistors, each of 1/5 Ω?

Answer: (b) 1/25 Ω

Resistance is the minimum when resistors are connected in parallel

1/ R = 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) = 25 Ω

Question 9: A person carries out an experiment and thus plots the V-I graph of three taken samples of nichrome wire with different resistances R 1, R 2 and R 3, respectively (Figure.12.5). Which one of the following is true?

(a) R 1 = R 2 = R 3

(b) R 1 > R 2 > R 3

(d) R 2 > R 3 > R 1

Answer 9: (c)

The graph’s slope is 1/R because the current ( I ) is plotted on the y-axis, and the potential difference ( V ) is plotted on the x-axis . It implies that the less resistance, the steeper the slope. R 1 will therefore be the minimum and R 3 the maximum.

Question 10: Two resistors of resistance 2 Ω and 4 Ω, when connected to a battery, will have

(a) the same potential difference across them when connected in series

(b) same current flows through them when connected in series

(d) different p

Answer: (b) same current flowing through them when connected in series

Since the resistor gets a common current in a series arrangement, the current is not split into branches.

Question 11: What does an electric circuit mean?

Answer: An electric circuit is a continuous, closed path or loop composed of electronic components through which an electric current flows. Conductors, cells, Switch, and Load are the components of a simple circuit.

Question 12: An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω resistances are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current that flows through it?

R 1 = 100 , R 2 = 50 , R 3 = 500

All the devices are in parallel, so

1/ R = 1/ R 1 + 1/ R 2 + 1/ R 3

1/ R = 1/ 100 + 1/ 50 + 1/ 500 = ( 5 + 10 + 1 )/500 = 16/ 500

R = 500/ 16 = 31.25

Current, through all the appliances

I = V/ R = 220 / 31.25 = 220 X 31.25 = 7.04 A

Now, if only electric iron is connected to the same source such that it takes as much current as all three appliances, i.e. I = 7.04 A, its resistance should be equal to 31.25 .

Question 13: How is the resistivity of alloys compared with those of pure metals from which they may have been formed?

Answer: An alloy often has a higher resistivity than the individual metals that make up the alloy.

Question 14: Write the relation between the resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it.

Answer: The relation between resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it can be represented as follows:

P = V 2 / R

Question 15: How does the use of a fuse wire protect electrical appliances?

Answer: Compared to the main wiring, the fuse wire has a high resistance. Whenever there is an abrupt surge in electric current, the circuit is broken by melting fuse wire. This keeps electrical equipment from being damaged.

Question 16: Why are copper wires used as connecting wires?

Answer: Copper wires are used as the connecting wires because, in the case of copper, the electrical resistivity for it is low. It is ductile, inexpensive and it is an excellent electrical conductor.

Question: Define the SI unit of current.

Answer: The SI unit of current is ampere. An ampere is defined by the flow of one coulomb of Charge per second.

Question 18: How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?

Answer: In order to get 4 Ω, resistance 2 Ω should be connected in series with the parallel combination of 3 Ω and 6 Ω.

1/ R CD = 1/ 3 + 1/ 6 = ( 2 + 1)/ 6

= 3/ 6 = 1/ 2

R CD = 2 , R AB = 2

R AD = R AB + R CD

= 2 + 2 = 4

Therefore, the total resistance of the circuit is R= 4

(b) In order to get 1 , all three resistors should be connected in parallel as

1/ R = 1/ 2 + 1/ 3 + 1/ 6 = ( 3 + 2+ 1)/ 6 = 1

Therefore, the net equivalent resistance of the circuit is R = 1

Question 19: A rectangular block of iron has dimensions L x L x b. What will be the resistance of the block measured between the two square ends? Given p resistivity.

Answer: We have given that a rectangular block of iron has dimensions l x l x b. We need to find the resistance of the block measured between the two square ends.

The resistance is given by the below formula as follows :

L is length of block

A is area of cross section

In this case,

Length of the rectangular block is l and area of block is l x b. So, resistance of the block measured between the two square ends is :

R = p b/l 2

So, the resistance of the block measured between the two square ends इस R = pb/l 2

Question 20: Ammeter burn out when connected in parallel. Give reasons.

Answer: When a low resistance wire is connected in parallel, a huge quantity of current travels through it, causing it to be either burned out or short-circuited.

Question 21: Should the resistance of an ammeter be low or high? Give reason(s).

Answer: The resistance of an ammeter should be zero, as the ammeter should not affect the flow of current in a circuit.

Question 22: Why does the connecting rod of an electric heater not glow, but the heating element does?

Answer: As the resistance of the connecting rod is lower than that of the heating element, the connecting rod of an electric heater does not glow. Thus, the heating element produces more heat than the connecting cord, and it glows.

Question 23: The power of a lamp is 60 W. Find the energy in joules consumed by it in 1s.

Answer: Here, given the power of the lamp, P = 60 W time,

So, energy consumed = power x time = (60 x 1) J = 60 J

Question 24: A wire of resistivity ‘p’ is stretched to double its length. What will be its new resistivity?

Answer: When a wire of resistivity p is stretched to double its length, then the new resistivity tends to remain the same because resistivity depends on the nature of the material.

Question 25: What is the resistance of any connecting wire?

Answer: The resistance of the connecting wire made of a good conductor is extremely low and they are assumed to have zero resistance. So, less heat is produced in them and they can be easily used in connections.

Question 26: A number of n resistors each of resistance ‘R’ are first connected in series and then in parallel connection. What is the ratio of the total effective resistance of the circuit in series combination and parallel combination?

Answer : Total effective resistance of the circuit in series combination is R s = nR

And for parallel combination is R p = R/ n and

R s/ R p = nR/ R/ n

The ratio will be n 2 .

Question 27: Calculate the total number of electrons constituting one coulomb of charge.

The Charge of an electron = 1.6 × 10 -19 C.

According to the concept of charge quantisation,

Q = nqe, where we suppose ‘ n’ is the number of electrons and similarly ‘ qe’ is the Charge of the electron.

Substituting these values in the said equation, the number of electrons constituting one coulomb of Charge can be calculated as follows:

1C = n X 1.6 X 10 -19

n= 1 1.6 X 10 -19 = 6.25 X 10 18

Therefore, the number of electrons in one coulomb of Charge = 6. 25 × 10 18 .

Question 28: How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts.

Here, V = 220V , R = 55

By Ohm’s law, V = IR

Therefore, 220 = 7 x 55 or I = 4A

The wattage of electric iron = Power

= V 2 R = (220) 2 55 = 880 W

Question 29: A current of 1 ampere flows in a circuit of series connection containing an electric lamp and a conductor of 5 Ω and connected to a 10 V battery. Calculate the resistance of the given electric lamp.

Therefore, if the resistance of 10 Ω is connected in parallel with this series combination, what type of change (if any) in current flowing through the 5 Ω conductor and potential difference across the lamp will take place? Give reasons.

Let R lamp represent the resistance of the lamp.

Current ( I ) = 1 A

Resistance of conductor (R conductor ) = 5 Ω

The potential difference of battery ( V) = 10 V

Given that the lamp and conductor are linked in series, the same amount of current 1 A will flow through them both.

Using Ohm’s law,

R net = V I

R net = 10 क्ष 1

We know, in series connection

R net = R lamp + R Conductor

10 = R lamp + 5

The potential difference across lamps,

R lamp = I x R lamp

= 1 x 5 = 5 V

When a resistor of 10 Ω resistor connected parallel to the series combination of lamp and conductor

( R net = 5 + 5 = 10 ) then the equivalent resistance,

1/ R eq = 1/ 10 + 1/ 10 = 2/ 10 = 1/ 5

Using Ohm’s law,

I’= V/ R eq

Equal distribution of current will occur in two parallel parts.

Thus, I’/2 = 1A current will pass through both the lamp and the resistor of 5 (because they are connected in series).

The potential difference across the lamp (R lamp = 5 ).

V’ lamp = 1×5 = 5 V

Therefore, the current flowing through the conductor of resistance 5 and the potential difference across the bulb won’t change.

Question 30: What is electrical resistivity? In a particular series electrical circuit comprising a resistor made up of a metallic wire, the ammeter generally reads 5 A. The previous reading of the ammeter decreases to half in case the length of the wire is doubled. Why?

Answer: Resistivity is a property of a conductor that prevents the flow of electric current. A specific material has a particular resistance. Resistance is inversely proportional to current flow and directly proportional to conductor length.

When the length is doubled, the resistance doubles and the current flow is reduced by half. This is what’s causing the ammeter value to drop.

Question 31: (i) List the three factors on which the resistance of a conductor depends.

(ii) Write the SI unit of resistivity.

(i) A conductor’s resistance is influenced by the following factors:

(1) Length of the conductor: The resistance (R) will increase as the conductor’s length (I) increases.

(2) Area of the cross-section of the conductor: (as the cross-sectional area of the conductor increases, the resistance decreases.

(3) Nature of conductor.

(ii) SI unit of resistivity is Ω m.

Question 32: An electric bulb which is connected to a 220 V generator and the current is 2.5 A. Calculate the power of the bulb.

Here, V= 220 V, I = 2.5 A

Given, Power of the bulb, P = VI = 220 × 2.5 W = 550 W

Question 33: Name a device that helps to maintain a potential difference across a conductor.

One of the devices that aid in maintaining a potential difference across a conductor is a battery, which can consist of one or more electric cells.

Question 34: What is the resistance of an ammeter?

An ammeter’s resistance generally is very minimal, and in an ideal ammeter, it is zero.

Question 35: What is the resistance of a voltmeter?

Answer: The resistance of a voltmeter is ideally infinite resistance.

Question 36: What is the commercial unit of electrical energy? Represent it in terms of joules.

The commercial unit of electrical energy is kilowatt/hr

1 kW/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 10 6 J

Question 37: Explain two disadvantages of series arrangement for a household circuit.

The two drawbacks of series circuits for household wiring are:

If one electrical appliance in a series circuit stops functioning for any reason, the entire circuit will break, and all other electrical appliances will also stop functioning.
Because there is only one switch for every electrical device in a series circuit, they cannot be turned on or off independently.

Question 38: What is meant by the saying that the potential difference between two points is 1 V?

The potential difference between two points is 1V when 1 J of work is done to move a 1 C of Charge from one location to the other.

Question 39: Two equal wires of equal cross-sectional area, one of copper and the other of manganin , have the same resistance. Which one will be longer?

Using the equation, = RA I , where is the resistivity, R is the resistance, and A is the area.

Resistance of Copper wire = 1 l 1 A

Resistance of Manganin wire = 2 l 2 A

1 l 1 = 2 l 2 (As l is constant)

Since 1 <<< 2

So, l 1 >>>> l 2

I.e. Copper wire would be longer.

Question 40: Three equal resistances are connected in series and then in parallel. What will be the ratio of their change in resistances?

Answer 40: When connected in series, Resistance R series = R+R+R= 3R

When connected in parallel, Resistance R parallel = R/3

Ratio of change in resistances= R series R parallel = 3R R/3

Therefore, the ratio of change in resistances is 9:1

Question 41: State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

According to Ohm’s law, the potential difference (voltage) across an ideal conductor is proportional to the current flowing through it at a given temperature.

I.e. V/I = R

Verification of Ohm’s law

Make the circuit indicated in Fig., which consists of four 1.5 V cells, an ammeter, a voltmeter, and a nichrome wire of length XY, say, 0.5 m. (The metals nickel, chromium, manganese, and iron make up the alloy known as nichrome.)

Start by using a single cell as the circuit’s source. Take note of the ammeter’s reading for current (I) and the voltmeter’s reading (V) for the potential difference across the nichrome wire ‘XY’ in the circuit. Add them to the table provided.

Connect two batteries to the circuit next, and then note the ammeter and voltmeter readings for the current flowing through the nichrome wire and the potential difference across the nichrome wire values, respectively.

Use three cells in the circuit first, then four cells, and repeat the process above for each group of cells.

Question 42: If there are 3 x 10 11 electrons flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron 1.6 x 10 19 C.

Using the equation, q = ne

= 3 x 10 11 x 1.6 x 10 – 19 C

= 4.8 x 10 8 C

= 4.8 x 10 8 2 x 60

= 4 x 10 7 A

The current flowing through the electric circuit is 4 x 10 7 A.

Question 43: Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series, and the entire combination is connected to a battery of 30 V. Ammeter and Voltmeter are connected in the circuit. Draw a circuit diagram to connect all the devices in the proper, correct order. What is the current flowing and potential difference across 10 Ω resistance?

Answer 43:

Given, Total resistance, R = R1 + R2 + R3 = 5 + 10 + 15 = 30Ω

Total potential difference, V = 30 volts

V = IR ⇒ I = V R = 30 30 = 1 ampere

∴ Current remains constant in this series,

∴ I1 = I2 = I3 = I;

I2 = 1amp;

As V2 = I2

R2 = 1 × 10 = 10 volts

∴ The potential difference across the 10 Ω is 10 volts.

Question 44: If an electric heater rated 800 W operates 6h/day. Find the Cost of energy to operate it for 30 days at ₹3.00 per unit of consumption.

Answer 44:

Here, the Power of the heater, P = 800 W;

Time, t = 6 hour/day;

No. of days, n = 30;

Cost per unit = ₹3.00;

Thus, Consumed in 1 day = 800 × 6 = 4800 Wh

And, Energy consumed in 30 days = 4800 × 30 = 144000 Wh

144000 1000 = kWh = 144 units

Now, the Cost of 1 unit = ₹3

Therefore, Cost of 144 units = 3 × 144 = ₹432

Question 45: What is the electrical resistivity of a given material? What is its unit? Discuss an experiment to study the factors on which the resistance of conducting wire depends.

Answer 45: Resistivity is an inherent property of a conductor that resists the flow of electric current. The resistivity of each material is unique.

The SI unit of resistance is Ω m.

Experiment to study the depending factors of the resistance of conducting wire.

A nichrome wire, a torch, a 10 W bulb, an ammeter (0–5 A range), a plug key, and some connecting wires are needed.

As illustrated in the figure, assemble the circuit by connecting four 1.5 V dry batteries in series with the ammeter, thereby leaving a gap XY in the circuit.

Observation:

Resistance depends on the length of the conductor, the material of the conductor, and the area of the cross-section.

Connecting the nichrome wire in the XY gap completes the circuit. Insert the key. The ammeter reading should be noted. From the plug, remove the key. [Remember: After measuring the current flowing through the circuit, always remove the key from the plug.]

Replace the nichrome wire in the circuit with the torch bulb, and then determine the current flowing through it by measuring the ammeter’s reading.

Repeat the previous process now using the 10 W bulb in the XY gap. Are there variations in the ammeter readings for the various components connected in the gap XY? What do the aforementioned observations suggest?

By leaving any material component in the gap, you are able to repeat this activity. Watch the ammeter values for each situation. Analyse the results.

Question 46: Calculate the resistance of a given metal wire of length 2m and area of cross-section 1.55 × 106 m² if the resistivity of the metal is taken to be 2.8 × 10-8 Ωm.

Answer 46: For the given metal wire,

Length, l= 2 m

Area of cross-section, A= 1.55 X 10 -6 m 2

Resistivity of the metal, p = 2.8 X 10 -8 m

Since, resistance, R = l A

So, R = ( 2.8 X 10 -8 X 2 1.55 X 10 -6 )

= 5.6 1.55 X 10 -2

= 3.6 X 10 -2

Therefore, R = 0.036

Question 47: When will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why?

Resistance is represented by the equation,

ρ is the resistivity of the wire material,

l is the wire

A is the cross-sectional area of the wire.

It is clear from the equation that the resistance is inversely proportional to the area of the wire cross-section. Therefore, the resistance increases with wire thickness and vice versa. Therefore, a thick wire conducts current more readily than a thin wire.

Question 48: What is represented by joule/coulomb?

Answer 48: The potential difference is represented by the joule/coulomb.

Question 49: A nichrome wire of resistivity 100 W m and copper wire of resistivity 1.62 ohm -m of the same length and same area of the cross-section are connected in series, and current is passed through them. Why does the nichrome wire get heated first?

Answer 49: Looking at the equation

Q= I 2 (pL / A)t

Nichrome wire gets heated first because it has a higher resistance than copper wire.

Question 50: Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and thus can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.

Answer 50:

P = Potential difference

I = Current

R = Resistance

For resistor A,

18 = I 2 x 2

I 2 = 18 2

This is the maximum known current flowing through resistor A.

The maximum known current flowing through the resistors B and C, I’= 3 x 1 2 = 1.5 A.

Question 51: How will you infer, with the help of an experiment, that the same current flows through each and every part of the circuit containing three resistances in series connected to a battery?

You can collect three resistors, R1, R2, and R3, in series to make the circuit.
Then use an ammeter to observe the changes in the overall current flow.
You can remove R1 and take the readings of the potential difference of R2 and R3.
You can remove R2 and take the reading of the potential difference between R1 and R3.

Since the ammeter reading was the same in each case, it can be assumed that the circuit’s current is constant. One can set up an ammeter in several places and watch the current flow to double-check.

Question 52: Calculate the estimated resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance of 20 ohms.

Answer 52: 3.14 X (10 -4 ) 2 m 2

length of the wire, l = 1 m,

radius of the wire, r = 0.01 cm = 1 × 10 -4 m and

given resistance, R = 20Ω

R = l A , where is the resistivity of the material of the wire.

20 = l r 2 = 1 m 3.14 X (10 -4 ) 2 m 2

Therefore, = 6.28 X 10 -7 m

Question 53: A charge of 2 C moves between two plates, maintained at a potential difference of IV. What is the energy acquired by the Charge?

Answer 53: The energy acquired by the Charge, W = QV

Therefore, the energy acquired is 2 J.

Question 54: Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Ohm’s law is used to calculate how the current flow changes through an electrical component.

Ohm’s law states that

I = V/R, which gives the current.

The potential difference is now divided in half while maintaining the same resistance,

Let V’ = V/2 be the new voltage .

Let R’ = R be the new resistance, and the new amount of current be I’.

Ohm’s law is thus used to calculate the current change as shown below:

I’ = V’ R’ = ( V 2 ) R = 1 2 V R = 1 2

As a result, the electrical component’s current is reduced by half,keeping resistance constant.

Question 55: What is the overloading of an electrical circuit? Explain two possible causes due to which overloading might occur in any household circuit. Explain one precaution, if any, that should be taken to avoid the overloading of a domestic electric circuit.

Overloading: The power ratings of the appliances being utilised at a given moment determine the current flowing in household wiring. Electrical appliances with high power ratings take a tremendous amount of electricity from the circuit if too many of them are turned on at once. The overloading of the circuit takes place. . The copper wires in residential circuits get heated up due to extremely high temperatures and can immediately catch fire as a result of heavy currents running through them.

Precaution: As a result, overloading can seriously harm buildings and electrical equipment. To prevent these damages, a fuse with the appropriate rating must be used. Such a fuse wire will melt before the heated circuit wire’s temperature rises to a point where it breaks the circuit.

Question 56: What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

According to the Joule’s heating effect, the heat produced in a resistor is known to be

(i) Directly proportional to the square of current for the given resistor.
(ii) Directly proportional to the resistance for a given current,
(iii) Directly proportional to the time of current flowing through the resistor.

It can be expressed as H = I2Rt

‘H’ is the heating effect, ‘I’ is the electric current, ‘R’ is resistance, and ‘t’ is time.

Experiment to demonstrate Joule’s law of heating

Take an immersion rod for water heating and attach it to a regulator-connected socket. It’s crucial to keep in mind that a regulator regulates how much current flows through a gadget.
Keep the regulator’s pointer at the lowest setting and time how long it takes the immersion rod to heat a specific volume of water.
Increase the regulator’s pointer to the following level. Time the same quantity of water heating with an immersion rod.
To measure higher amounts of the regulator, repeat the previous step.

It has been observed that it takes less time to heat the same amount of water with an increasing electric current. This illustrates the Joule’s Law of Heating.

Application:

Electric appliances like toasters, ovens, kettles, and heaters operate using the leafing effect of current.

Question 57: Why are the coils of electric toasters and irons made of an alloy rather than any pure metal.Give reason(s).

Due to its high resistivity, an alloy has a substantially higher melting point than a pure metal. Alloys are resistant to melting when temperatures are high. As a result, alloys are utilised in heating devices like electric toasters and irons.

Question 58: Name a device that helps to maintain a potential difference across a conductor in a circuit. When do we say that the potential difference across a conductor is 1 volt? Calculate the amount of work done in shifting a charge of 2 coulombs from a point A to B having potential +10V and -5V, respectively.

Answer 58: In a circuit, a battery (or cell) aids in maintaining the potential difference across a conductor.

If 1 joule of labour is expended in transporting 1 coulomb of electrical charge from one location to the other, the potential difference between the two points is said to be 1 volt.

Given, Charge, Q = 2C

Potential at A = +10 V, Potential at B = -5V

Potential difference, (V) = +10 – (-5) = 10 + 5 = 15 volts

W = 15 × 2 = 30 J

Question 59: Which is a better conductor among iron and mercury?

Answer 59: Iron is a better conductor than mercury because the resistivity of mercury is more than the resistivity of iron.

Question 60: Which has more resistance, 100 W bulb or 60 W bulb?

Answer 60: As it is clearly known that R 1 P , the resistance of the 60 W bulb is more.

Question 61: Find the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 106 Ω, (b) 1 Ω, 103 Ω, and 106 Ω.

Answer 61: (a) When 1 Ω and 106 Ω when connected in parallel gives the 10 6 equivalent resistance as follows:

1 R = 1 1 + 1 10 6

R= 10 6 1+ 10 6 10 6 10 6 = 1

Therefore, the equivalent resistance is 1 Ω. 1+ 10 6

(b) When 1 Ω, 103 Ω, and 106 Ω are in parallel, the equivalent resistance is given by

1 R = 1 1 + 1 10 3 + 1 10 6

Solving, we get

R = 10 6 + 10 3 +1 10 6 = 1000000 1000001 = 0.999

Therefore, the equivalent resistance is 0.999 Ω.

Question 62: What are the benefits of connecting electrical devices in parallel with the battery instead of connecting them in series?

There is no voltage division among the appliances when the electrical devices are connected in parallel. The supply voltage is equal to the potential difference across the devices. Devices connected in parallel lower the circuit’s effective resistance as well.

Question 63: Why does the cord of an electric heater not glow while the heating element does?

Answer 63: An electric heater’s heating element is constructed from a high-resistance alloy. The heating element glows red and gets excessively hot when the electricity passes through it. Typically, copper or aluminium, which have low resistance, is used to make the rope. Consequently, the cord doesn’t glow.

Question 64: Discuss the heat generated while transferring 96000 coulombs of Charge in one hour through a potential difference of 50 V.

Answer 64: According to Joule’s law, the heat produced can be calculated as follows:

where assuming,

voltage, V = 50 V

I will be current

t be the time in seconds, 1 hour = 3600 seconds

The amount of current is calculated as follows:

Amount of Current = Amount of Charge Time flow of Charge

Substituting the value, we get

I = 96000 3600 = 26.66 A

Now, to find the heat generated

H = 50 X 26.66 X 3600 = 4.8 X 10 6 J

Therefore, the heat generated is 4.8 X 10 6 J

Question 65: An electric iron of resistance 20 Ω draws a current of 5 A. Calculate the heat developed in 30 s.

Answer 65: The Joule’s law of heating, which is represented by the equation, can be used to determine how the heat is produced as follows:

Putting the data in the above equation, we get,

H = 100 × 5 × 30 = 15,000 J

The amount of heat produced by the electric iron in 30 s is 15,000 J.

Question 66: What factors determine the rate at which energy is delivered by a current?

Answer 66: Electric power is the rate at which electric equipment uses electricity. Therefore, the power of the appliance is defined as the rate at which energy is delivered by a current.

Question 67: How is the connection of a voltmeter made in the circuit to measure the potential difference between two points?

Answer 67: The voltmeter should be connected in parallel to each of the two points in order to measure the voltage between any two points.

Question 68: Draw a circuit taking an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the resistor of 12 ohm. What would be the new reading in the ammeter and the voltmeter?

Answer 68:

Let us take the total resistance of the circuit = R

The equivalent resistance R is equal to the total resistance because all three resistors are connected in series.

R = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Therefore,

V = 2V + 2V + 2V = 6V

I = V R = 6 25 = 0.24 A

The reading of the voltmeter across R’ = 12 Ω is

= 0.24 X 12 = 2.88 V

Question 69: Find the following in the electric circuit given in Figure 12.9

(a) Effective resistance of the two 8 Ω resistors in the given combination

(b) Current flowing through the resistor of 4 Ω

(d) Power dissipated in a resistor of 4 Ω (e) Difference in ammeter readings, if any

Answer 69: (i) Since two 8 resistors are in parallel, then their effective resistance R p is given by

1 R p = 1 R 1 + 1 R 2 = 1 8 + 1 8 = 1 4

(ii) Total resistance in the circuit

R = 4 + R p = 4 + 4 + 8

Current, through the circuit,

I = V R = 8 8 = 1 A

Thus, the current through the 4 resistor is 1 A as 4 and R p are in series and the same current flows through them.

(iii) Potential difference across 4 resistor is potential drop by the 4 resistor.

i.e. V = IR = 1 X 4 = 4 V

(iv) Power dissipated in 4 resistor

P = I 2 R = 1 2 X 4 = 4 W

(v) There is no difference in the reading of ammeters A 1 and A 2 as the same current flows through all elements in a series current.

Question 70: A copper wire having a diameter of 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of the wire to make its resistance 10 Ω? How much will the resistance change if the diameter is doubled?

Answer 70: The formula provides the resistance of a copper wire with a cross-sectional area of m2 and a length in metres.

The area of the cross-section of the wire is calculated as follows

A = ( Diameter 2 ) 2

Substituting the values in the formula, we get

l = RA = 10 X 3.14 X ( 0.0005 2 2 ) 1.6 X 10 -18 = 10 X 3.14 X 25 4 X 1.6 = 122.72 m

The new diameter of the wire is 1mm, or 0.001m when the wire’s diameter is doubled. Therefore, the resistance can be calculated as follows:

R = l A = 1.6 X 10 -18 X 122.72 m ( 0.001 2 ) 2 = 250.2 X 10 -2 = 2.5

The new resistance is 2.5 , and the wire’s length is 122.72 m.

Question 71: Difference features between Overloading and Short-circuiting in Domestic circuits

Answer 71:

Overloading: Overloading occurs when a circuit is used by too many electrical devices with high power ratings that are switched on simultaneously.

The copper wire used in domestic wiring heats up to an exceedingly high temperature as a result of an excessively high current running through the circuit, and a fire may subsequently ignite.

Short-circuiting: Short-circuiting is a direct result of touching bare live and neutral wires. Because the circuit’s resistance is so low in this situation, a lot of current passes through it, heating the wires to a high temperature and possibly igniting a wire. .

Question 72: When a battery of 12 V is connected across an unknown resistor, there is a current flow of 2.5 mA in the electric circuit. Find the resistance of the resistor.

Answer 72: Using Ohm’s Law, the resistor’s value can be determined as follows:

Putting the data in the equation, we get

R = 12 2.5 X 10 -8 = 4.8 X 10 3 = 4.8

Question 73: Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance.

Here, R 1 = 10 , R 2 = 15 , R 3 = 5

In a parallel connection, equivalent resistance ( R eq ) is given by

1 R eq = 1 R 1 + 1 R 2 + 1 R 3

So, 1 R eq = 1 10 + 1 15 + 1 5

1 R eq = 3+ 2+6 30 = 11 30

Therefore, R eq = 30 11 = 2.73

Question 74: A battery of 9 V is connected in a series system with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. What quantity of current would flow through the 12 Ω resistor?

Answer 74: There is no existing division in a series connection. An equal amount of current travels across each resistor.

We apply Ohm’s law to determine the amount of current passing through the resistors.

Let’s

first determine the equivalent resistance in the manner described below:

R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Using Ohm’s law,

I = V R = 9V 13.4 = 0.671 A

The current flowing across the 12 Ω resistor is 0.671 A.

Question 75: Suppose the resistance of an electrical component remains constant, and the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?

Answer 75:

Knowing that,

If, V’ = V 2

If I’ = V ‘ R = V 2R = 1 2

As a result, an electrical component’s current reduces by half of what it was.

Question 76: Two same resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in both cases.

Answer 76:

Let the resistance of each resistor be R.

For series combination,

R s = R 1 + R 2

So, R s = R + R = 2R

For parallel combination,

1 R p = 1 R 1 + 1 R 2 or R p = R 1 R 2 R 1 + R 2

So, R p = R x R R + R = R 2

Required Ratio = R s R p = 2R R/2 = 4 : 1

Question 77: Explain the use of an electric fuse. What type of material is used for fuse wire and why?

Answer 77: Electric fuses guard against the very high electric current by blocking it from flowing into circuits and appliances. It is composed of a wire formed of a metal or alloy with an appropriate melting point, such as lead, copper, iron, or aluminium. The temperature of the fuse wire rises if a current more than the allowed amount runs through the circuit. The fuse wire melts, as a result, breaking the circuit.

Question 78: When an electric current flows through a conductor, it tends to become hot. Justify.List the factors on which the heat produced in a conductor depends. State Joule’s law of heating. How will the heat produced in an electric circuit be affected by this if the resistance in the circuit is doubled for the same current?

Answer 78: A conductor heats up when an electric current is carried through it. This is referred to as the current heating effect. The electrical energy is converted into heat energy to produce the heating effect of current. An electrical energy source is a cell or battery. The potential difference between the two terminals of the cell is created by the chemical reaction within, which causes the electrons to move and for current to flow through a resistor. The source must continue using up its energy. While maintaining the current, some of the source energy may be used for productive activity, while the remaining source energy may be used to generate heat.

Question 79: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer 79: Let ‘x’ be the number of resistors required.

The equivalent resistance of the resistor R in the parallel combination is given by

1 R = x X 1 176 = R = 176 x

Now, using Ohm’s law. The number of resistors can be calculated as follows:

Substituting the values, we get

176 x = V I

x = 176 X 5 220 = 4

The number of resistors required is 4.

Question 80: Two wires of the same material and same length have radii R and r. Compare their resistances.

Answer 80: Suppose R and r are resistances, then R = r as p and I are the same.

Question 81: Several electric bulbs designed and are supposed to be used on a 220 V electric supply line are rated 10 W. How many such lamps can be connected in parallel together across the two wires of a 220 V supply line if the maximum allowable current is 5 A?

Answer 81: The resistance of the bulb is calculated as follows:

R = (220) 2 10 = 4840

The resistance of x number of electric bulbs is calculated as follows:

R = V I = 220 5 = 44

The resistance of each electric bulb is 4840 .

The equivalent resistance of x bulbs is given by

1 R = 1 R 1 + 1 R 1 + 1 R 1 + ………up to x times

1 R = 1 R 1 X x

x = R1 R = 4840 44 = 110

Hence, 110 lamps can be connected together in parallel.

Question 82: A fuse wire melts at 5 A. If it is desired that the fuse wire of the same material melt at 10 A, then in your opinion whether the new fuse wire should be of a smaller or larger radius than the earlier one? Give reasons for your answer.

Answer 82: Let R be the resistance of the wire; the heat produced in the fuse at 5A is

H = (5) 2 R (H – I 2 R t)

Fue melts at (5) 2 R joules of heat

Let R’ be the resistance of the new wire

So, the heat produced in 1 second =(10) 2 R’

To prevent it from melting

(5)2 R = (10)2R’ or R’ = R 4

Therefore, the cross-sectional area of the new fuse wire is four times the first fuse.

Now, A = r 2 , so the new radius is twice as large as the old one. The new fuse wire, which is the same material and length as the old one, has a greater radius at 10 A.

Question 83: How many bulbs of 81 should be joined in parallel to draw a current of 2 A from a battery of 4V?

Answer 83:

Let n be the number of bulbs.

1/R = 1/R 1 + 1/R 2 +………………+ 1/R n = n 8

The number of bulbs is 4.

Question 84: When will current flow more easily: through a thick wire or a thin wire of the same material, when connected to the same electric source? Why?

Answer 84: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.

Question 85: A hot plate of an electric oven connected to a 220 V supply line has two resistance coils such as A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What will be the currents in the three cases?

Case (i) When coils are used separately

By using Ohm’s law, we will be able to find the current flowing through each coil as follows:

I = 220 V 24 = 9.166 A

When used individually, each resistor allows 9.166 A of current to pass through it.

Case (ii) When the coils are connected in series

The total resistance is 24 Ω + 24 Ω = 48 Ω in the series circuit

The current flowing through this series circuit is calculated as follows:

I = V R = 220 V 48 = 4.58 A

Therefore, a current of 4.58 A will flow through the circuit in series.

Case (iii) When the coils are in parallel

connection,

the equivalent resistance is calculated as follows:

R = 24 X 24 24 + 24 = 576 48 = 12

By using Ohm’s law, the current flowing through the parallel circuit is given by

I = V R = 220 12 = 18.33 A

The current is 18.33 A in the parallel circuit.

Question 86: What happens to the current in a circuit if its resistance is doubled?

Answer 86: As current and resistance are inversely proportional, the current is reduced to half of its previous value.

Question 87: Let the resistance of a component of electric current remain constant while the potential difference across the two ends of the component decreases to half of its earlier value. What type of change will occur in the current through it?

Answer 87: A thick wire connected to the same source conducts current more easily than a short wire made of the same material. As thickness reduces, resistance increases.

Question 88: Compare the power consumed in the 2 Ω resistor in each of the following circuit conditions: (i) a 6 V battery in series connection with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel connection with 12 Ω and 2 Ω resistors.

Answer 88: (i) Since the resistors 1 Ω and 2 Ω are connected in series, and there is a 6 V potential difference, their equivalent resistance is given by 1 Ω + 2 Ω = 3 Ω. Using Ohm’s law, the following formula is used to determine the circuit’s current:

I = V R = 6 3 = 2 A

2 A current will flow across all the components in the circuit because there is no division of current in a circuit of series connection.

The power in the 2 resistor is calculated as follows:

P = I 2 R = (2) 2 X 2 = 8 W

Thus, the power consumed by the 2 Ω resistor is 8 W.

(ii) The voltage between the resistors stays constant when 12 and 2 resistors are linked in parallel. Given that a 2 Ω resistor has a 4 V voltage across it, we can use the formula below to determine how much power is used by the resistor: V 2 4 2

P = V 2 R = 4 2 2 = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Question 89: Two cubes, A and B, are made of the same material. The side of B is thrice that of A. Find the ratio R A /R B .

The value of R A = L A and

R B = 3L 9 A

R A : R B = 3: 1

Question 90: What happens to the resistance of a circuit if the current through it is doubled?

Resistance is unchanged since the circuit’s resistance is independent of the current flowing through it.

Question 91: Two metallic wires, A and B, are connected. Wire A has lengths l and radius r, while B has lengths 2l and 2r. If both the wires are of the same material then find the ratio of total resistances of series combination and the resistance of wire A.?

Answer 91 . Here, the Resistance of metallic wire A, R 1 = l A

Resistance of metallic wire B , R 2 = 2l 4 r 2

The total resistance in series can be expressed as R = R 1 + R 2

= l A + 2l 4 r 2

= 3 l 2 r 2

The ratio of the total resistance (R) in series to the resistance of A (R1) is

R R 1 = 3 l 2 r 2 l r 2

The ratio of the total resistance (R) in series to the resistance of wire A is 3 2 .

Question 91: Illustrate how you would connect given three resistors, each of whose resistance is 6 Ω so that the combination has a resistance of (i) 9 Ω or (ii) 4Ω

Answer 91:

(i) When we connect R1 in series with the parallel combination of R2 and R3, as shown in Fig. (a).

The equivalent resistance is

R = R 1 + R 2 R 3 R 2 + R 3 = 6 + 6 X 6 6 + 6

= 6 + 3 = 9

(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is

R = 12 x 6 12 + 6 = 72 18 = 4

Question 92: How does the resistance of a wire depend upon its radius?

Answer 92: As R 1 A

The resistance of the above-mentioned wire is directly proportional to its given radius.

Question 93: Which of the two uses more energy, a 250 W TV set in 1 hr or a 1200 W toaster in 10 minutes?

The total energy consumed by the electrical devices is represented by the equation

H = Pt , where the power of the appliance is P and t is the time

This formula is used to determine how much energy a TV with a 250 W power rating uses:

H = 250 W × 3600 seconds

= 9 × 105 J

In the same way, the energy consumed by a toaster with a 1200 W power rating is

H = 1200 W × 600 s = 7.2 × 105 J

From the part of the above calculation, it can be said that the energy consumed by the TV is greater than the toaster.

Question 94: Two wires are of the same length and same radius, but one of them is of copper, and the other is of iron. Which will have more resistance.

Since, R = 1 A

But A and I have the same value. It is absolutely determined by the resistivity; hence iron has a higher resistance.

Question 95: An electric heater of 8 Ω resistance draws 15 A of current from the service mains supply for 2 hours. Calculate the rate at which heat is produced in the given heater.

The rate at which the heat production takes place in the heater is thus calculated using the following formula

P = (15A) 2 × 8 Ω = 1800 watt

The electric heater produces heat at the rate of 1800 watt

Question 96: The resistance of a given wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm metre , find the estimated length of the wire.

Answer 96:

Here, r= 0.01 cm = 10 -4 m, p = 50 x 10 -8 m and R = 10

As, R = l A

Or, I = RA = R ( r 2 )

So, I = 10 50 X 10 -8 3.14 X (10 -4 ) 2

I = 0.628 m = 62.8 cm

Question 97: Explain the following.

Why is tungsten used almost exclusively for the filament of electric lamps?
Why are the conductors of electrical heating devices, like bread-toasters and electric irons, mostly made of an alloy rather than a pure metal?
Why is the series arrangement not used in domestic circuits?
How does the resistance of a wire vary with its area of cross-section?
Why are copper and aluminium wires usually employed for electricity transmission?
Tungsten has a very high resistance and melting point. This characteristic prevents it from burning easily when heated. At high temperatures, electric bulbs are operated. As a result, tungsten is a popular metal choice for electric lamp filaments.
Due to their high resistivity, alloys are used as the conductors in electric heating equipment like bread-toasters and electric irons. Because of its high resistance, it generates a lot of heat.
Because each component in the circuit only receives a tiny voltage as a result of the voltage being divided into a series circuit, when one component fails, the circuit is broken and none of the components work. . Because of this, domestic circuits do not employ series circuits.
The relationship between resistance and cross-sectional area is inversely proportional. This means the resistance decreases as the cross-sectional area increases and vice versa.
Copper and aluminium are frequently used for the transmission of electricity because they are effective conductors of electricity and have low resistance.

Question 98: How will you justify that the same potential difference (voltage) will exist across three resistors connected in a parallel arrangement to a battery?

You can take three resistors, R1, R2 and R3, and connect them in parallel to make a circuit, as shown in the figure.

Then use a voltmeter to take the reading of the potential difference of the three resistors in parallel combination.

Now, you can remove the resistor R1 and take the reading of the potential difference of the remaining resistors in combination.

Then, you can remove the resistor R2 and take the reading of the potential difference of the remaining resistor.

In each case, the Voltmeter reading appears to be the same, which shows that the same potential difference tends to exist across three resistors connected in a parallel arrangement.

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Q.1 An object is placed in front of a convex lens at a distance of

where f is the magnitude of the focal length of the lens. Prove that the magnification produced by the lens is n. Also find the two values of the object distances for which a convex lens of power 2.5 D will produce an image that is four times as large as the object.

Marks: 5 Ans

Given: Object distance , | u | = f f n ; Focal length = f; Image distance, v = ByLens’Formula , 1 f = 1 v 1 u Object distance is always negative. 1 f = 1 v + 1 f f n 1 v = 1 f 1 f f n 1 v = f f n f f f f n 1 v = 1 n f f n v= f f n 1 n Now, magnification, | m | = v u |m|=n Hence, proved. Now, u= f f n Butf = 1 2.5 m = 0.4 m ˆµ f = 1 P u = 0.4 0.4 4 ( ˆµ n=4 ) u= ( 0.4 0.1 ) u = 0.5mor 0.3 m

Q.2 An object is placed at 10cm in front of a concave mirror of focal length 15cm. Find the position, nature and size of the image.

Given u = 10 cm ; f = 15 cm ; v = ; m = 1 v + 1 u = 1 f 1 v + 1 10 = 1 15 1 v = 1 15 + 1 10 1 v = 2+3 30 = 1 30 v = 30 cm Positive sign indicates that the image is virtual and erect and form 30 cm behind the mirror.

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Chapter 1 - chemical reactions and equations.

Chapter 2 - Acids, Bases and Salts

Chapter 3 - metals and non-metals, chapter 4 - carbon and its compounds, chapter 5 - periodic classification of elements, chapter 6 - life processes, chapter 7 - control and coordination, chapter 8 - how do organisms reproduce, chapter 9 - heredity and evolution, chapter 10 - light reflection and refraction, chapter 11 - human eye and colourful world, chapter 13 - magnetic effects of electric current, chapter 14 - sources of energy, chapter 15 - our environment, chapter 16 - management of natural resources, faqs (frequently asked questions), 1. why is learning about electricity important for the students in school.

In order to be thorough with the most important commodities in modern Sciences , one must understand electricity. A fundamental understanding of electricity is also necessary for many technical occupations in order to create the technologies and goods that we use in our every -day lives. For this purpose, Extramarks has curated this important questions Class 10 Science Chapter 12

2. How has electricity made our life easier in present times?

Nowadays, electricity is almost necessary for our daily work functioning. It is utilized for residential purposes such as operating fans, computer, electric stoves, air conditioning, and lighting up rooms. It is used in factories to operate heavy machines. Electricity is also used to produce various products, including food, clothing, and paper.

3. Is Electricty Important chapter for CBSE Class 10 Exam?

Yes, electricity chapter is important for CBSE Class 10 Exam

4. Where can I get all study materials related to CBSE Class 10?

You can get all the study materials of CBSE Class 10 at Extramarks

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Extra Questions for Class 10 Science Chapter 12 Electricity

Get extra questions for Class 10 Science Chapter 12 Electricity with PDF. Our subject expert prepared these solutions as per the latest NCERT textbook. These extra questions will be helpful to revise the important topics and concepts. You can easily download all the questions and answers in PDF format from our app.

Electricity Class 10 Science Extra Questions with Answers

Question 1: Write S.I. unit of resistivity.

Answer: Ohm-metre (Ωm).

Question 2: Name a device that helps to maintain a potential difference across a conductor.

Answer: Cell or battery

Question 3: Write relation between heat energy produced in a conductor when a potential difference V is applied across its terminals and a current I flows through for ‘t’

Answer: Heat produced, H = VIt

Question 4: State difference between the wire used in the element of an electric heater and in a fuse wire.

Answer: The wire used in the element of electric heater has a high resistivity and have a high melting point, i.e. even at a high temperature element do not burn while fuse wire have a low melting point and high resistivity.

Question 5: How is an ammeter connected in a circuit to measure current flowing through it?

Answer: In series.

Question 6: What happens to resistance of a conductor when its area of cross-section is increased?

Answer: Resistance decreases as R∝1/A

Question 7: Two resistors of 10 Ω and 15 Ω are connected in series to a battery of 6 V. How can the values of current passing through them be compared?

Answer: In series, same current flows through each resistor. So, ratio of current is 1:1.

Question 8: How much current will an electric bulb draw from 220 V source if the resistance of the bulb is 1200Ω? If in place of bulb, a heater of resistance 100 Ω is connected to the sources, calculate the current drawn by it

Class 10 Chapter 12 Electricity Extra Questions 8

Question 9: Draw a schematic diagram of an electric circuit comprising of 3 cells and an electric bulb, ammeter, plug-key in the ON mode and another with same components but with two bulbs in parallel and a voltmeter across the combination.

Class 10 Chapter 12 Electricity Extra Questions 9

Question 10: Out of the two wires X and Y shown below, which one has greater resistance? Justify your answer.

Class 10 Chapter 12 Electricity Extra Questions 10

Answer: Wire ‘Y’ has greater resistance as it has more length than wire ‘X’. It is because resistance of wire is directly proportional to the length of wire.

Question 11: A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

Class 10 Chapter 12 Electricity Extra Questions 11

Question 12: (a) What do the following circuit symbols represent?

(b) The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. Find the resistance of heater when in use.

Class 10 Chapter 12 Electricity Extra Questions 12

Answer: (a) (i) Wires crossing without touching each other. (ii) Rheostat/Variable resistor

(b) Given: V = 60 V, I = 4 A, R = ? From Ohm’s law, V = IR ⇒ 60 = 4 × R ⇒ R = 15 Ω

Question 13: The charge possessed by an electron is 1.6 × 10 -19 coulombs. Find the number of electrons that will flow per second to constitute a current of 1 ampere.

Class 10 Chapter 12 Electricity Extra Questions 13

Question 14: Explain the role of fuse in series with any electrical appliance in an electric circuit. Why should a fuse with defined rating for an electric circuit not be replaced by one with a larger rating?

Answer: Fuse wire is a safety device connected in series with the live wire of circuit. It has high resistivity and low melting point. It melts when a sudden urge of large current passes through it and disconnects the entire circuit from the electrical supply. But, in case if we use a larger rating instead of a defined rating, then it will not protect the circuit as high current will easily pass through it and it will not melt.

Question 15: The wattage of a bulb is 24 W when it is connected to a 12 V battery. Calculate its effective wattage if it operates on a 6 V battery (Neglect the change in resistance due to unequal heating of the filament in the two cases).

Class 10 Chapter 12 Electricity Extra Questions 15

Question 16: In an experiment to study the relation between the potential difference across a resistor and the current through it, a student recorded the following observations:

Potential difference (V)	1.0	2.2	3.0	4.0	6.4
Current (A)	0.1	0.2	0.6	0.4	0.6

On examine the above observations, the teacher asked the student to reject one set of readings as the values were out of agreement with the rest. Which one of the above sets of readings can be rejected? Calculate the mean value of resistance of the resistor based on the remaining four sets of readings.

Answer: The third reading for V = 3.0 volt and I = 0.6 A will be rejected as it has larger deviation from the rest of the readings. The value of resistance in the other four observations will be I (using R = V/I) 10Ω, 11 Ω, 10 Ω and 10.67 Ω. So, the mean value of resistance = 41.67/4 = 10.417 = 10.42 Ω

Question 17: What is an electric circuit? Distinguish between an open and a closed circuit.

Answer: An arrangement for maintaining the continuous flow of electric current by the electrical energy source through the various electrical components connected with each other by conducting wires is termed as electric circuit. An open circuit does not carry any current, while a closed circuit carries current.

Question 18: Study the following electric circuit and find (i) the current flowing in the circuit and (ii) the potential difference across 10 Ω resistor.

Class 10 Chapter 12 Electricity Extra Questions 18 i

Answer: 10 Ω and 20 v are connected in series, their equivalent resistance is R S = R 1 + R 2 ⇒ R S = 10+20 ⇒ R S = 30 Ω

Class 10 Chapter 12 Electricity Extra Questions 18 ii

Question 19: Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.

Answer: When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt. Potential difference between two points is

Class 10 Chapter 12 Electricity Extra Questions 19

Question 20: V-I graph for two wires A and B are shown in the figure. If both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Give justification for your answer.

Class 10 Chapter 12 Electricity Extra Questions 20 i

Answer: Greater than slope of V-I graph, greater will be the resistance of given metallic wire. In the given graph, wire A has greater slope then B. Hence, wire A has greater resistance. For the wires of same length and same thickness, resistance depends on the nature of material of the wire, i.e.,

Class 10 Chapter 12 Electricity Extra Questions 20 ii

Hence, wire ‘A’ is made of a material of high resistivity.

Question 21: The figure below shows three cylindrical copper conductors along with their face areas and lengths. Discuss in which geometrical shape the resistance will be highest.

Class 10 Chapter 12 Electricity Extra Questions 21

Question 22: An electric bulb of resistance 200Ω draws a current of 1 Ampere. Calculate the power of the bulb the potential difference at its ends and the energy in kWh consumed burning it for 5h.

Answer: Power of the bulb,

P = 1 2 × 200

Energy consumed by bulb in 5h in burning = Power × Time

= 200 × 5

= 1000 Wh

= 1KWh

Question 23: Two identical wires one of nichrome and other of copper are connected in series and a current (I) is passed through them. State the change observed in the temperatures of the two wires. Justify your answer. State the law which explains the above observation.

Answer: The resistivity of nichrome is more than that of copper so its resistance is also high. Therefore, large amount of heat is produced in the nichrome wire for the same current as compared to that of copper wire. Accordingly, more change in temperature is observed in the nichrome wire. This is explained by Joule’s law of heating.

Joule’s law of heating: It states that the amount of heat produced in a conductor is

(i) directly proportional to the square of current flowing through it, i.e., H ∝ I 2 (ii) directly proportional to the resistance offered by the conductor to the current, ie., H ∝ R

(iii) directly proportional to the time for which current is flowing through it, i.e., H ∝ t Combining these, we get H ∝ I 2 Rt ⇒ H = KI 2 Rt Where K is proportionality constant and in SI system, it is equal to one.

Question 24: An electric bulb is rated at 60 W, 240 V. Calculate its resistance. If the voltage drops to 192 V, calculate the power consumed and the current drawn by the bulb. (Assume that the resistance of the bulb remain unchanged.)

Class 10 Chapter 12 Electricity Extra Questions 24

Question 25: A torch bulb is rated 2.5 V and 750 mA. Calculate (i) its power, (ii) its resistance and (iii) the energy consumed, if this bulb is lighted for four hours.

Class 10 Chapter 12 Electricity Extra Questions 25

Question 26: Series arrangements are not used for domestic circuits. List any three reasons.

Answer: Series arrangements are not used for domestic circuit because

The electrical appliances need current of widely different values to operate properly.
In series arrangement, when one component fails, the circuit is broken and none of the components works.
All electrical appliances work at a constant voltage. But in series circuit, the current is constant throughout the electric circuit and potential is different across the different components. So, series arrangement is not suitable for domestic circuits.

Question 27: Name the physical quantity which is (i) same (ii) different in all the bulbs when three bulbs of: (a) same wattage are connected in series. (b) same wattage are connected in parallel. (c) different wattage are connected in series. (d) different wattage are connected in parallel.

Answer: (a) For identical bulbs in series- same current, same potential difference. (b) For identical bulbs in parallel- same potential difference, different current. (c) For unidentical bulbs in series- same current, different potential difference. (d) For unidentical bulbs in parallel- different current, same potential difference.

Question 28: Two devices of rating 44 W, 220 V and 11 W, 220 V are connected in series. The combination is connected across a 440V mains. The fuse of which of the two devices is likely to burn when the switch is ON? Justify your answer.

Class 10 Chapter 12 Electricity Extra Questions 26

Question 29: Two resistors with resistances 5Ω and 10 Ω are to be connected to a battery of emf 6 V so as to obtain: (i) minimum current (ii) maximum current (a) How will you connect the resistances in each case? (b) Calculate the strength of the total current in the circuit in the two cases.

Answer: (a) As current is inversely proportional to resistance for the same voltage. So, to get maximum current, the equivalent resistance has to be less. This means the resistors must be connected in parallel. To get minimum current, the equivalent resistance has to be greater as I∝1/R. This means the resistors must be connected in series.

Class 10 Chapter 12 Electricity Extra Questions 29

Question 30: (a) Define the term ‘volt’.

(b) State the relation between work, charge and potential difference for an electric circuit. Calculate the potential difference between the two terminals of a battery if 100 J of work is required to transfer 20 C of charge from one terminal of the battery to the other.

Answer: (a) When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt.

(b) Potential difference, V = Work done on unit charge =W/q Work is 100J, q=20C Potential difference, V= W/q = 100/20 = 5V

Question 31: Define the term ‘coulomb’.

Answer: When 1 A current flows across the wire in 1 second, the charge transfer across its ends is said to be 1 coulomb.

Question 32: (a) How is the direction of electric current related to the direction of flow of electrons in a wire?

(b) Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes.

Answer: (a) Conventional direction of electric current is opposite to the direction of flow of electrons in a wire.

Class 10 Chapter 12 Electricity Extra Questions 32

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Important Questions for CBSE Class 10 Science Chapter 15 - Our Environment 2024-25

Class 10 Important Question
Chapter 15: Our Environment

CBSE Class 10 Science Chapter-15 Important Questions with Answers - Free PDF Download

Class 10 is a very important year for students as it is the first time students are attempting to write their board examinations. Their 10th board examination marks play a very important role in determining their future grades as well. In this article, we will take a detailed look into class 10 science chapter 15 important questions so that students can prepare better for their examinations. It is very important that students must utilize this time to study hard and these important questions will help students to revise better and go over the main points in the chapter. It is of importance that students be conscious of the important questions which have a high potential of coming within the exams. Students who don’t understand the topic alright must study these questions of sophistication for class 10 so that students can have a better understanding of the chapter and also know what are the important topics to specialize in.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Download CBSE Class 10 Science Important Questions 2024-25 PDF

Also, check CBSE Class 10 Science Important Questions for other chapters:

CBSE Class 10 Science Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15	Our Environment
16	Chapter 16

Related Chapters

Study Important Questions for Class 10 Chapter 15- Our Environment

Very Short Answer Questions (1 Mark)

1. A food chain always starts with

respiration

photosynthesis

decay

nitrogen fixation

Ans: Photosynthesis

2. Ozone layer is damaged by-

methane

carbon-dioxide

Sulphur-dioxide

3. Which of the following limits the number of trophic levels in a food chain?

water

polluted air

deficient food supply

decrease in energy at higher trophic levels

Ans: Decrease in energy at higher energy levels.

4. Name the main source of energy in self-sustaining ecosystem?

5. Write an aquatic food chain.

Ans: An aquatic food chain is as follows:

Phytoplankton $\to $ Zooplankton $\to $ Small fish$\to $ large fish.

6. Which of the following is non-biodegradable-?

paper

cloth

Ans: plastic

7. Which of the following is not a terrestrial ecosystem-?

forest

desert

aquarium

Ans: Aquarium

8. What will happen if deer is missing in the given food chain?

Grass $\to $ Deer $\to $ Tiger

The population of tiger decreases and the population of grass increases

The population of grass decreases

Tiger will start eating grass

The population of tiger increases

Ans: Population of tiger decreases and population of grass increases.

9. What is trophic level?

Ans: The position that an organism occupied in a food chain is called trophic level.

10. Write a fresh water food chain?

Ans: Phytoplankton $\to $ zooplankton $\to $ small fish $\to $ large fish.

11. The decomposers in an ecosystem-

convert organic material to inorganic forms

convert inorganic material to simpler forms

convert inorganic material into organic compound

do not break down organic compound

Ans: Convert organic material to inorganic forms.

12. The second trophic level is always of-

herbivores

autotrophs

carnivores

Ans: Herbivores

13. The percentage of solar radiation absorbed by all the green plants for the process of photosynthesis is about-

14. Which of the following belong to the same trophic level: grasshopper, spider, grass, hawk, and lizard?

Ans: Grasshoppers and spider

15. What is acid rain?

Ans: Any form of precipitation that contains high levels of nitric and sulfuric acids is called acid rain. It can also occur in the form of snow, fog, and tiny bits of dry material that settle to Earth.

16. The ecosystem of earth is known as-

biome

community

biosphere

association

Ans: Biosphere

17. Which of the following constitute a food chain?

Grass, goat and human

Goat, cow and elephant

Grass fish and goat

Grass, wheat and mango

Ans: Grass, goat, human.

18. Flow of energy in an ecosystem is always-

Unidirectional

bidirectional

multidirectional

no specific direction.

Ans: Unidirectional

19. Name the main source of energy in any self-sustaining system.

20. Which of the following in not a part of biotic component of an ecosystem: water, algae, fish, bacteria?

21. Which of the following limits the of trophic levels in a food chain-

polluted

Ans: Decrease in energy at higher trophic levels.

22. In natural ecosystems, decomposers include-

only bacteria and fungi

only microscopic animals

herbivores and carnivores

both (b) and (c)

Ans: Only bacteria and fungi

23. All living organisms of the earth constitute a-

biotic community

24. What are the various steps of food chains called?

Ans: Trophic levels

25. Which one is not biodegradable: paper, plastic, sewage?

Ans: Plastic

26. Which of the following groups contain only biodegradable items?

Grass, flowers and leather

Grass, wood and plastic

Fruit peels, cake and lime-juice

Cake, wood and grass

Ans: Groups (a), (c) and (d).

27. Which of the following constitute a food chain?

Grass, wheat and mango

Grass, fish and goat.

Ans: Grass, goat, human

28. Which of the following are environment-friendly practices?

Carrying cloth-bag to put purchases in while shopping.

Switching off unnecessary lights and fans.

Walking to school instead of getting your mother to drop you on her scooter.

All of the above.

Ans: All of the above.

29. Construct a food chain composing the following Snake, Hawk, Rats, Plants.

Ans: Plants $\to $ Rats $\to $ Snake $\to $ Hawks

30. Name the process that is a direct outcome of excessive burning of fossil fuels?

Ans: Global warming is a direct outcome of excessive burning of fossil fuels.

31. Using Kulhads as disposable cups to serve tea in trains, proved to be a bad idea. Why?

Ans: Yes, because making Kulhads on large scales leads to the loss of top soil.

32. Why is plastic not degraded by bacteria?

Ans: Plastic is not degraded by bacteria because they do not have enzymes which is used to degrade plastic.

Short Answer Questions (2 Marks)

1. Give scientific terms for the following-

(a) The process of eating and being eaten

Ans: Food chain

(b) The relationship between abiotic and biotic component

Ans: Ecosystem

2. What is meant by environment? Name its components.

Ans: All the surroundings which have an impact on human lives is called an environment. It has two components-

(a) Abiotic component (non-living)

(b) Biotic component (living)

3. What is 10% law? Give an example

Ans: Only 10% of energy is available at the next trophic level in 10% law. For example-Suppose 1000 Joules of light energy emitted by the sun falls on the plants. Then the plants or first trophic level has 10 joules of energy in it. Now according to 10 percent law, only 10% of 10 joules of energy (which is 1 joule) will be available for transfer to the next trophic level, so that the herbivore will have only 1 joule of energy stored as food at the second trophic level. 10% of the remaining 1 joule will be transferred to third trophic level of carnivore. So, the energy available in the lion as food will be only 0.1 joule.

4. What is artificial ecosystem? Give two examples.

Ans: Ecosystem which are made by humans are known as artificial ecosystem. For example- Dams, parks.

5. Energy transfer is said to be unidirectional whereas biochemical transfer is said to be cyclic. Why?

Ans: Energy transfer is unidirectional because when the energy is absorbed by autotrophs from the sun, it is never reabsorbed by it. And when consumers eat up the producers directly or indirectly the energy transferred in this process can never be reversed in the food chain. In biogeochemical transfer is cyclic because chemical elements move from environment to organism and back to the environment.

6. Why is there a need to ban the use of polythene bags?

Ans: Polythene bags need to be banned because-

They are non-biodegradable.

Cannot be able to decompose them.

One of the cause of land pollution.

7. What is the significance of food chain?

Ans: Significance of food chain is-

It is used to transmit energy from one organism to the next.

It is the method by which a particular organism collects its food.

It is a way of depicting the flow of energy.

8. How would you dispose the following wastes?

(a) domestic wastes like vegetables peels

Ans: Domestic wastes should be disposed off in a pit.

(b) industrial wastes

Ans: Industrial wastes should be treated first to remove poisonous chemicals and then disposed off in water resources.

9. Why vegetarian food habit helps us in getting more energy?

Ans: Vegetarians obtain food directly from plants so due to 10 percent rule only 10% of energy is available at the successive level than previous level thus, vegetarian food habit helps us in getting more energy.

10. Write a food chain having two trophic levels.

Ans: Grass $\to $ deer

11. Diagrammatically represent the transfer of energy in a food chain.

12. Consider the following food chains-

Plants $\to $ mice $\to $ hawks

If energy available at the producer level in both the food chains is 100J. In which case will hawks get more energy and how much?

Ans: Hawk get more energy in food chain having three trophic levels.

Plants $(100J)$ $\to $ mice $(10J)$ $\to $ hawks $(1J)$

Thus, Energy available to hawk is $\text{1J}$.

13. Why is there a need to ban the use of polythene bags?

14. What are the two functions of ecosystem?

Ans: The two functions of ecosystem are-

Ecosystem regulates essential ecological processes and life support systems and renders stability.

It regulates and maintains itself and resists any stresses or disturbances upto a certain limit. This is known as cybernetic system.

15. What percentage of solar energy is trapped and utilized by plants?

Ans: 1% of solar energy is trapped and utilized by plants.

16. What are the harmful effects of acid rain?

Ans: The harmful effects of acid rain are-

It effects human nervous system, respiratory system and digestive system.

It can also leach aluminium from the soil.

It affects soil fauna and lead to reduced forest productivity.

It may cause extensive damage to materials and terrestrial ecosystems such as water, fish, vegetation, soils, building etc.

17. Differentiate between abiotic and biotic components of ecosystem.

Ans : The difference between abiotic and biotic components of ecosystem is as follows:

Abiotic components	Biotic components
Abiotic factors are the non-living things of an ecosystem.	Biotic factors are the living things of an ecosystem.
It cannot adapt as per the environmental conditions	It can adapt to the changes in the environment
Sunlight, temperature, water are the examples of abiotic components.	Plants, trees, and animals are examples of biotic components.

18. Give any two methods reducing the problem of waste disposal.

Ans: The two methods reducing the problem of waste disposal are-

Practicing the 3 R's: Reduce, Reuse and Recycle.

By throwing biodegradable and non-biodegradable waste into separate dustbins so that recycling can be done easily.

19. Give reason: “Life on earth depends on the sun.”

Ans: All living beings needs energy to be alive and they get energy in the form of food. And the food directly or indirectly comes from green plants. Sun plays an important role to produce food in the process of photosynthesis.

20. What are trophic levels? Give an example of a food chain and state the different trophic level in it.

Ans: The position that an organism occupied in a food chain is called trophic level. An example of a food chain and state the different trophic level in it is as follows:

Plants (Producer) $\to $ Deer (Primary consumer) $\to $ Lion (Secondary consumer)

21. What is the role of decomposers in the ecosystem?

Ans: Decomposers play a critical role in the flow of energy through an ecosystem. They break apart dead organisms into simpler inorganic materials, making nutrients available to primary producers.

22. What is ozone and how does it affect any ecosystem?

Ans: Ozone is a gas composed of three atoms of oxygen. Its molecular formula is ${{O}_{3}}$ . It f orms a layer in the upper atmosphere. It is very essential for the life on this planet. It shields the surface of the earth from ultraviolet radiation (UV) coming from sun as these radiations are very harmful causing skin cancer and cataract in humans. It also does harm to the crops.

23. How can you help in reducing the problems of waste disposal? Give any two methods.

24. What will happen if we kill all the organisms in one trophic level?

Ans: If all the organisms of one trophic level are killed, the food chain will become completely imbalanced. The organisms in the immediate higher trophic level will die out due to unavailability of food. Thus, the whole food chain will collapse.

25. Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?

Ans: Yes, the impact of removing all the organisms in a trophic level will be different for different trophic levels. It will not be possible to remove any organism in any trophic level without causing damage to the ecosystem.

26. If all the wastes we generate is biodegradable, will this have no impact on the environment?

Ans: If all the wastes we generate is biodegradable, it will have impact on the environment-

The production of harmful gases causes pollution.

Microbes will not be able to decompose all the biodegradable waste.

27. What are the problems caused by non-biodegradable wastes that we generate?

Ans: The problems caused by non-biodegradable wastes that we generate are-

Since, waste cannot be broken down into simpler forms hence they keep on accumulating which leads to pollution.

They cause diseases.

It also causes biological magnification.

28. What limits the number of trophic levels in a food chain.

Ans: The loss of energy from one trophic level to the next limits the number of trophic levels in a food chain.

29. What is the harm of clay cups?

Ans: The c lay cups contain a chemical known as styrene that causes side effects such as fatigue, irritation and many more health problems. It causes depletion of top fertile soil.

30. State one reason to justify the position of man at the apex of most food chains?

Ans: As humans are an intelligent organism so they can take advantage of position by manipulation. This is the main reason the position of man at the apex of most food chains.

31. Which food chains are advantageous in terms of energy?

Ans: Short food chains i.e two step chains are more efficient in terms of energy.

32. If all the wastes we generate is bio-degradable what impact may this have on the environment?

Ans: If all the wastes we generate is biodegradable, it will have impact on the environment-

33. Write the harmful effect of ozone depletion.

Ans: The harmful effects of ozone depletion are-

Causes skin cancers

Causes eye cataracts

Leads to immune deficiency disorders.

Affect plant growth

Reduces agricultural productivity.

34. Which of the following will have the maximum concentration of harmful chemicals in its body? Peacock, frog, Grass, Snake, Grasshopper

Ans: Peacock will have maximum concentration of harmful chemicals in its body among Peacock, frog, Grass, Snake, Grasshopper.

35. Why energy of herbivores never come back to the autographs?

Ans: Energy of herbivorous never comes back to autographs because flow of energy is always unidirectional in food chain.

36. Give the correct sequence of various & trophic levels in a food chain.

Ans: The correct sequence of various & trophic levels in a food chain is as follows:

Sequence of Various and Trophic levels in a food Chain

37. What is biological magnification and give its causes?

Ans: Gathering of various unimportant and harmful substances by organisms at different levels of a food chain is known as biological magnification. Its causes are the excessive use of pesticides which enter our food chain. As The bottom feeders of a food chain consume these and gradually it is carried to the top of that particular food chain.

38. DDT has entered food chain. Which food habit is safer- vegetarian or non-vegetarian?

Ans: Vegetarian habit is safer. Because less DDT will accumulate in our body. And due to Bio magnification higher level of DDT in higher trophic levels is found.

39. Aquarium requires regular cleaning whereas lakes normally do not. Why?

Ans: Lake has more diverse forms of life due to that they have larger number of food chains which leads to natural cleaning. Thus, the ecosystem is more stable. The aquarium has a very limited number of food chains and unable to sustain itself. That’s why Aquarium requires regular cleaning whereas lakes normally do not.

40. How will accumulation of bio degradable waste effect our environment?

Ans: The accumulation of bio degradable waste effect our environment because-

Short Answer Questions (3 Marks)

1. DDT that was sprayed in minute amount on food plants was detected in high concentration in man? How did it happen? Explain.

Ans: The phenomenon behind this is biological magnification because when DDT are used to protect crops from diseases and pests. They enter the soil. From soil these are absorbed by plants. And then consumed by organisms. They get accumulated at different trophic levels. As the human beings occupy the top position in any food chain, maximum concentration of such harmful chemicals get accumulated in the bodies of man.

2. Describe how ozone layer is formed?

Ans: Formation of the ozone layer is as follows-

During the origin of life Earth, some of underwater micro-organisms were to photosynthesize due to molecular oxygen was released in atmosphere.

This oxygen is released to stratosphere where it began to react with ultraviolet radiations from sun to form free oxygen (O).

Free oxygen combines with molecular oxygen to form ozone the reaction is given as-

${{\text{O}}_{\text{2}}}\text{+u}\text{.v light O+O}$

$\text{2O+2}{{\text{O}}_{\text{2}}}\text{2}{{\text{O}}_{\text{3}}}\text{(ozone)}$

3. What are the major components of environment?

Ans: The major components of environment are as follows-

Lithosphere - It is the solid outer section of Earth which includes Earth's crust. It extends from the surface of Earth to a depth of about $70-100km$ .The main component of lithosphere is earth’s tectonic plates.

Hydrosphere- It comprises of all forms of water bodies on earth including marine (oceans, seas) freshwater (rivers, lakes, ponds, streams) and groundwater resources etc. It covers $71\%$ of earth’s surface. $97\%$ of water found on Earth is in the oceans in the form of salt water. Only $3\%$ of water on Earth is freshwater. Out of this, $30.8\%$ is available as groundwater and $68.9\%$ is in frozen forms as in glaciers. Amount of $0.3\%$ is available in rivers, reservoirs and lakes and is easily accessible to man.

Atmosphere – It is gaseous layer enveloping the Earth. The atmosphere with oxygen in abundance is unique to Earth and sustains life. . It mainly comprises $78.08\%$ nitrogen, $20.95\%$ oxygen, $0.93\%$ argon, $0.039\%$ carbon dioxide, and traces of hydrogen, helium, and noble gases. The amount of water vapor present is variable $\left( 0-3 \right)\%$ Earth's atmosphere has a series of layers, each with its own specific traits.

Biosphere- It refers to all the regions on Earth where life exists. The ecosystems that support life could be in soil, air, water or land.

4. Why are some substances biodegradables and some non-biodegradable?

Ans: Some substances are biodegradable and some non-biodegradable because some substances can be decomposed by microorganisms and some cannot as he micro-organisms like bacteria and other decomposers organisms (called saprophytes) present in our environment specific in their action. They break down the materials or products made from natural materials (paper) as they have some peculiar enzymes for this process. But as enzymes are specific in their action, these cannot break down many man-made materials likes plastic. These can be acted upon by physical processes but not by biological processes. Therefore, these types of substances persist for long time and cannot be decomposed into simpler substances.

5. Explain why a food chain consists of few steps only? Write a food chain having five steps.

Ans: This is because of 10% law as only 10% of energy is available at the next trophic level. As If a food chain has six or more than six steps, energy is not sufficient for the survival of organism at that trophic level. A food chain having five steps is as follows:

Grass $\to $ insects $\to $ frog $\to $ snake $\to $ eagle

6. What is the difference between food chain and food web?

Ans: Food chain- The food chain describes which organism in the environment eats another organism. In ecology, the food chain is the series of transfer of matter and energy from organisms to organisms in the form of food. It is the sequence of events in an environment or ecosystem in which one living organism eats another living organism and another larger organism eats that organism later. It is a part of food web.

Food web- Food web means, mutually, many food chains via which energy flows into the ecosystem. The Food web is an interconnection of the various food web. A food web is just like the food chain except that the food web is larger than the food chain. Rarely, one organism is eaten by multiple predators, or it consumes many other organisms. The food webs are more complex.

7. What is biological magnification? Illustrate with the help of example.

Ans: Biological magnification refers to the process where toxic substances move up the food chain and become more concentrated at each level. For example-

Water $\to $ Phytoplankton $\to $ Fish $\to $ Bird

$ \text{(0}\text{.02 ppm}\text{of harmful chemical)}$ $\left( 5.0ppm \right)$ $\left( \text{240ppm} \right)$ $\left( \text{1600ppm} \right)$

8. What are the ill effects of ozone layer depletion?

Ans: The ill effects of ozone layer depletion are-

Human health: -

Agriculture and plant life: -

Marine environment

Photosynthesizing phytoplankton presents in the sea which also help in reducing the

global warming.

The lives of many plastics have been found to be shortened due to exposure to UV radiations.

9. What is the significance of food chains?

Ans: The significance of food chains is-

It is a means of transfer of food from one trophic level to another.

It provides information about the living component of an ecosystem.

It helps us in understanding the interactions and interdependence amongst different organism in an ecosystem.

It is a pathway for the flow of energy in any ecosystem.

10. How Garbage pollution can be controlled?

Ans: Garbage pollution can be controlled as-

By practicing 3 R’s .

By recycling of certain wastes products like plastic and paper.

By making use of biodegradable products as much as we can.

By producing biogas from the organic wastes.

Proper separation of biodegradable and non-bio-degradable waste during disposal.

By making the compost of biodegradable wastes by burying them under soil.

11. What are the components of an ecosystem? Explain with examples

Ans: An ecosystem has two major components-

Biotic Components- It includes producers (plants), consumers (animals) and decomposers (bacteria and fungi).

Producers- Organisms which are able to photosynthesis are called producers. It includes all green plants.

Consumers- Organisms which depends upon other are called consumers. It is of few types-

Herbivores- Animals which directly depends upon plants.

Carnivores- These animals eat herbivores.

Secondary carnivores- Animals which depends upon carnivores.

Tertiary carnivores- Largest animals which depends upon secondary carnivores.

Decomposers- These organisms depend upon dead plants and animals. They change complex organic substances into simple inorganic substances.

Abiotic components- Non-living components. It include physical and chemical factors such as light, water, soil, air temperature, oxygen, carbon, nitrogen and other nutrients.

12. Write any three activities which are eco-friendly.

Ans: The three activities which are eco-friendly.

S eparation of biodegradable and non-biodegradable substances.

Rainwater harvesting.

13. Energy transfer is said to be unidirectional whereas biochemical transfer is said be cyclic. Why?

14. Give difference between producers and consumers. Mention one example of each.

Ans: The difference between producers and consumers is as follows:

Producers	Consumers
They are autotrophs.	They are heterotrophs
They occupy first trophic level.	They occupy second to third trophic level.
They synthesises their own food.	They cannot synthesises their own food.

15. There are no predators for tiger or lion. Why?

Ans: Lions and tigers are at the highest trophic level. They are largest animals which feed upon the secondary carnivores like wolves etc. they are not killed and eaten by other animals.

16. What are the measures to protect ozone depletion?

Ans: The measures to protect ozone depletion are-

Avoid the consumption of gases dangerous to the ozone layer, due to their content or manufacturing process. Some of the most dangerous gases are CFCs (chlorofluorocarbons), halogenated hydrocarbon, methyl bromide and nitrous oxide.

Minimize the use of cars. The best transport option is urban, bicycle, or walking. If you use a car to a destination, try to carpool with others to decrease the use of cars in order to pollute less and save.

Do not use cleaning products that are harmful to the environment and to us. Many cleaning products contain solvents and substances corrosive, but you can replace these dangerous substances with non-toxic products such as vinegar or bicarbonate.

Buy local products. In this way, you not only get fresh products but you avoid consuming food that has travelled long distances. As the more distance travelled, the more nitrous oxide is produced due to the medium used to transport that product.

Maintain air conditioners, as their malfunctions cause CFC to escape into the atmosphere.

17. Describe three biotic components of ecosystem. Also give examples.

Ans: The three biotic components of ecosystem are-

Producers- All the green plants have a unique capability to synthesis organic substance such as sugar and starch by the process of photosynthesis. Therefore, they are called producers.

Consumers- These are the living organisms which depend directly or indirectly on plants for their food. Consumers may be herbivore, carnivores, and omnivores. Example- Lion, tiger.

Decomposers- Decomposers are the organisms which depend upon the dead and decaying organisms their waste material. They form important link between living and non-living components. Example-algae.

18. What is the role of decomposers in an ecosystem?

Ans: Decomposers play a critical role in the flow of energy through an ecosystem. They break apart dead organisms into simpler inorganic materials, making nutrients available to primary producers. they are the important link between living and non-living components of environments.

19. What will happen if we kill all the organisms in one trophic level?

If the herbivores are killed, then the carnivores would not able be to get food and would die.

If carnivores are killed, then the population of herbivores would increase to unsustainable level.

If producers are killed, then the nutrient cycle in the area would not be completed.

Thus, the whole food chain will collapse.

20. What is Ozone? How does it affect any ecosystem?

Ans: Ozone is a gas composed of three atoms of oxygen. Its molecular formula is ${{O}_{3}}$ . It f orms a layer in the upper atmosphere. It is very essential for the life on this planet. It shields the surface of the earth from ultra-violet radiation (UV) coming from sun as these radiations are very harmful causing skin cancer and cataract in humans. It also does harm to the crops.

21. Why are some substances biodegradable and some non-biodegradable?

22. Give any two ways in which biodegradable substances would affect the environment.

(a) They will serve as breeding ground for flies and mosquitoes which are carriers of 3disease like cholera, malaria etc. Give any two ways in which non-biodegradable substances would affect the environment.

Ans: Excess use of non-biodegradable pesticide and fertilizers run off with rain water to water bodies cause water pollution.

(b) They produce foul smell, thus causing air pollution.

Ans: They may choke the sever system of city or town that may overflow over roads.

23. What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?

Ans: Biological magnification refers to the process where toxic substances move up the food chain and become more concentrated at each level. The concentration of harmful chemicals will be different at different trophic levels. It will be lowest in the first trophic level and highest in the last trophic level of the food chain.

24. Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?

Ans: Ozone layer stops ultraviolet radiations from the Sun from reaching the earth. Ultraviolet rays cause cancer, cataract and damage to the immune system of human beings.

To limit this damage following steps should be taken-

We should minimize the use of vehicles.

We should not encourage the burning of fossil fuels.

It is now mandatory for all the manufacturing companies to make CFC- free refrigerators throughout the world.

In $1987$, United Nations Environment Programme (UNEP) succeeded I forging an agreement between nations to freeze chlorofluorocarbons (CFCs) production to $1986$ levels. CFCs are the main cause of ozone layer depletion.

25. Why some substances are degraded and others not?

Ans: Some substances are degraded and others not because different components of food are changed to simpler substances by digestive enzymes and these enzymes are very much specific in nature and action. Similarly, substances are broken down by bacteria and saprophytes. They are also very specific in action and breakdown of the particular substance. Therefore, some substances are biodegradable and other are non-biodegradable.

25. What will happen if all the carnivores are removed from the earth?

Ans: If all the carnivores are removed from the earth, the population of herbivores will increase. Large population of herbivores will overgraze. And all plants will disappear from the earth surface and ultimately the earth may become a desert. The biosphere will get disturbed which will lead to end of life on earth.

26. What will happen to grasslands if all the grazers are removed from there?

Ans: If all the grazers are removed from grassland, then because carnivores keep the populations of other carnivores and herbivores in check. If there were no carnivores, the herbivore populations would explode and they will rapidly consume large amounts of plants and fungi, growing until there is not enough food to sustain them. Eventually, the herbivores would starve, leaving only those plants that were distasteful or poisonous to them. Species diversity would, therefore, drop dramatically.

27. The number of malarial patients in a village increase tremendously, when a large number of frogs were exported from the village. What could be the cause for it? Explain the help of food chain.

Ans: Phytoplankton $\to $Zooplankton $\to $ Mosquito larva $\to $ Frog

In the absence of frog, more mosquito larva survives, giving rise to large number of mosquitoes which cause increase incidence of malaria.

28. What are decomposers and what is the importance of them in the ecosystem?

Ans: Decomposers are the organisms which depend upon the dead and decaying organisms their waste material. They form important link between living and non-living components. They are important because Decomposers decompose the complex substances into simple ones so that plants can use it again.

29. Why food chains consist of three or four steps only?

Ans: Food chains consist of three or four steps only is because of 10% law as only 10% of energy is available at the next trophic level. As If a food chain has six or more than six steps, energy is not sufficient for the survival of organism at that trophic level.

30. What will happen if decomposers are not there in the environment?

Ans: If decomposers are not there in the environment, then dead leaves, dead insects, and dead animals would pile up everywhere. So, presence of decomposers is essential for the replenishment of soil and biogeochemical cycle of elements or substances.

31. Are plants actually producers of energy?

Ans: No, plants are not actually producers of energy, they can trap the energy of sun and can convert solar energy into chemical energy in the form of carbohydrates and other food materials so they are called transducers.

32. Look at the following figures. Choose the correct one and give reason for your Choice.

An ecosystem Represent in a Pyramid Shape

Ans: Figure A is correct.

In an ecosystem, the number of individuals at producer level is maximum. This number reduces at each successive level. Therefore, the shape is a pyramid with broader base and tapering apex.

On an average 10% of the food changes into body mass and is available for the next level of consumers.

33. It is the responsibility of the government to arrange for the management and disposal of waste. As an individual you have no role to play. Do you agree? Support your answers with two reasons.

Ans: I do not agree. As an individual, I also have the responsibility and can contribute in the following ways:

Practising 3 R’s.

Make compost pit for bio degradable waste.

Disposal of garbage only at appropriate places.

Cut down waste generation.

Recycle non-biodegradable waste.

Important Question of Our Environment Class 10 - Free PDF Download

The chapter on our environment is a very crucial chapter for students as it gives students insight into how the environment serves every organism in the animal kingdom. Students need to understand the chapter as it shows students the functioning of the food chain system in the environment. Students will also learn about the various functions of the ecosystem and how it serves various organisms. The differentiation between biotic and abiotic components of the ecosystem. Students will learn how to define an environment and all of its components. Chapter 15 of class 10 science can be found easily on the Vedantu website. The website allows for easy access to its important questions and also it is easily downloadable. Students will find these notes to be helpful during the preparation of their examinations, these important questions are easily downloadable in a pdf format. Students who find it difficult to study on the screen can also print this material so they don't waste too much time on the internet.

Chapter 15 Class 10 Science Important Concepts

This chapter on “our environment” holds importance in various aspects as it can help students understand the chapter as it shows students the functioning of the food chain system in the environment. There are some important concepts students should be familiar with when studying this subject and this includes the following

Photosynthesis

The process where green plants convert solar energy into chemical energy is known as photosynthesis. Inorganic molecules like carbon dioxide and water are used to synthesize food like starch. The process starts with the trapping of sunlight which is done by the chlorophyll present in green plants. Raw materials such as water and carbon dioxide are the prime ingredients for photosynthesis. Water is absorbed from the soil and carbon dioxide is taken from the atmosphere. The sunlight is the catalyst here in this process which converts the carbon dioxide and water into starch and oxygen. Starch is used up by the plant whereas the oxygen is left out to the atmosphere for other organisms to utilize.

Ozone Layer

The ozone layer is a layer that protects us from the extremely harmful rays of the sun. The ozone layer is a layer that is invisible in the atmosphere but can protect us from the harmful UV rays that the sun puts out. This chapter focuses on how the ozone layer protects us and what damages it. CFCs damage the ozone layer and create holes in the layer, this can only be avoided if we start taking better care of the environment and stop polluting so much. The more carbon dioxide there in the air the more it gets affected.

Energy Transfer is Said to Unidirectional

Energy transfer is said to be unidirectional because any energy that is lost to heat is said to the environment can’t be reutilized by plants for photosynthesis. Energy decreases from each trophic level by around 10 percent and thus it cannot be used again.

Significance of Food Chain

It means the transfer of food from one trophic level to another. It provides information about the various living components in our ecosystem. we can get a better understanding of the relation of different organisms that are present in our ecosystem. It is a pathway of flow from one ecosystem to another.

Harmful Effects of Acid Rain

It affects historical monuments and old buildings especially those made from marble. For example the taj mahal

There are certain bacteria that are good for maintaining soil fertility and this can be killed in acid rain.

It makes the water in lakes, ponds, and other water bodies bad for aquatic life.

Acid rain destroys the fertility of the soil which makes it hard to grow cereal crops and trees.

Life on Earth Depends on the Sun

For the earth, the sun is the ultimate source of energy. On various trophic levels, the plants convert the solar energy into chemical energy which is also transferred to various organisms. The energy that is also stored in various fossil fuels is basically transferred to solar energy because fossil fuel is made up of different plants and animals. Therefore we can see how solar energy is transferred into various forms of energy to consume.

Energy Received from Vegetarian Meals

Any organism that consumes vegetarian food is close to the product level that gets the maximum level of energy as compared to an organism of a higher trophic level because only 10 percent of the energy available is available at successive levels than any previous level.

Class 10 Science Chapter 15 Important Questions

Why do vegetarian food habits help us get more energy?

Write a food chain that has two trophic levels.

What percentage of solar energy is trapped and utilized by plants?

What are the harmful effects of acid rain?

Differentiate between biotic and abiotic components of the ecosystem.

Name the process that is a direct outcome of excessive burning of fuel.

Why is plastic not degraded by bacteria?

What is meant by the environment? Nam its components.

What is the 10% law? Name its components.

What is an artificial ecosystem? Name its components.

Why is there a need to ban polythene bags?

Diagrammatically represent the transfer of food chains

What are the two functions of the ecosystem?

Benefits of Important Question of Our Environment Class 10

Science is a very vast subject and can be quite difficult. So by using the right guide material like the ones found in Vedantu students can utilize it to their maximum and score the best marks possible.

Students studying this chapter on the environment will find that it helps in higher studies and help them understand the basics in any bio-based subject in the future.

Students can utilize and practice the important questions so that they can ace their examinations.

This is a fundamental subject for high school kids and plays an important role in higher studies

.It provides students with a structure with which they will study for his or her upcoming examinations.

Students don’t need to worry about the relevance of these questions as they're all cross-checked and updated consistently with the newest CBSE guidelines and rules . therefore the information in Vedantu is genuine and reliable.

Students can use this text to use their time wisely, it helps boost their confidence after consistent practice and students can plan their preparation accordingly.

Conclusion

This chapter takes into account various components of the ecosystem. It starts from how the food chain works and how the various organisms in the environment help each other, the chapter talks about all the harms in the environment and all the energy that can be consumed from the sun. This article on chapter 15 class 10 science important questions will help guide students through their preparations. Students will be able to study more efficiently by using these important questions. These questions will help students to understand the topics that have more weightage than others and thus students will find that they don't have to waste time studying irrelevant topics. This article is important for students to understand the various concepts in the chapter. Students can work more diligently towards their goals and strive for higher marks as well. These important questions make sure that students are aware of the various topics in the chapter and with constant practice students will learn to tackle the difficult questions in the examinations.

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CBSE Class 10 Science Study Materials

FAQs on Important Questions for CBSE Class 10 Science Chapter 15 - Our Environment 2024-25

Q1. How can we protect our environment according to Chapter 15 of Class 10 Science?

Ans: We disturb and destroy our environment by adding waste to it. Everything that we do or don't do affects our environment. We share our environment with thousands of other living and non-living organisms. To protect our environment, we can reduce the amount of waste that we produce. We can reduce the usage of products that have a short shelf life and produce waste after their disposal such as plastics. We can recycle various products such as paper and use them again. We can limit the usage of products that emit CFCs into the environment such as refrigerators and air conditioners.

Q2. What is the environment as mentioned in Chapter 15 of Class 10 Science?

Ans: The environment is a very broad term. It includes human beings, plants, animals, books around us, the soil and everything else around us. The environment provides us with the land on which we live, the air we breathe, the food we eat and the clothes we wear. Environment makes life possible on Earth. It contains various essential nutrients such as oxygen, carbon dioxide and nitrogen. Water circulates in the environment, thereby, replenishing our rivers and groundwater and providing sufficient water and moisture to our crops.

Q3. What are the types of consumers mentioned in Chapter 15 of Class 10 Science?

Ans: The food we eat provides us with the energy to function. The three categories of consumers are primary consumers, secondary consumers and tertiary consumers. Primary consumers are animals that feed on the producers (the green plants). They are called herbivores, for example, deer. The secondary consumers are those that feed on primary consumers and are called carnivores. The third level of the food chain is occupied by tertiary consumers. Primary and secondary consumers are their source of food such as vultures and pythons. To read more about this chapter visit Vedantu’s official website (vedantu.com) and download the study materials free of cost.

Q4. Why did the United Nations act to control the production of chlorofluorocarbons (CFC) used in refrigerators according to Chapter 15 of Class 10 Science?

Ans: UNEP reached an agreement in 1987 to freeze the production of CFCs at the 1986 level. Three atoms of oxygen combine to form a molecule of the zone. The presence of the ozone layer hundreds of kilometres above the surface protects us from the harmful ultraviolet rays of the sun. If these rays reach us they can cause serious damage to organisms including skin cancer in humans. The layer of ozone in the atmosphere had been depleting because of the increase in the concentration of CFCs in the environment. Acknowledging the seriousness of the issue, the UN finalized the agreement.

Q5. What role does our environment play according to Chapter 15 of Class 10 Science?

Ans: Earth has a unique place in our solar system because it has elements that make life possible. Everything that surrounds us is our environment. The resources we need to live- oxygen and water- are provided by the environment. The same environment recycles life-giving resources as they are not present in unlimited quantities. The environment protects us from the harmful rays of the sun. The sunrays that reach us after travelling through various layers of the environment enable plants to make food and produce oxygen.

CBSE Class 10 Science Important Questions

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Case Based Questions Test: Electricity - Class 10 MCQ

10 questions mcq test - case based questions test: electricity, observe the graph and answer the question. the v-i graph for a conductor is as shown in figure. what do you infer from this graph .

A. V ∝ 1 / I
D. V ∝ 1 / I 2

Observe the graph and answer the question. The V-I graph for a conductor is as shown in figure. Which of the following law justify the above graph:

A. Faraday's Law
B. Ohm’s Law
C. Faraday's Law
D. Joule’s Law

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Observe the graph and answer the question. The V-I graph for a conductor is as shown in figure. Name the physical quantity represented by the slope of this graph

B. Resistance
C. Potential difference
D. None of the above

Observe the graph and answer the question. The V-I graph for a conductor is as shown in figure.

Ohm is the SI unit of:

A. Potential difference
D. Resistivity

When an electric current of one ampere passes through a component across which a potential difference (voltage) of one volt exists, then the resistance of that component is one ohm.

Resistance of a conductor depends on:

A. length of conductor
B. area of cross-section
C. temperature
D. all of the above

Read the passage and answer the questions:

Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series and the combination is connected to the battery of 30 V. Ammeter and voltmeter are connected in the circuit.

Q. Which of the following is the correct circuit diagram to connect all the devices in proper correct order.

Q. The device used to measure the current:

B. Galvanometer
C. Voltmeter
D. None of these

Q. How much is the total resistance?

Q. Which of the following is connected in series in circuit:

B. Voltmeter
C. Both of these

Q. Two students perform experiments on two given resistors R 1 and R 2 and plot the following V-I graphs. If R 1 > R 2 , which of the diagrams correctly represent the situation on the plotted curves?

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MCQ Questions for Class 10 Science Chapter 12 Electricity with Answers

We have compiled the NCERT MCQ Questions for Class 10 Science Chapter 12 Electricity with Answers Pdf free download covering the entire syllabus. Practice MCQ Questions for Class 10 Science with Answers on a daily basis and score well in exams. Refer to the Electricity Class 10 MCQs Questions with Answers here along with a detailed explanation.

Electricity Class 10 MCQs Questions with Answers

1. When electric current is passed, electrons move from: (a) high potential to low potential. (b) low potential to high potential. (c) in the direction of the current. (d) against the direction of the current.

2. The heating element of an electric iron is made up of: (a) copper (b) nichrome (c) aluminium (d) iron

3. The electrical resistance of insulators is (a) high (b) low (c) zero (d) infinitely high

4. Electrical resistivity of any given metallic wire depends upon (a) its thickness (b) its shape (c) nature of the material (d) its length

MCQ Questions for Class 10 Science Electricity World with Answers 1

6. Electric power is inversely proportional to (a) resistance (b) voltage (c) current (d) temperature

MCQ Chapter Electricity Class 10 Question 7. What is the commercial unit of electrical energy? (a) Joules (b) Kilojoules (c) Kilowatt-hour (d) Watt-hour

8. Three resistors of 1 Ω, 2 ft and 3 Ω are connected in parallel. The combined resistance of the three resistors should be (a) greater than 3 Ω (b) less than 1 Ω (c) equal to 2 Ω (d) between 1 Ω and 3 Ω

9. An electric bulb is connected to a 220V generator. The current is 0.50 A. What is the power of the bulb? (a) 440 W (b) 110 W (c) 55 W (d) 0.0023 W

Answer: b Explaination: Here, V = 220 V, I = 0.50 A ∴ Power (P) = VI = 220 x 0.50 = 110 W

10. The resistivity of insulators is of the order of (a) 10 -8 Ω -m (b) 10 1 Ω -m (c) 10 -6 Ω -m (d) 10 6 Ω -m

11. 1 kWh = ……….. J (a) 3.6 × 10 -6 J (b) $\frac{1}{3.6}$ × 10 6 J (c) 3.6 × 10 6 J (d) $\frac{1}{3.6}$ × 10 -6 J

12. Which of the following gases are filled in electric bulbs? (a) Helium and Neon (b) Neon and Argon (c) Argon and Hydrogen (d) Argon and Nitrogen

13. 100 J of heat is produced each second in a 4Ω resistor. The potential difference across the resistor will be: (a) 30 V (b) 10 V (c) 20 V (d) 25 V

14. Electric potential is a: (a) scalar quantity (b) vector quantity (c) neither scalar nor vector (d) sometimes scalar and sometimes vector

Electricity Question 15. 1 mV is equal to: (a) 10 volt (b) 1000 volt (c) 10 -3 volt (d) 10 -6 volt

Electricity MCQ Question 16. Coulomb is the SI unit of: (a) charge (b) current (c) potential difference (d) resistance

Fill in the Blanks

1. The SI unit of current is ……… . 2. According to ……… Law, the potential difference across the ends of a resistor is directly proportional to the ……… through it, provided its remains constant. 3. The resistance of a conductor depends directly on its ……… , inversely on its ……… and also on the ……… of the conductor. 4. The SI unit of resistivity is ……… . 5. If the potential difference across the ends of a conductor is doubled, the current flowing through it, gets ……… .

1. ampere 2. Ohm’s, current, temperature 3. length, area of cross-section, material 4. ohm-metre (Ω m) 5. doubled

Hope the information shed above regarding NCERT MCQ Questions for Class 10 Science Chapter 12 Electricity with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 10 Science Electricity MCQs Multiple Choice Questions with Answers, feel free to reach us so that we can revert back to us at the earliest possible.

2 thoughts on “MCQ Questions for Class 10 Science Chapter 12 Electricity with Answers”

It was very helpful for me … it helped me to understand the chapter in a better way …the whole chapter was covered in this questions ..Now I think that I am fully prepared for the test …thanks allot

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Delineating bacterial genera based on gene content analysis: a case study of the Mycoplasmatales-Entomoplasmatales clade within the class Mollicutes

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Genome-based analysis allows for large-scale classification of diverse bacteria and has been widely adopted for delineating species. Unfortunately, for higher taxonomic ranks such as genus, establishing a generally accepted approach based on genome analysis is challenging. While core-genome phylogenies depict the evolutionary relationships among species, determining the correspondence between clades and genera may not be straightforward. For genotypic divergence, percentage of conserved proteins (POCP) and genome-wide average amino acid identity (AAI) are commonly used, but often do not provide a clear threshold for classification. In this work, we investigated the utility of global comparisons and data visualization in identifying clusters of species based on their overall gene content, and rationalized that such patterns can be integrated with phylogeny and other information such as phenotypes for improving taxonomy. As a proof of concept, we selected 177 representative genome sequences from the Mycoplasmatales-Entomoplasmatales clade within the class Mollicutes for a case study. We found that the clustering patterns corresponded to the current understanding of these organisms, namely the split into three above-genus groups: Hominis, Pneumoniae, and Spiroplasma-Entomoplasmataceae-Mycoides (SEM). However, at the genus level, several important issues were found. For example, recent taxonomic revisions that split the Hominis group into three genera and Entomoplasmataceae into five genera are problematic, as those newly described or emended genera lack clear differentiations in gene content from one another. Moreover, several cases of mis-classification were identified. These findings demonstrated the utility of this approach and the potential application for other bacteria.

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Case Study Questions Class 10 Science
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Case Study Questions Class 10 Science Chapter 12 Electricity
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CBSE 10th Science Electricity Case Study Questions With Solution 2021
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Delineating bacterial genera based on gene content analysis: a case
Genome-based analysis allows for large-scale classification of diverse bacteria and has been widely adopted for delineating species. Unfortunately, for higher taxonomic ranks such as genus, establishing a generally accepted approach based on genome analysis is challenging. While core-genome phylogenies depict the evolutionary relationships among species, determining the correspondence between ...

Case Study Questions Class 10 Science Chapter 12 Electricity

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கணக்குப்பதிவியல்

English Subjects

Computer Science

Business Maths and Statistics

Accountancy

Computer Applications

Computer Technology

11th Standard stateboard question papers & Study material

9th Standard stateboard question papers & Study material

Social Science

சமூக அறிவியல்

10th Standard stateboard question papers & Study material

7th Standard stateboard question papers & Study material

8th Standard stateboard question papers & Study material