case study class 10 electricity

Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity, case study: 1, case study: 2, case study:3.

Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point.

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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

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

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

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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Class 10 Science: Case Study Chapter 12 Electricity PDF Download

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.

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:

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

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

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

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

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Class 10 Science Chapter 11 Case Based Questions - Electricity

Case Study - 1

When electric current flows through the circuit this electrical energy is used in two ways, some part is used for doing work and remaining may be expended in the form of heat. We can see, in mixers after using it for long time it become more hot, fans also become hot after continuous use. This type of effect of electric current is called as heating effect of electric current. If I is the current flowing through the circuit then the amount of heat dissipated in that resistor will be H = VIt This effect was discovered by Joule, hence it is called as Joule’s law of heating. Also, we can write, H = I 2 Rt Thus, heat produced is directly proportional to the square of the electric current, directly proportional to the resistance of the resistor and the time for which electric current flows through the circuit. This heating effect is used in many applications. The heating effect is also used for producing light. In case of electric bulb, the filament produces more heat energy which is emitted in the form of light. And hence filament are made from tungsten which is having high melting point. In case of electric circuit, this heating effect is used to protect the electric circuit from damage. The rate of doing work  or rate of consumption of energy is called as power. Here, the rate at which electric energy dissipated or consumed in an electric circuit is called as electric power. And it is given by P= VI The SI unit of electric power is watt.

Q1: What is the SI unit of electric energy? Ans:  The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h. Q2: How heating effect works to protect electric circuit? Ans:  In case of electric circuit fuse is connected in series with the circuit which protects the electric devices by stopping the extra current flowing through them. When a large amount of current is flowing through the circuit the temperature of the fuse wire increases and because of that fuse wire melts which breaks the circuit.

Q3: 1KW h = ? Ans: 1kW h = 3.6*10 6  joule   Q4: If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb? Ans:  Given that, V = 200V, I = 1A Then, P = VI = 200*1 = 200 J/s = 200 W

Case Study - 2

Resistance is the opposition offered by the conductor to the flow of electric current. When two or more resistors are connected in series then electric current through each resistor is same but the electric potential across each resistor will be different. If R1, R2 and R3 are the resistance connected in series then current through each resistor will be I but potential difference across each resistor is V1, V2 and V3 respectively. Thus, the total potential difference is equal to the sum of potential difference across each resistor. Hence, V= V1 + V2 + V3 Again, IR = IR1 + IR2 + IR3 Thus, R = R1 + R2 + R3 Hence in case of series combination of resistors, the total resistance is the sum of resistance of each resistor in a circuit. Now, in case of parallel combination of resistors electric current through each resistor is different but the potential difference across each resistor is same. If resistors R1, R2 and R3 are connected in parallel combination then potential difference across each resistor will be V but current through each resistor is I1, I2 and I3 respectively. Thus, total current through the circuit is the sum of current flowing through each resistor. I = I1 + I2 + I3 Again, V/R= V/R1 + V/R2 + V/R3 Thus, 1/R = 1/R1 + 1/R2 + 1/R3 Hence, in case of parallel combination of resistors, the reciprocal of total resistance is the sum of reciprocal of each resistance connected in parallel.

Q1: In which case the equivalent resistance is more and why? Ans: In case of parallel combination of resistors the equivalent resistance is less than the individual resistance connected in parallel. Since, 1/R = 1/R1 + 1/R2 + 1/R3 +…. Q2: In our home, which type of combination of electric devices is preferred? Why? Ans:  At our home, we are connecting electrical devices in parallel combination because in parallel combination equivalent resistance is less and also we can draw an electric current according to the need of electric devices. Q3: If n resistors of resistance R are connected in parallel then what is the equivalent resistance? Ans:  If n resistors of resistance R are connected in parallel then equivalent resistance is given by, 1/Re = 1/R + 1/R + 1/R +….n times 1/R Thus, 1/Re = n/R Hence, Re= R/n is the required equivalent resistance of the given combination.

Case study - 3

We can see that, as the applied voltage is increased the current through the wire also increases. It means that, the potential difference across the terminals of the wire is directly proportional to the electric current passing through it at a given temperature. Thus, V= IR Where R is the proportionality constant called as resistance of the wire. Thus, we can say that the resistance of the wire is inversely proportional to the electric current. As the resistance increases current through the wire decreases. The resistance of the conductor is directly proportional to length of the conductor, inversely proportional to the area of cross section of the conductor and also depends on the nature of the material from which conductor is made. Thus R= qL/A, where q is the resistivity of the material of conductor. According resistivity of the material they are classified as conductors, insulators and semiconductors. It is observed that the resistance and resistivity of the material varies with temperature. And hence there are vast applications of these materials based on their resistivity. The SI unit of resistance is ohm while the SI unit of electric current is ampere. The potential difference is measured in volt. Conductors are the materials which are having less resistivity or more conductivity and hence they are used for transmission of electricity. Alloys are having more resistivity than conductors and hence they are used in electric heating devices. While insulators are bad conductors of electricity.

Q1: What is SI unit of resistivity? Ans:  The SI unit of resistivity is ohm meter. Q2: What is variable resistance? Ans: The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance. Q3: Why tungsten is used in electric bulbs? Ans:  Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high. Q4: 1M ohm = ? Ans: 1M ohm = 10 6  ohm 

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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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Class 10 Science Chapter 12 Electricity

Class 10 Chapter 12 Electricity as the name suggests, covers everything about electricity in detail. The constitution of electricity, the flow of electricity in the circuit, how electricity can be regulated, and much more. The chapter also includes Ohm’s law, resistors, and heating effects of electric circuits. The questions constitute 7 marks in the CBSE Class 10 exams. The inclusion of CBSE Electricity Chapter 12 is to help students create a strong foundation especially when students want to pursue the field of science and technology. 

The understanding of concepts and topics included in the NCERT Chapter 12 can be done with the help of study materials like notes of electricity class 10 CBSE, question bank, mind maps, and support materials. Preparing the right study material can help in scoring good marks in the final examination.

CBSE Class 10 Electricity Notes

Below we have provided the links to downloadable PDFs of class 10 ch 12 science notes and get an in-depth explanation and understanding of the chapter.

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CBSE Class 10 Electricity DoE Worksheet

Below, we have provided the links to downloadable PDFs of DoE Worksheets for Electricity Class 10 to practice more questions. 

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<red> ➜   <red> worksheet 17, <red> ➜   <red> worksheet 18, <red> ➜   <red> worksheet 19, <red> ➜   <red> worksheet 20, <red> ➜   <red> worksheet 21, cbse class 10 electricity experiential activities.

Below, we have provided the links to downloadable PDFs of Experiential Learning Activity for ch 12 class 10 Science to help students implement their acquired knowledge in the real world.

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Cbse class 10 electricity important questions.

Below, we have provided Class 10 Science Important Questions that cover all the important questions in Electricity. 

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Cbse class 10 electricity mind maps.

Below, we have provided Class 10 Science Mind maps that include mind maps of the related concepts in Electricity.

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Cbse class 10 electricity question bank.

Below, we have provided Class 10 Science Question Banks that cover every typology question with detailed explanations from various resources in one place

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<red> ➜   <red> kendriya vidyalaya question bank, cbse class 10 electricity support material.

Below, we have provided Class 10 Science Support Materials that cover Case Study-based questions from the various concepts explained in Science NCERT chapters.

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Why download these chapter-wise pdfs.

Science Class 10 Electricity chapter can include both objective and subjective questions related to Ohm’s law, SI unit of current, and magnetic effects of currents. The study materials are exam-centric and with the help of visual study materials like mind maps can help in connecting the knowledge they have acquired. With the right preparation, working strategically can help students build a strong base and score at least 7 marks in the final exams. The chapter-wise study materials are effective for both teachers and students. 

• Creating a study timetable and including these chapter-wise PDFs can help students prepare the chapters in a strategic and organized manner. Just allot sufficient time for understanding and revising the concepts.
• Students wouldn’t have to juggle between numerous sites to find various study material that suits their learning style. The chapter-wise study material can be accessed in one place.
• Download and browse these chapter-wise PDFs on any device in the comfort of your place by using an Internet connection.
• After downloading these PDFs, students can get these educational materials printed and prepare accordingly. 

How Can This Chapter-wise Material Help Students?

The Science Electricity chapter-wise materials can help in completing the chapter from the 10th NCERT textbook in addition to the extra study materials. Students may efficiently prepare for the chapter by downloading chapter notes, DoE worksheets, question banks, key questions, and a plethora of additional study resources.

• Students could structure and focus on a certain less strong area of the subject at hand by using science class 10 electricity notes.
• When it comes to making quick and efficient preparations or refining the flow of concepts under a certain topic that you might have overlooked, mind maps are a valuable tool.
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CBSE 10th Standard Science Subject Electricity Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

QB365 - Question Bank Software

Cbse 10th standard science subject electricity case study questions 2021.

10th Standard CBSE

Final Semester - June 2015

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?

(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)  \(\begin{equation} \text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000} \end{equation}\)

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Cbse 10th standard science subject electricity case study questions 2021 answer keys.

(i) (c) : The equivalent resistance in the parallel combination is lesser than the least value, of the individual resistance. (ii) (b ): Resistance of each piece  \(\begin{equation} =\frac{12}{3}=4 \Omega \end{equation}\) \(\begin{equation} \frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega \end{equation}\) (iii) (a): All the three resistors are in parallel. \(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6} \end{equation}\) \(\begin{equation} R_{P}=\frac{6}{9}=\frac{2}{3} \Omega \end{equation}\) (iv) (a): Voltage is same across each resistance. So, I 1  x 5 = I 2 X 10 = 15 x I 3 I 1  = 2I 2 = 3I 3 (v) (d): All are in parallel. \(\begin{equation} \begin{array}{l} \frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \\ I=\frac{3}{3}=1 \mathrm{~A} \end{array} \end{equation}\) So,current in each resistor  \(\begin{equation} I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A} \end{equation}\)

(i) (b) (ii) (c): In series combination, resistance is maximum and in parallel combination, resistance is mcm. (iii) (c) :  R 1  = r 1 + r 2 \(\begin{equation} \begin{array}{l} R_{2}=\frac{r_{1} r_{2}}{r_{1}+r_{2}} \\ \frac{R_{1}}{R_{2}}=\frac{\left(r_{1}+r_{2}\right)^{2}}{r_{1} r_{2}} \end{array} \end{equation}\) (iv) (c): In the given circuit, 3 \(\begin{equation} \Omega \end{equation}\)  resistors are in series. R S  = 3 + 3 = 6  \(\begin{equation} \Omega \end{equation}\) Now, R S  and 6 \(\begin{equation} \Omega \end{equation}\)  are parallel. \(\begin{equation} \frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \end{equation}\) (v) (a): V = 0.2 x 10= 2 V So, total voltage supplied is same as 2 V.

(i) (b) (ii) (c) (iii) (a): Given: H = I 2 Rt \(\begin{equation} \mathrm{So}, H^{\prime}=(2 I)^{2} \cdot \frac{R}{2} t=2 H \end{equation}\) (iv) (b): Given: 1= 5 A, resistance = R. Let r be the new radius. Now,H=I 2 Rt Also H' = I' 2 R' t From (i) and (ii),  \(\begin{equation} 5^{2} \times \rho \frac{L}{\pi r^{2}} t=10^{2} \times \rho \frac{L}{\pi r^{\prime 2}} \cdot t \end{equation}\) \(\begin{equation} \frac{25}{r^{2}}=\frac{100}{r^{\prime 2}} \Rightarrow \frac{r^{\prime}}{r}=2 \Rightarrow r^{\prime}=2 r \end{equation}\) (v) (c): Given:  \(\begin{equation} I=0.5 \mathrm{~A}, R=10 \Omega, t=5 \mathrm{~min} \end{equation}\) \(\begin{equation} \begin{array}{l} H=I^{2} R t=0.5 \times 0.5 \times 10 \times 5 \times 60 \\ H=750 \mathrm{~J} \end{array} \end{equation}\)

(i) (a) :  \(\begin{equation} E \propto t \end{equation}\) (ii) (c) : Given: P = 60W, t = 1 min E = 60 x 1 x 60 = 3600J (iii) (b) : Given: P = 400 \(\begin{equation} \Omega \end{equation}\) , t = 8 hour E = 400 x 8 = 3200Wh = 3.2kWh Cost = 3.2 x 5 =  \(\begin{equation} ₹ \end{equation}\) 16 (iv) (a) : Given: I= 5 A, R = 2  \(\begin{equation} \Omega \end{equation}\) , t = 30 min E = I 2 Rt = 5 x 5 x 2 x 30 x 60 E = 90000J = 90 kJ (v) (a): 1watt hr = 3600J

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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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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 10 Science Chapter 12 Electricity

September 27, 2019 by Rama Krishna

NCERT Solutions For Class 10 Science Chapter 12 Electricity : In this article, we will provide you all the necessary information regarding NCERT solutions for class 10 science physics chapter 12 electricity. Working on CBSE class 10 physics electricity questions and answers will help candidates to score good marks in-class tests as well as in the CBSE Class 10 board exam.

Electricity class 10 NCERT solutions will not only help in board exam preparation but also helps in clearing the competitive exams like Engineering. Also, candidates can find  electricity class 10 numericals with solutions which helps candidates solving their assignments. Read on to find out everything NCERT Solutions For Class 10 Science Chapter 12 Electricity .

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 12 Electricity, let’s have an overview of the list of topics and subtopics under Electricity class 10 NCERT solutions :

• Electricity
• Electric Current And Circuit
• Electric Potential And Potential Difference
• Circuit Diagram
• Factors On Which The Resistance Of A Conductor Depends
• Resistance Of A System Of Resistors
• Heating Effect Of Electric Current
• Electric Power

Free download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• विद्युत कक्षा 10 विज्ञान हिंदी में
• Class 10 Electricity Important Questions
• Electricity Class 10 Notes
• Electricity NCERT Exemplar Solutions

Class 10 Science Electricity Mind Map

Ncert solutions for class 10 science chapter 12 intext questions.

Page Number: 200

Question 1 What does an electric circuit mean ? Answer: A continuous and closed path along which an electric current flows is called an electric circuit.

Question 2 Define the unit of current. Answer: Unit of current is ampere. If one coulomb of charge flows through any section of a conductor in one second then the current through it is said to be one ampere. I = \(\frac { Q }{ t }\) or 1 A = I C s -1

Question 3 Calculate the number of electrons constituting one coulomb of charge. Answer: Charge on one electron, e = 1.6 x 10 -19 C Total charge, Q = 1 C Number of electrons, n = \(\frac { Q }{ e }\) = \(\frac { 1C }{ 1.6x{ 10 }^{ -19 } }\) = 6.25 x 10 18

Page Number: 202

Question 1 Name a device that helps to maintain a potential difference across a conductor. Answer: A battery.

Question 2 What is meant by saying that the potential difference between two points is IV? Answer: The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Question 3 How much energy is given to each coulomb of charge passing through a 6 V battery ? Answer: Energy given by battery = charge x potential difference or W = QV = 1C X 6V = 6J.

Page Number: 209

Question 1 On what factors does the resistance of a conductor depend ? OR List the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE2018] Answer: The resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.

Question 2 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ? Answer: The current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Therefore, smaller is the resistance of the conductor.

Question 3 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 ? Answer: When potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm’s law i.e., V ∝ I.

Question 4 Why are coils of electric toasters and electric irons are made of an-alloy rather than a pure metal ? OR Why are alloys commonly used in electric heating devices? Given reason. [CBSE 2018] Answer: The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of a pure metal, and (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.

Question 5 Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following : (i) Which among iron and mercury is a better conductor ? (ii) Which material is the best conductor ? Answer: (i) Resistivity of iron = 10.0 x 10 -8 Ω m Resistivity of mercury = 94.0 x 10 -8 Ω m. Thus iron is a better conductor because it has lower resistivity than mercury. (ii) Because silver has the lowest resistivity (= 1.60 x 10 -8 Ω m), therefore silver is the best conductor.

Page Number: 213

Page Number: 216

Question 1 Judge the equivalent resistance when the following are connected in parallel : (i) 1 Ω and 106 Ω, (if) 1 Ω and 103 Ω and 106 Ω. Answer: When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance. (i) Equivalent resistance < 1 Ω. (ii) Equivalent resistance < 1 Ω.

Question 3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ? Answer: Advantages of connecting electrical devices in parallel with the battery are :

• In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.
• In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.
• In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.
• In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.

Page Number: 218

Question 1 Why does the cord of an electric heater not glow while the heating element does ? Solution: Heat generated in a circuit is given by I 2 R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

Question 2 Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V. Solution: Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 10 6 J

Question 3 An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s. Solution: Here, R = 20 Ω, i = 5 A, t = 3s Heat developed, H = I 2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 10 4 J

Page Number: 220

Question 1 What determines the rate at which energy is delivered by a current ? Answer: Resistance of the circuit determines the rate at which energy is delivered by a current.

Question 2 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. Answer: Here, I = 5 A, V = 220 V, t = 2h = 7,200 s Power, P = V I = 220 x 5 = 1100 W Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 10 5 J

NCERT Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions

Question 1 A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is : (a) \(\frac { 1 }{ 25 }\) (b) \(\frac { 1 }{ 5 }\) (c) 5 (d) 25 Answer: (d) 25

Question 2 Which of the following terms does not represent electrical power in a circuit? (a) I 2 R (b) IR 2 (c) VI (d) \(\frac { { v }^{ 2 } }{ 2 }\) Answer: (fa) IR2

Question 3 An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be : (a) 100 W (b) 75 W (c) 50 W (d) 25 W Answer: (d) 25 W

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

Question 5 How is a voltmeter connected in the circuit to measure the potential difference between two points ? Answer: A voltmeter is connected in parallel to measure the potential difference between two points.

Question 8 When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor. Solution: Here, V = 12 V and I = 2.5 mA = 2.5 x 10 -3 A ∴ Resistance, R = \(\frac { V }{ I }\) = \(\frac { 12V }{ 2.5\times { 10 }^{ 3 }A }\) = 4,800 Ω = 4.8 x 10 -3 Ω

Question 9 A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor? Solution: Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω Potential difference, V = 9 V Current through the series circuit, I = \(\frac { V }{ R }\) = \(\frac { 12V }{ 13.4\Omega }\) = 0.67 A ∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

Question 11 Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω Solution: Here, R 1 = R 2 = R 3 = 6 Ω.

Question 12 Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ? Solution: Here, current, I = 5 A, voltage, V = 220 V ∴ Maxium power, P = I x V = 5 x 220 = 1100W Required no. of lamps \( =\frac { Max.Power }{ Power\quad of\quad 1\quad lamp } \quad =\quad \frac { 1100 }{ 10 } =110\) ∴ 110 lamps can be connected in parallel.

Question 16 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes ? Solution: Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min = 1200 x \(\frac { 10 }{ 60 }\) = 200 Wh 60 Thus, the TV set uses more energy than the toaster.

Question 17 An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. Solution: Here, R = 8 Ω, 1 = 15 A, t = 2 h The rate at which heat is developed in the heater is equal to the power. Therefore, P = I 2 R = (15) 2 x 8 = 1800 Js -1

Question 18 Explain the following: (i) Why is tungsten used almost exclusively for filament of electric lamps ? (ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ? (in) Why is the series arrangement not used for domestic circuits ? (iv) How does the resistance of a wire vary with its area of cross-section ? (v) Why are copper and aluminium wires usually employed for electricity transmission? Answer: (i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted. (ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature. (iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken. (iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R ∝ (1/πr 2 ). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor. (v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

Electric current, potential difference and electric current, Ohms law, Resistance, Resistivity factors on which the resistance of a conductor depends; Series combination of resistors, parallel combination of resistors; and its application on daily life; Heating effect of Electric current, electric Power, Interrelation between P, V, and R.

Formulae Handbook for Class 10 Maths and Science

Question 1: Name a device that helps to maintain a potential difference across a conductor. Answer: Cell or battery eliminator.

Question 2: What is meant by saying that the potential difference between two points is 1 V? Answer: As we know that V = W / q Thus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.

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Question 1: On what factors does the resistance of a conductor depend Answer: Resistance of a conductor depends upon: (i) Resistivity of the material. (ii) Length of the conductor. (iii) Cross-sectional area of the conductor.

Question 2: Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why Answer: The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R ∝ 1/A.

Download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF

Question 4: Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal Answer: The coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons; (i) The resistivity of an alloy is higher than that of a pure metal. (ii) It has high melting point and does not oxidise.

Question 5: Use the data in Table 12.2 of NCERT book to answer the following: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor? ‘ Answer: (a) Iron because its resistivity is less than mercury. (b) Silver is the best conductor as it has least resistivity.

Question 1: Judge the equivalent resistance when the following are connected in parallel. (a) 1 Ω and 10 6 Ω (b) 1 Ω , 10 3 Ω and 10 6 Ω Answer: Equivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit. Hence, in both the given cases, the equivalent resistance is less than 1 Ω.

Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ? Answer: Advantages of connecting electrical devices in parallel:

• When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.
• Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
• If one appliance fails to work, other will continue to work properly.

Question 1: Why does the cord of an electric heater not glow while the heating element does? Answer: The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced ‘H’ is H = I 2 Rt Thus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.

Question 3: An electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s. Answer: Given R = 20 Ω, I = 5 A, t = 30 s H = I 2 Rt = (5) 2 x 20 x 30 = 15000 J = 1.5 x 10 4 J

Question 1: What determines the rate at which energy is delivered by a current? Answer: Electric power determines the rate at which energy is delivered by a current.

Question 2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h. Answer: Given I = 5 A, V = 220 V, t = 2 h Power, p = VI = 220 x 5 = 1100 W Energy consumed = Vlt = Pt = 1100 x 2 = 2200 Wh

Textbook Questions

Question 2: 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 Answer: (b) P = V 2/ R = I 2 R = VI Option (b) does not represent electrical power.

Question 5: How is a voltmeter connected in the circuit to measure the potential difference between two points? Answer: A voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential of the source.

Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes? Answer: Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh. Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh. ∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Question 18: Explain the following. (a) Why is the tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminum wires usually employed for electricity transmission? Answer: (a) It has high melting point and emits light at a high temperature. (b) It has more resistivity and less temperature coefficient of resistance. (c) (i) All appliances do not get same potential in series arrangement. (ii) All appliances cannot be individually operated. (d) R ∝ =1 / Area of cross – section. (e) They are very good conductors of electricity.

Short Answer Type Questions

Question 2: Should the resistance of an ammeter be low or high? Give reason. Answer: The resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.

Question 3: How does use of a fuse wire protect electrical appliances? Answer: The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.

Question 6: Why is parallel arrangement used in domestic wiring? Answer: Parallel arrangement is used in domestic wiring because (i) Each appliance gets the same voltage as that of the mains supply. (ii) If one component is switched off, others can work properly. (iii) Fault in any branch of the circuit can be easily identified.

Question 7: B 1, B 2 and B 3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

Long Answer Type Questions

Multiple Choice Questions (MCQs) [1 Mark each]

Question 4. An ammeter has 20 divisions between 0 mark and 2A mark on its scale. The least count of ammeter is (a) 0.01A (b) 0.2A (c) 0.1A (d) 1A Answer: (c) Number of divisions = 20 Maximum reading of ammeter = 2 A Least count of ammeter = 2/20 = 1/10 = 0.1 A

Question 5. A student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter. The least count of voltmeter is (a) 0.1 V (b) 0.01 V (c) 0.05 V (d) 1.0 V Answer: (c) Number of divisions = 20 Maximum reading of the voltmeter = 1 V Least count of voltmeter = 1/20 = 0.05 V

Question 7. Which of the following is the correct method to connect the ammeter and voltmeter with resistance in the circuit to verify Ohm’s law? [CCE 2012] (a) Ammeter and voltmeter in series (b) Ammeter in series and voltmeter in parallel (c) Ammeter in parallel and voltmeter in series (d) Ammeter and voltmeter in parallel Answer: (b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.

Question 8. In an experiment on studying the dependence of the current I flowing through a given resistor on the potential difference V applied across it, a student has to change the value of the current. For doing this, he should change the (a) number of cells used (b) resistor itself (c) ammeter used in the circuit (d) Voltmeter used in the circuit Answer: (a) If we change the number of cells in electric circuit, the potential difference will change and as a result current flowing in the circuit changes.

Question 9. A milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions. If the pointer of the milliammeter is indicating the seventh graduation after 300 mark, the current flowing in the circuit is (a) 335 mA (b) 330 mA (c) 331mA (d) 340 mA Answer: (a) Number of divisions = 20 Least count of milliammeter = (100-0) / 20 = 5 mA Milliammeter reading = 300 + 7 x 5 = 335 mA

Question 10. If a student while studying the dependence of current on the potential difference keeps the circuit closed for a long time to measure the current and potential difference, then (a) ammeter’s zero error will change (b) ammeter will give more reading (c) voltmeter will show constantly higher readings (d) resistor will get heated up and its value will change Answer: (d) If the circuit is closed for a long time, then current flows in it for a long time which results that the resistor is heated.

Question 12. When parallel resistors are of three different values, the potential difference across its terminals is [CCE 2015] (a) greatest across smallest resistance (b) greatest across largest resistance (c) equal across each resistance (d) least across the smallest resistance Answer: (c) Potential difference across each resistor is same in parallel combination of resistors.

NCERT Solutions for Class 10 Science Chapter 12 Electricity (Hindi Medium)

Electricity Study of Electric Charges at Rest and in Motion

Charge Something associated with the matter due to which it produces and experiences electric and magnetic effects. Resides on the outer surface of the conductor. Q = ne S.I. unit coulomb (C)

Electric Current (I) The time rate of flow of charge (Q) through any cross-section I = \(\frac{Q}{t}\) S.I. unit ampere (A)

Types of Current Direct Current Current whose magnitude and direction does not vary with time.

Alternating Current Current whose magnitude and direction periodically changes with time.

Electric Potential Work done per unit charge V = \(\frac{W}{Q}\) S.I. unit volt

Ohm’s law: If the physical conditions remain same, current I ∝ V => V = IR R-electric resistance Substances which obey ohm’s law called ohmic and that do not obey called non-ohmic substances.

Dependence of Resistance

On length (l) and area of cross-section (A) R ∝ l ∝ \(\frac{1}{A}\) R = \(\rho \frac{l}{A}\) ρ = resistivity Resistivity depends on the material of the conductor only.

On Temperature R t = R 0 ( 1 + αt) α = temperature coefficient of resistance

Resistance (R): Obstruction offered to flow of electrons. SI unit ohm Resistance, R ∝ \(\frac{\ell^{2}}{m}\) l = length and m = mass of conducting wire

After stretching, if length increases by n times then resistance will increase by n 2 times i.e., R 2 = n 2 R 1 . Similarly if radius be reduced to \(\frac{1}{n}\) times then area of cross-section decreases \(\frac{1}{n^{2}}\) times so the resistance becomes n 4 times i.e.. R 2 = n 4 R 1 After stretching, if length of a conductor increases by x%, then resistance will increase by 2x% (valid only if x< 10%).

• Using n conductors of equal resistance, the number of possible combinations is 2 n-1 .
• If the resistances of n conductors are totally different, then the number of possible combinations will be 2 n .
• If n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by \(\frac{R_{s}}{R_{p}}=\frac{n^{2}}{1}\)
• If a wire of resistance R is cut in n equal parts and then these parts are collected to form a bundle, then equivalent resistance of combination will be \(\frac{\mathrm{R}}{\mathrm{n}^{2}}\)
• If equivalent resistance of R 1 and R 2 in series and parallel be R s and R p respectively, then R 1 = \(\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}+\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]\) and R 2 = \(=\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}-\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]\)

Grouping of Resistances

Series Grouping of Resistances Equivalent resistance, resistance, R s = R 1 + R 2 + … + R n In this case same current flows through each resistance but potential difference in the ratio of resistance

Parallel Grouping of Resistances \(\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}\) In this case same potential across each resistance but current distributes in the reverse ratio of their resistances

Heating Effect of Electric Current As current flows through a conductor, the free electrons lose energy which is converted into heat. Joule’s heating law H ∝ I 2 H ∝ R H ∝ t H = I 2 Rt = VIt

Practical Applications

• Electric heater, electric iron and water heater, etc. work on the principle of heating effect of current
• Electric bulb glows when electric current flows through the filament of the bulb

Electric Power Rate at which electric energy is dissipated or consumed in a circuit, P = VI , or P = I 2 R = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)

Watt is a smaller unit of power, its other bigger units are kilowatt (KW), Megawatt (MW) and Horsepower (HP) 1 KW = 10 3 W 1 MW = 10 6 W 1 hp =746 W The commercial unit of electrical energy is 1 Kwh. 1 Kwh = 3.6 × 10 6 J

LED It is a device which glows even if a weak electric current is allowed to flow through it

Now that you are provided all the necessary information regarding Electricity class 10 NCERT solutions and we hope this detailed article on NCERT Solutions For Class 10 Science Chapter 12 Electricity is helpful. If you have any doubt regarding this article or  NCERT Solutions For Class 10 Science Chapter 12 Electricity, leave your comments in the comment section below and we will get back to you as soon as possible.

NCERT Solutions for Class 10 Science All Chapters

• 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
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• Chapter 8 How do Organisms Reproduce?
• Chapter 9 Heredity and Evolution
• Chapter 10 Light Reflection and Refraction
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• Chapter 14 Sources of Energy
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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

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

(a) maximum in (i)

(b) maximum in (ii)

(c) maximum in (iii)

(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

(a) same in all the cases

(b) minimum in case (i)

(c) maximum in case(ii)

(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

(c) its shape

(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.

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

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

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

(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?

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(c) R3 > R2 > R1

(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

(c) the shape of the resistor 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

(c) The brightness of bulb B will be more than that of A

(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

(c) the same potential difference across 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

(c) watt-second

(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.

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?

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.

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.

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.

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.

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

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.

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

(c) Potential difference across 4 Ω resistance

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

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.

With BYJU’S, students can excel in their studies and can score better marks in the board examination. Class 10 is an important stage of a student’s life, as it consists of topics which are necessary to understand thoroughly for future entrance exams. To help you grasp the concepts clearly, BYJU’S brings you notes , sample papers , and animation videos. For a customised learning experience, visit BYJU’S website or download BYJU’S – The Learning App.

Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Science Chapter 12

List out the topics included in chapter 12 of ncert exemplar solutions for class 10 science., how can we demonstrate joule’s heating effect according to chapter 12 of ncert exemplar solutions for class 10 science, are ncert exemplar solutions for class 10 science chapter 12 the best study materials for students.

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CBSE Class 10 Science Chapter Wise Important Case Study Questions

Chapter wise important case study questions cbse class 10 science: cbse class 10 science board exam 2024 is just around the corner and students are working hard to score maximum marks. check these case study questions from class 10 science to ace your examination this year also download the solutions from the pdf attached towards the end. .

CBSE Class 10 Science Chapter Wise Important Case Study Questions: While the CBSE Board exam for Class 10 students are ongoing, the CBSE Class 10 Science board exam 2024 is to be held on March 2, 2024. With the exams just a  few days away, CBSE Class 10th Board exam candidates are rushing to prepare the remaining syllabus, practising their weak portions, trying to revise the important questions from the past year papers, practise questions, etc. 

Why are CBSE Class 10 Science Case Study Questions Important?

• Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each.  
• Section B : 6 Very Short Answer type questions carrying 2 marks each. Answers to these questions should be in the range of 30 to 50 words.  
• Section C : 7 Short Answer type questions carrying 3 marks each. Answers to these questions should be in the range of 50 to 80 words.  
• Section D : 3 Long Answer type questions carrying 5 marks each. Answers to these questions should be in the range of 80 to 120 words.
• Section E : 3 Case Based/ Source Based units of assessment (4 marks each) with sub-parts.  

How to solve case study questions in CBSE Class 10 Science?

• Read the case given and the associated questions carefully.
• Read the questions attentively and analyse what they are asking.
• Apply your subject knowledge and theories in the given case to decide what the correct answers should be.

1.A chemical reaction is a representation of chemical change in terms of symbols and formulae of reactants and products. There are various types of chemical reactions like combination, decomposition, displacement, double displacement, oxidation and reduction reactions. Reactions in which heat is released along with the formation of products are called exothermic chemical reactions. All combustion reactions are exothermic reactions.

(i) The massive force that pushes the rocket forward through space is generated due to the

(a) combination reaction

(b) decomposition reaction

(c) displacement reaction

(d) double displacement reaction

(ii) A white salt on heating decomposes to give brown fumes and yellow residue is left behind. The yellow residue left is of

(a) lead nitrate

(b) nitrogen oxide

(c) lead oxide

(d) oxygen gas

(iii) Which of the following reactions represents a combination reaction?

(a) CaO (s) + H2O (l) → Ca (OH)2 (aq)

(b) CaCO3 (s) → CaO (s) + CO2(g)

(c) Zn(s) + CuSO4 (aq) → ZnSO4 (aq) + Cu(s)

(d) 2FeSO4(s) → Fe2O3 (s) +SO2(g) + SO3(g)

(iv) Complete the following statements by choosing correct type of reaction for X and Y.

Statement 1: The heating of lead nitrate is an example of ‘X’ reaction.

Statement 2: The burning of magnesium is an example of ‘Y’ reaction.

(a)X-Combination,Y-Decomposition

(b)X-Decomposition,Y-Combination

(c)X-Combination,Y-Displacement

(d) X- Displacement, Y-Decomposition

2.The earlier concept of oxidation and reduction is based on the addition or removal of oxygen or hydrogen elements so, in terms of oxygen and hydrogen, oxidation is addition of oxygen to a substance and removal of hydrogen from a substance. On the other hand, reduction is addition of hydrogen to a substance and removal of oxygen from a substance. The substance which gives oxygen to another substance or removes hydrogen from another substance in an oxidation reaction is known as oxidising agent, while the substance which gives hydrogen to another substance or removes oxygen from another substance in a reduction reaction is known as reducing agent. For example, 

(i) A redox reaction is one in which

(a) both the substances are reduced

(b) both the substances are oxidised

(c) an acid is neutralised by the base

(d) one substance is oxidised while the other is reduced.

(ii) In the reaction, H2S+Cl2⟶S+2HCl

(a) H2S is the reducing agent. 

(b) HCl is the oxidising agent.

(c) H2S is the oxidising agent. 

(d) Cl2 is the reducing agent.

(iii) Which of the following processes does not involve either oxidation or reduction?

(a) Formation of slaked lime from quicklime.

(b) Heating mercuric oxide.

(c) Formation of manganese chloride from manganese oxide (MnO2).

(d) Formation of zinc from zinc blende.

(iv) Mg+CuO⟶MgO+Cu

Which of the following is wrong relating to the above reaction?

(a) CuO gets reduced

(b) Mg gets oxidised.

(c) CuO gets oxidised. 

(d) It is a redox reaction.

3.A copper vessel gets tarnished due to formation of an oxide layer on its surface. On rubbing lemon on the vessel, the surface is cleaned, and the vessel begins to shine again. This is due to the fact that which reacts with the acid present in lemon to form a salt which is washed away with water. As a result, the layer of copper oxide is removed from the surface of the vessel and the shining surface is exposed.

1.Which of the following acids is present in lemon?

(a) Formic acid

(b) Acetic acid

(c) Citric acid

(d) Hydrochloric acid

2.The nature of copper oxide is

d) amphoteric

3.Name the salt formed in the above reaction

a) copper carbonate

b) copper chloride

c)copper citrate

d) copper citrate

4.The phenomenon of copper getting tarnished is

a) corrosion

b) rancidity

c) displacement

d)none of these

4.Metals as we know, are very useful in all fields, industries in particular. Non-metals are no less in any way. Oxygen present in air is essential for breathing as well as for combustion. Non-metals form a large number of compounds which are extremely useful, e.g., ammonia, nitric acid, sulphuric acid, etc. Non-metals are found to exist in three states of matter. Only solid non-metals are expected to be hard however, they have low density and are brittle. They usually have low melting and boiling points and are poor conductors of electricity.

i.____________ is a non-metal but is lustrous

A.Phosphorus

ii.Which of the following is known as 'King of chemicals'?

C. Sulphuric acid

D. Nitric acid

iii.Which of the following non-metals is a liquid?

iv.Hydrogen is used

A.for the synthesis of ammonia

B. for the synthesis of methyl alcohol

C.nitrogenous fertilizers

D. all of these

5.Nisha observed that the bottoms of cooking utensils were turning black in colour while the flame of her stove was yellow in colour. Her daughter suggested cleaning the air holes of the stove to get a clean, blue flame. She also told her mother that this would prevent the fuel from getting wasted.

a) Identify the reasons behind the sooty flame arising from the stove.

b) Can you distinguish between saturated and unsaturated compounds by burning them? Justify your answer.

c) Why do you think the colour of the flame turns blue once the air holes of the stove are cleaned?

6.Blood transport food, Oxygen and waste materials in our bodies. It consists of plasma as a fluid medium. A pumping organ [heart] is required to push the blood around the body. The blood flows through the chambers of the heart in a specific manner and direction. While flowing throughout the body, blood exerts a pressure against the wall or a vessel.

• Pulmonary artery
• Pulmonary vein
• Very narrow and have high resistance
• Much wide and have low resistance
• Very narrow and have low resistance
• Much wide and have high resistance
• It is a hollow muscular organ
• It is four chambered having three auricles and one ventricle.
• It has different chambers to prevent O2 rich blood from mixing with the blood containing CO2
• Both A & C
• Blood = Plasma + RBC + WBC + Platelets
• Plasma = Blood – RBC
• Lymph = Plasma + RBC
• Serum = Plasma + RBC + WBC

7.A brain is displayed at the Allen Institute for Brain Science. The human brain is a 3-pound (1.4-kilogram) mass of jelly-like fats and tissues—yet it's the most complex of all known living structures The human brain is more complex than any other known structure in the universe. Weighing in at three pounds, on average, this spongy mass of fat and protein is made up of two overarching types of cells—called glia and neurons— and it contains many billions of each. Neurons are notable for their branch-like projections called axons and dendrites, which gather and transmit electrochemical signals. Different types of glial cells provide physical protection to neurons and help keep them, and the brain, healthy. Together, this complex network of cells gives rise to every aspect of our shared humanity. We could not breathe, play, love, or remember without the brain.

1)Animals such as elephants, dolphins, and whales actually have larger brains, but humans have the most developed cerebrum. It's packed to capacity inside our skulls and is highly folded. Why our brain is highly folded?

• b) Learning

3)Which among these protects our brain?

a)Neurotransmitter

b) Cerebrospinal fluid

d) Grey matter

4.Ram was studying in his room. Suddenly he smells something burning and sees smoke in the room. He rushes out of the room immediately. Was Ram’s action voluntary or involuntary? Why?

8.Preeti is very fond of gardening. She has different flowering plants in her garden. One day a few naughty children entered her garden and plucked many leaves of Bryophyllum plant and threw them here and there in the garden. After few days, Preeti observed that new Bryophyllum plants were coming out from the leaves which fell on the ground.

1.What does the incident sited in the paragraph indicate?

(a). Bryophyllum leaves have special buds that germinate to give rise to new plant.

(b). Bryophyllum can propagate vegetatively through leaves.

(c). Bryophyllum is a flowering plant that reproduces only asexually

(d). Both (a) and (b).

2.Which of the following plants can propagate vegetatively through leaves like Bryophyllum?

3.Do you think any other vegetative part of Bryophyllum can help in propagation? If yes, then which part?

(c) Flowers

4.Which of the following plant is artificially propagated (vegetatively) by stem cuttings in horticultural practices?

(b)Snakeplant

(d)Water hyacinth

9.The growing size of the human population is a cause of concern for all people. The rate of birth and death in a given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of the population.

1) What are common signs of sexual maturation in boys?

a) Broadening of shoulders

b) Development of mammary glands

c) Broadening of waist

d) High pitch of voice

2) Common sign of sexual maturation in girls is

a) Low pitch voice

b) Appearance of moustache and beard

c) Development of mammary glands

d) Broadening of shoulders

3) Which contraceptive method changes the hormonal balance of the body?

b) Diaphragms

c) Oral pills

d) Both a) and b)

4) What should be maintained for healthy society?

a) Rate of birth and death rate

b) Male and female sex ratio

c) Child sex ratio

d) None of these

10.Pea plants can have smooth seeds or wrinkled seeds. One of the phenotypes is completely dominant over the other. A farmer decides to pollinate one flower of a plant with smooth seeds using pollen from a plant with wrinkled seeds. The resulting pea pod has all smooth seeds.

i) Which of the following conclusions can be drawn?

(1) The allele for smooth seeds is dominated over that of wrinkled seeds.

(2) The plant with smooth seeds is heterozygous.

(3) The plant with wrinkled seeds is homozygous.

(b) 1 and 2 only

(c) 1 and 3 only

(d) 1, 2 and 3

ii) Which of the following crosses will give smooth and wrinkled seeds in same proportion?

(a) RR X rr

(b) Rr X rr

(d) rr X rr

iii) Which of the following cross can be used to determine the genotype of a plant with dominant phenotype?

(a) RR X RR

(b) Rr X Rr

(c) Rr X RR

(d) RR X rr

iv) On crossing of two heterozygous smooth seeded plants (Rr), a total of 1000 plants were obtained in F1 generation. What will be the respective number of smooth and wrinkled seeds obtained in F1 generation?

(a) 750, 250

(b) 500, 500

(C) 800, 200

(d) 950, 50

11.Food chains are very important for the survival of most species.When only one element is removed from the food chain it can result in extinction of a species in some cases.The foundation of the food chain consists of primary producers.Primary producers or autotrophs,can use either solar energy or chemical energy to create complex organic compounds,whereas species at higher trophic levels cannot and so must consume producers or other life that itself consumes producers. Because the sun’s light is necessary for photosynthesis,most life could not exist if the sun disappeared.Even so,it has recently been discovered that there are some forms of life,chemotrophs,that appear to gain all their metabolic energy from chemosynthesis driven by hydrothermal vents,thus showing that some life may not require solar energy to thrive.

1.If 10,000 J solar energy falls on green plants in a terrestrial ecosystem,what percentage of solar energy will be converted into food energy?

(d)It will depend on the type of the terrestrial plant

2.Matter and energy are two fundamental inputs of an ecosystem. Movement of

(a)Energy is by directional and matter is repeatedly circulating

(b)Energy is repeatedly circulating and matter is unidirectional

(c)Energy is unidirectional and matter is repeatedly circulating

(d)Energy is multidirectional and matter is bidirectional

3.Raj is eating curd/yoghurt. For this food intake in a food chain he should be considered as occupying

(a)First trophic level

(b)Second trophic level

(c)Third trophic level

(d)Fourth trophic level

4.Which of the following, limits the number of trophic levels in a food chain

(a)Decrease in energy at higher trophic levels

(b)Less availability of food

(c)Polluted air

5.The decomposers are not included in the food chain. The correct reason for the same is because decomposers

(a) Act at every trophic level at the food chain

(b) Do not breakdown organic compounds

(c) Convert organic material to inorganic forms

(d) Release enzymes outside their body to convert organic material to inorganic forms

12.Shyam participated in a group discussion in his inter school competition on the practical application of light and was very happy to win an award for his school. That very evening his father gave treat to celebrate Shyam’s win. Shyam while sitting saw an image of a person sitting at his backside in his curved plate and could see that person’s mobile drop in the flower bed. Person was not aware until Shyam went and informed him. He thanked Shyam for his clever move.

a)From which side of his plate Shyam observed the incident –

i)outward curved

ii)inward curved

iii)plane surface

b)Part of plate from which Shyam observed the incident acted like a-

i)concave mirror

ii)convex mirror

iii)plane mirror

c)The nature of the size of the image formed in above situation is –

i)real, inverted and magnified

ii)same size , laterally inverted

iii)virtual, erect and diminished

iv)real , inverted and diminished

d)Magnification of the image formed by convex mirror is –

more than 1

iii)equal to 1

iv)less than 1

• The location of image formed by a convex lens when the object is placed at infinity is

(a) at focus

(c) at optical center

• When the object is placed at the focus of concave lens, the image formed is

(a)real and smaller

(b) virtual and smaller

(c) virtual and inverted

• The size of image formed by a convex lens when the object is placed at the focus ofconvex lens is

(a) highly magnified

(b) point in size

• When the object is placed at 2F in front of convex lens, the location of image is

(b) between F and optical center

(c) at infinity

(d) none of the above

14.One of the wires in domestic circuits supply, usually with a red insulation cover, is called live wire. with black insulation is called neutral wire. The earth wire, which has insulation of green colour, is usually connected to a metal plate deep in the earth near the house appliances that has a metallic body. Overloading contact, in such a situation the current in the circuit abruptly increases. circuit prevents damage to the appliances and the circuit due to overloading.

1 When do we say that an electrical appliance

2 Mention the function of earth wire in electrical line

3 How is an electric fuse connected in a domestic circuit?

4 When overloading and short circuiting are said to occur?

5 What is a live wire?

15.Light of all the colours travel at the same speed in vacuum for all wavelengths. But in any transparent medium(glass or water), the light of different colours travels at different speeds for different wavelengths, which means that the refractive index of a particular medium is different for different wavelengths. As there is a difference in their speeds, the light of different colours bend through different angles. The speed of violet colour is maximum and the speed of red colour is minimum in glass so, the red light deviates least and violet colour deviates most. Hence, higher the wavelength of a colour of light, smaller the refractive index and less is the bending of light.

(i)Which of the following statements is correct regarding the propagation of Light of different colours of white light in air?

(a) Red light moves fastest.

(b) Blue light moves faster than green light.

(c) All the colours of the white light move with the same speed.

(d) Yellow light moves with the mean speed as that of the red and the violet light.

(ii)Which of the following is the correct order of wavelength?

(a) Red> Green> Yellow

(b) Red> Violet> Green

(c) Yellow> Green> Violet

(d) Red> Yellow> Orange

(iii)Which of the following is the correct order of speed of light in glass?

(a) Red> Green> Blue

(b) Blue> Green> Red

(c) Violet> Red> Green

(d) Green> Red> Blue

(iv)Which colour has maximum frequency?

16.The region around a magnet where magnetism acts is represented by the magnetic field.The force of magnetism is due to moving charge or some magnetic material. Like stationary charges produce an electric field proportional to the magnitude of charge, moving charges produce magnetic fields proportional to the current. In other words, a current carrying conductor produces a magnetic field around it. The subatomic particles in the conductor, like the electrons moving in atomic orbitals, are responsible for the production of magnetic fields. The magnetic field lines around a straight conductor (straight wire) carrying current are concentric circles whose centres lie on the wire.

1)The magnetic field associated with a current carrying straight conductor is in anti- clockwise direction. If the conductor was held horizontally along east west direction,what is the direction of current through it?

2)Name and state the rule applied to determine the direction of magnetic field in a straight current carrying conductor.

3)Ramus performs an experiment to study the magnetic effect of current around a current carrying straight conductor with the help of a magnetic compass. He reports that

a)The degree of deflection of magnetic compass increases when the compass is moved away from the conductor.

b)The degree of deflection of the magnetic compass increases when the current through the conductor is increased.

Which of the above observations of the student appears to be wrong and why?

Case Study Questions Class 10 Science CBSE Chapter Wise PDF

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NCERT Solutions for Class 10 Science Chapter 12 Electricity

The Class 10 NCERT Solutions for Science Chapter 12 Electricity includes all the intext and exercise questions. Class 10 Science Chapter 12 Electricity NCERT questions and answers help students to clear their doubts and to obtain good marks in Class 10 board exam. All the solutions provided in this article are strictly based on the CBSE syllabus and curriculum.

Class 10 Science Chapter 12 NCERT Questions and Answers

Class 10 Science Chapter 12 Electricity NCERT Questions and Answers are prepared by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 10 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 10 Science Chapter 12 Intext Questions

Intext Question (Page No. 200)

Question 1: What does an electric circuit mean?

Answer: A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric power source, wires, switches and electric devices like resistors bulbs etc.

Question 2: Define the unit of current.

Answer: When one-coulomb charge flows through an electric device in a circuit in one second, then the current flowing through the device is said to be one ampere.

Question 3: Calculate the number of electrons constituting one coulomb of charge.

Answer:   Charge on one electron, 𝑒 = 1.6 × 10 − 19 C

Intext Question (Page No. 202)

Question 1: Name a device that helps to maintain a potential difference across a conductor.

Answer: Cell or battery maintain potential difference across a conductor.

Question 2: What is meant by saying that the potential difference between two points is 1 V?

Answer: When 1 J of work is done to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.

Question 3: How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer: Potential difference created by battery Δ𝑉 = 6 V

Potential difference between two points in a circuit is defined as energy required (or work done) in moving one coulomb of charge from one point to the other.

Intext Question (Page No. 209)

Question 1: On what factors does the resistance of a conductor depend?

Answer:   The resistance of a resistor depends on

• Properties of the material of conductor, or resistivity of a material
• Length of conductor
• Area of cross-section of the conductor
• The resistivity of a material depends on temperature. Therefore, the resistance of a resistor also depends on the temperature of the conductor.

Question 2: Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer: Resistance of a conductor depends on resistivity 𝜌, length 𝑙 and on the area of cross-section 𝐴 as 

∴ when connected to the same power source, current through thick wire is more compared to thin wire of same material.

Question 3: 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?

Answer: The change in the current flowing through the electrical component can be determined by Ohm’s Law. According to Ohm’s Law, the current is given by I = V/R Now, the potential difference is reduced to half keeping the resistance constant, Let the new voltage be V’ = V/2

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

The change in the current can be determined using Ohm’s law as follows:

Therefore, the current flowing the electrical component is reduced by half.

Question 4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer: The resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. 

Question 5: Use the data in Table 12.2 to answer the following – Table 12.2 Electrical resistivity of some substances at 20°C

(a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?

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

(b) Among all the materials listed in the table, silver is the best conductor because the resistivity of silver is lowest among all, i.e., 1.60 × 10 –8 .

Intext Question (Page No. 213)

Question 1: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series. 

Question 2: Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Resisters are connected in series. So, the net resistance in the circuit = 5 Ω + 8 Ω + 12 Ω = 25 Ω Net potential = 6 V

Now for the 12 Ω resistor, current = 0.24 A

So, using Ohm’s law V = 0.24 × 12 V = 2.88 V

Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.

Intext Question (Page No. 216)

Question 1: Judge the equivalent resistance when the following are connected in parallel – 

From the above two problems, when very low resistance is connected to very high resistance, the resistance of the combination will be close to low resistance.

Question 2: An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω 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 through it?

To draw the same current resistance of electric iron required is 31.25 Ω and current through it is 7.04 A

Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? 

Answer: To operate properly, different electric devices need a different amount of current. In series combination all the devices get the same current, whereas in parallel combination potential difference across all the resistors is same and current will be distributed according to resistance.

In series combination, if one device fails circuit becomes broken and other devices stop working. Whereas in parallel combination all devices are independently connected to mains, even if one device fails other devices continue to work.

Question 4: 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: Case 1: The circuit diagram below shows the connection of three resistors

From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows: R eq = 2 Ω +2 Ω = 4 Ω Hence, the total resistance of the circuit is 4 Ω.

Case 2: When all resistors are connected in parallel, the equivalent resistance is,

Question 5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer: Highest resistance is possible when all resistors are connected in series.

R = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω ∴ highest resistance possible is 48 Ω

Lowest resistance is possible when all resistors are connected in parallel  

∴ Lowest resistance possible is 2 Ω

Intext Question (Page No. 218)

Question 1: Why does the cord of an electric heater not glow while the heating element does? 

Answer: When the same current flows through the conducting wire and heating element, the heat generated (I 2 Rt) is very high in heating element compared to conducting wire. This is due to the resistance of the heating element is very high compared to the resistance of conducting wire. Thus, when same current flows through the conducting wire and heating element, heating element gets hot and glows.

Question 2: Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.

Answer: From Joule’s law of heating, the heat generated can be written as H = V × I × t where, V is the voltage, V = 50 V I is the current t is the time in seconds,  The amount of current can be calculated as follows:

Therefore, the heat generated when 96000 coulomb of charge flows through a potential difference of 50 V is 4.8 × 106 J

Question 3: An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:  From Joule’s law of heating, the heat generated can be written as

Therefore, the heat generated in 30 s is 15 × 10 3 J

Intext Question (Page No. 220)

Question 1: What determines the rate at which energy is delivered by a current? 

Answer: Electric power is the rate of consumption of electrical energy by electric appliances. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Question 2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

NCERT Solutions for Class 10 Science Chapter 12 Exercise Questions

Question 1:   A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –

Question 2: Which of the following terms does not represent electrical power in a circuit? 

Question 3: An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 𝑊 (b) 75 𝑊 (c) 50 𝑊 (d) 25 𝑊

Answer: (d) 25 𝑊

The resistance of the bulb remains constant if the supply voltage is reduced to110 V. If the bulb is operated on 110 V, then the energy consumed by it is given by the expression for power

Hence, the option (d) is correct.

Question 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be – 

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer: (c) 1:4

Therefore, the ratio of heat produced in series and parallel combinations is 1:4. Hence, the option (c) is correct.

Question 5: How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer: Voltmeter is always connected in parallel with circuit element to measure the potential difference across it.

Question 6: A copper wire has a diameter 0.5 mm and resistivity of 1.6 × 10 – 8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Question 7:   The values of current  flowing in a given resistor for the corresponding values of potential difference 𝑉 across the resistor are given below –

Plot a graph between 𝑉 and 𝐼 and calculate the resistance of that resistor. 

Answer: The plot between voltage and current is called VI characteristic. The voltage is plotted on x-axis and current is plotted on y-axis.

Question 8: When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

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

Answer:   In series combination current through all the resistors will be the same. Resistance of resistors when connected in series is given by

So, the current through the 12 Ω resistor will be same as 0.67 A.

Question 10: How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? 

Answer: Let the required number of resistors be 𝑛. Given, Current I = 5A and Potential difference V = 220V Now, from Ohm’s law, V = IR

Now for 𝑛 number of resistors of resistance 176 Ω, the equivalent resistance of the resistors connected in parallel is 44 Ω.

Therefore 4 resistors of 176 Ω are required to draw 5 A current on 220 V line.

Question 11: Show how you would connect three resistors, each of resistance 6 Ω so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω. 

Answer : If we connect all the three resistors in series, their equivalent resistor would 6 Ω + 6 Ω + 6 Ω =18 Ω, which is not the desired value. Similarly, if we connect all the three resistors in parallel, their equivalent resistor would be

which is again not the desired value. We can obtain the desired value by connecting any two of the resistors in either series or parallel.

If two resistors are connected in parallel, then their equivalent resistance is

The third resistor is in series, hence the equivalent resistance is calculated as follows: R  = 6 Ω + 3 Ω = 9 Ω

When two resistors are connected in series, their equivalent resistance is given by

R  = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is calculated as follows:

Question 12: Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Question 13: A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases? 

Question 14: Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Question 15: Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Question 16: Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:   Energy consumed by an electrical appliance is given by H = Pt Power rating of TV set is 250 W Energy consumed by TV set 1 hour = 250 × 60 × 60 = 900000 J Power rating of toaster is 1200 W Energy consumed by toaster in 10 minutes = 1200 × 10 × 60 =720000 J Hence, TV set uses more energy than toaster.

Question 17: An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Question 18: Explain the following. 

(i) Why is tungsten used almost exclusively for filament of electric lamps?

(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

(in) Why is the series arrangement not used for domestic circuits?

(iv) How does the resistance of a wire vary with its area of cross-section?

(v) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: (i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted.

(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.

(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working because the whole circuit is broken.

(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R ∝ (1/πr 2 ). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor.

(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

Topics covered under Class 10 Science Chapter 12 Electricity

Below we have listed the topics discussed in NCERT Solutions for Class 10 Science Chapter 12. The list gives you a quick look at the different topics and subtopics of this chapter.

NCERT Solutions for Class 10 Science Chapter 12 – A Brief Discussion

Chapter Overview: Electricity has an important place in modern society. It is a controllable and convenient form of energy for a variety of uses in homes, schools, hospitals, industries and so on. In this chapter students will learn about electric current, electric circuit, circuit diagram. This chapter also covers Ohm’s law, resistor etc. 

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Evergreen Science Solutions for Class 10 Chapter 12 Electricity

• Last modified on: 1 year ago
• Reading Time: 13 Minutes

Class 10 Science Evergreen Solutions are a comprehensive set of solutions that cover all the chapters in the CBSE Science curriculum for class 10. These solutions are designed to help students build a strong foundation in science and prepare effectively for their exams.

Chapters Covered in Evergreen Science for Class 10

• Chemical Reactions and Equations
• Acid Base & Salts
• Metals And Non Metals
• Carbon And Its Compounds
• Periodic Classification of Elements
• Life Processes
• Control and Coordination
• How do Organisms Reproduce
• Heredity And Evolution
• Light-Reflection and Refraction
• Human Eye and Colourful World
• Electricity
• Magnetic Effects of Electric Curren
• Source Of Energy
• Our Environment
• Sustainable Management of Natural Resources

Why ‘Evergreen Science’ Useful for Class 10 Students?

An evergreen science book for CBSE class 10 is useful for several reasons:

• Comprehensive coverage: These books cover all the essential topics in a subject, providing students with a complete understanding of the subject matter.
• Simplified language: Evergreen science books are written in a clear and concise manner, making it easy for students to understand complex concepts.
• Illustrations and diagrams: These books contain numerous illustrations and diagrams that help students visualize concepts and make it easier for them to remember.
• Exam preparation: Evergreen science books contain a wide range of practice questions, solved examples, and exercises that help students prepare for exams.
• Long-term relevance: These books are considered “evergreen” because their content is relevant and useful even after several years. This makes them a valuable resource for students even beyond their class 10 exams.

In summary, evergreen science books for CBSE class 10 are an essential resource for students who want to build a strong foundation in science and excel in their exams.

Who should practice from class 10 science evergreen solutions?

Class 10 science evergreen solutions are recommended for students who are studying science in class 10, especially those who are preparing for the CBSE board exams. These solutions are helpful for students who want to:

• Build a strong foundation: Evergreen solutions provide a comprehensive coverage of all the topics in the science curriculum, helping students to build a strong foundation in the subject.
• Improve understanding: By providing detailed explanations, examples, and illustrations, evergreen solutions can help students to better understand complex scientific concepts.
• Practice questions: Evergreen solutions include a variety of practice questions, which can help students to test their understanding and practice answering questions in the format that they will encounter on the exam.
• Prepare for exams: Evergreen solutions are designed to help students prepare for exams, with practice questions and solved examples that closely resemble the types of questions that they will encounter on the exam.

Evergreen Class 10 Science Chapter Description

Here are some descriptions about the solutions for each chapter:

Chapter 1: Chemical Reactions and Equations

• The solutions provide a detailed explanation of different types of chemical reactions, balanced chemical equations, and their applications.
• The solutions also include numerous solved examples and practice questions to help students understand the concepts and prepare for exams.

Chapter 2: Acids, Bases and Salts

• The solutions explain the properties and uses of acids, bases, and salts in detail.
• The solutions also include a variety of practice questions, including multiple-choice questions, fill in the blanks, and short answer questions.

Chapter 3: Metals and Non-metals

• The solutions provide a detailed explanation of the properties, occurrence, and uses of metals and non-metals.
• The solutions include a variety of practice questions, including numerical problems and multiple-choice questions.

Chapter 4: Carbon and Its Compounds

• The solutions explain the properties, occurrence, and uses of carbon and its compounds in detail.
• The solutions include a variety of practice questions, including short answer questions and numerical problems.

Chapter 5: Periodic Classification of Elements

• The solutions provide a detailed explanation of the periodic table, the classification of elements, and their properties.
• The solutions also include a variety of practice questions, including fill in the blanks, short answer questions, and multiple-choice questions.

Chapter 6: Life Processes

• The solutions explain the different life processes and their importance in living organisms.
• The solutions also include a variety of practice questions, including short answer questions, multiple-choice questions, and true/false questions.

Chapter 7: Control and Coordination

• The solutions provide a detailed explanation of the nervous system, endocrine system, and their role in controlling and coordinating body functions.
• The solutions include a variety of practice questions, including fill in the blanks, short answer questions, and multiple-choice questions.

Chapter 8: How do Organisms Reproduce?

• The solutions explain the different modes of reproduction, sexual and asexual reproduction, and their importance.
• The solutions also include a variety of practice questions, including true/false questions, fill in the blanks, and short answer questions.

Chapter 9: Heredity and Evolution

• The solutions provide a detailed explanation of genetics, DNA, and the process of evolution.
• The solutions include a variety of practice questions, including short answer questions, fill in the blanks, and multiple-choice questions.

Chapter 10: Light – Reflection and Refraction

• The solutions explain the concepts of reflection and refraction of light and their applications in detail.
• The solutions also include a variety of practice questions, including numerical problems, fill in the blanks, and short answer questions.

Chapter 11: Human Eye and Colourful World

• The solutions provide a detailed explanation of the structure and functioning of the human eye, and the concept of colours.

Chapter 12: Electricity

• The solutions explain the concepts of electric current, circuits, and electromagnetism in detail.

Chapter 13: Magnetic Effects of Electric Current

• The solutions explain the concepts of magnetic fields, electromagnetic induction, and their applications in detail.

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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

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

Mastery app thanks

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