## 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

(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 ohm, the amount of heat produced is

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

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

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

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

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

## 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

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

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

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

## 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

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

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

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

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## Case Study Questions Class 10 Science Electricity

Case study questions class 10 science chapter 12 electricity.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

## Case study: 1

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.

1) What is SI unit of resistivity?

2) What is variable resistance?

3) Why tungsten is used in electric bulbs?

4) 1M ohm = ?

1) The SI unit of resistivity is ohm meter.

2) The electric component which is used to regulate the electric current without changing voltage source is called as variable resistance.

3) Tungsten filament are used in electric bulbs because the resistivity of Tungsten is more and it’s melting point is also high.

4) 1M ohm = 10 6 ohm

## 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.

Questions:

1) In which case the equivalent resistance is more and why?

2) In our home, which type of combination of electric devices is preferred? Why?

3) If n resistors of resistance R are connected in parallel then what is the equivalent resistance?

1) 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 +….

2) 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.

3) 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

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.

1) What is the SI unit of electric energy?

2) How heating effect works to protect electric circuit?

3) 1KW h = ?

4) If a bulb is working at a voltage of 200V and the current is 1A then what is the power of the bulb?

1) The SI unit of electric energy is watt hour. And the commercial unit of electric energy is kW h.

2) 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.

3) 1kW h = 3.6*10 6 joule

4) Given that,

V = 200V, I = 1A

Then, P = VI = 200*1 = 200 J/s = 200 W

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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 $$$$\Omega$$$$  and 3 $$$$\Omega$$$$ 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 $$$$\Omega$$$$ , 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 $$$$₹$$$$ 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 $$$$\Omega$$$$  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)  $$$$\text { Energy (in kWh) }=\frac{V(\text { volt }) \times I(\text { ampere }) \times t(\text { sec })}{1000}$$$$

## *****************************************

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  $$$$=\frac{12}{3}=4 \Omega$$$$ $$$$\frac{1}{R_{P}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4} \Rightarrow R_{p}=\frac{4}{3} \Omega$$$$ (iii) (a): All the three resistors are in parallel. $$$$\frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{3}+\frac{1}{1}=\frac{1+2+6}{6}=\frac{9}{6}$$$$ $$$$R_{P}=\frac{6}{9}=\frac{2}{3} \Omega$$$$ (iv) (a): Voltage is same across each resistance. So, I 1  x 5 = I 2 X 10 = 15 x I 3 I 1  = 2I 2 = 3I 3 (v) (d): All are in parallel. $$$$\begin{array}{l} \frac{1}{R_{p}}=\frac{1}{12} \times 4=\frac{1}{3} \Rightarrow R_{p}=3 \Omega \\ I=\frac{3}{3}=1 \mathrm{~A} \end{array}$$$$ So,current in each resistor  $$$$I^{\prime}=\frac{3}{12}=\frac{1}{4} \mathrm{~A}$$$$

(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 $$$$\begin{array}{l} R_{2}=\frac{r_{1} r_{2}}{r_{1}+r_{2}} \\ \frac{R_{1}}{R_{2}}=\frac{\left(r_{1}+r_{2}\right)^{2}}{r_{1} r_{2}} \end{array}$$$$ (iv) (c): In the given circuit, 3 $$$$\Omega$$$$  resistors are in series. R S  = 3 + 3 = 6  $$$$\Omega$$$$ Now, R S  and 6 $$$$\Omega$$$$  are parallel. $$\frac{1}{R_{p}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R_{p}=3$$\Omega$$$$ (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 $$$$\mathrm{So}, H^{\prime}=(2 I)^{2} \cdot \frac{R}{2} t=2 H$$$$ (iv) (b): Given: 1= 5 A, resistance = R. Let r be the new radius. Now,H=I 2 Rt Also H' = I' 2 R' t From (i) and (ii),  $$$$5^{2} \times \rho \frac{L}{\pi r^{2}} t=10^{2} \times \rho \frac{L}{\pi r^{\prime 2}} \cdot t$$$$ $$$$\frac{25}{r^{2}}=\frac{100}{r^{\prime 2}} \Rightarrow \frac{r^{\prime}}{r}=2 \Rightarrow r^{\prime}=2 r$$$$ (v) (c): Given:  $$$$I=0.5 \mathrm{~A}, R=10 \Omega, t=5 \mathrm{~min}$$$$ $$$$\begin{array}{l} H=I^{2} R t=0.5 \times 0.5 \times 10 \times 5 \times 60 \\ H=750 \mathrm{~J} \end{array}$$$$

(i) (a) :  $$$$E \propto t$$$$ (ii) (c) : Given: P = 60W, t = 1 min E = 60 x 1 x 60 = 3600J (iii) (b) : Given: P = 400 $$$$\Omega$$$$ , t = 8 hour E = 400 x 8 = 3200Wh = 3.2kWh Cost = 3.2 x 5 =  $$$$₹$$$$ 16 (iv) (a) : Given: I= 5 A, R = 2  $$$$\Omega$$$$ , 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 for Class 10 Science Chapter 12 Electricity

Question 1:

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

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

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

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

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

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

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

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## Class 10 Science 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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CBSE NCERT Solutions

## 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

## Class 10 Computer Science Notes and Questions

• NCERT Exemplar
• Science Exemplar Class 10
• Electricity

## 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.

## 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

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.

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?

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

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

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

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

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

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.

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

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, leave a comment cancel reply.

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## myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

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

## What are case study questions?

• Sample Papers with Case Study questions
• Class 10 Science Case Study question examples
• 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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## 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.
• 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

(b) nitrogen 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?

b) Development of mammary glands

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

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?

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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• Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

Class 10 is an essential stage in every student's career. Their performance in Class 10 acts as a base for their future studies. So it's important to score well in Class 10. The most important and challenging subject in Class 10 is Science and many students seem to get confused here and lose marks. If students want a successful career in future, then they can't afford to lose marks in this stage of their life. It's important to understand every chapter in science thoroughly to score good marks. Chapter 12 of Class 10 Science which is about electricity is one of the difficult chapters. A student who is incapable of understanding this chapter must practice Important Questions for Class 10 Science Chapter 12 . These important questions of Electricity Class 10 can make the students through on the concepts of this chapter. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

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

## Study Important Questions for Class 10 Science Chapter 12 – Electricity

Very Short Answer Questions (1 Mark)

1. Which two circuit components are connected in parallel in the following circuit diagram?

(a) ${{\text{R}}_{\text{1}}}$ and ${{\text{R}}_{\text{2}}}$ only

(b) ${{\text{R}}_{\text{1}}}{\text{,}}{{\text{R}}_{\text{2}}}$ only

(c) ${{\text{R}}_{\text{2}}}$ and ${\text{V}}$ only

(d) ${{\text{R}}_{\text{1}}}$ and ${\text{V}}$only

Ans: (a) The two circuit components that are connected parallel in the circuit diagram is  and ${R_2}$ only.

2. A metallic conductor has loosely bound electrons called free electrons. The metallic conductor is

(a) negatively charged

(b) positively charged

(c) neutral

(d) Either positively charged or negatively charged

Ans: (c) The metallic conductor is neutral.

3. Which of the following expressions does not represent the electric power in the circuit?

(a) ${\text{VI}}$.

(b) ${{\text{I}}^{\text{2}}}{\text{/R}}$

(c) ${{\text{V}}^{\text{2}}}{\text{/R}}$

(d)  ${{\text{I}}^{\text{2}}}{\text{R}}$

Ans: (b) The expression which does not represent the electric power in the circuit is ${I^2}/R$ .

4. Resistivity of a metallic wise depends on

(a) its length

(b) its shape

(c) its thickness

(d) nature of material

Ans: (d) Resistivity of a metallic wire depends on the nature of the material.

5. If the current I through a resistor is increased by ${\mathbf{100}}\%$ the increased in power dissipation will be (assume temperature remain unchanged)

(a) ${\mathbf{100}}\%$

(b) ${\mathbf{200}}\%$

(c) ${\mathbf{300}}\%$

(d) ${\mathbf{400}}\%$

Ans: (c) The increase in power dissipation will be $300\%$ .

6. For the circuit arrangement shown below, a student would observe.

(a) Some reading in both ammeter and voltmeter.

(b) No reading in either the ammeter or the voltmeter.

(c) Some reading in the ammeter but no reading in the voltmeter.

(d) Some reading in the voltmeter but no reading in the ammeter.

Ans: (c) A student will observe some reading in the ammeter but no reading in the voltmeter.

7. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected in parallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{{\text{1}}}{{\text{5}}}$

(b) ${\text{5}}$

(c) $\dfrac{{\text{1}}}{{{\text{25}}}}$

(d) ${\text{25}}$

Ans: (d)The ratio $\dfrac{R}{{R'}}$ is $25$ .

8. The resistance of the conductor is $R$ . If its length is doubled, then its new resistance will be

Ans: (c) The new resistance is $4R$ .

9. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances ${R_1},{R_2},{R_3}$ respectively as shown in the figure. Which of the following is live?

(a) ${R_3} > {R_2} > {R_1}$

(b) ${R_2} > {R_3} > {R_1}$

(c) ${R_1} > {R_2} > {R_3}$

(d) ${R_1} = {R_2} = {R_3}$

Ans: (a)According to the graph , ${R_3} > {R_2} > {R_1}$

10. The nature of the graph between potential difference and the electric current flowing through a conductor is

(a)parabolic

(b) circle

(c) straight line

(d) hyperbolic

Ans: (c)The nature of the graph between potential difference and the electric current flowing through a conductor is a straight line.

11. An electric heater is salted at $1500$ w. How much heat is produced per hour?

(i) $5400$ J

(ii) $54000$ J

(iii) $5.4 \times {10^5}$

(iv) $5.4 \times {10^6}$.

Ans: (iv) The electric heater produces $5.4 \times {10^6}$ J per hour.

12. A student says that the resistance of two wires of the same length and same area of  cross section is the same. This statement is correct if

(a) Both wires are of different materials

(b) Both wires are made of the same material and are at different temperatures.

(c) Both wires are made of the same material and are at the same temperature.

(d) Both wires are made of different materials and are at the same temperature.

Ans: This statement is correct if (c)The resistance of two wires of the same length and same area of cross section is the same if both wires are made of the same material and are at the same temperature.

13. In an experiment ohm, s law a student obtained a graph as shown in the diagram. The value of resistance of the resistor is

(a) $0.1\Omega$

(b) $1.0\Omega$

(c) $10\Omega$

(d) $100\Omega$

Ans: (d) The value of resistance of the resistor is $100\Omega$ .

14. Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is

(a) $1$ Joule

(b) $10$ Joule

(c)  zero

(d) $100$ Joule

Ans: (c) Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is zero.

15. Three resistors are shown in the figure. The resistance of the combination is

(a)$3\Omega$

(b) $6\Omega$

(c) $9\Omega$

(d) $7\Omega$

Ans: (c) The resistance of the combination is $9\Omega$ .

16. Name a device that helps to maintain a potential difference between across a conductor.

Ans: A device that helps to maintain a potential difference between conductors is the battery.

17. What determines the rate at which energy is delivered by a current?

Ans: The rate at which energy is delivered by a current is determined by electric power.

18. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is

(a) $\dfrac{1}{5}$

(c) $\dfrac{1}{{25}}$

Ans: (d)A wire of resistance ${\text{R}}$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are ${\text{R'}}$. In this cases, the ratio$\dfrac{R}{{R'}}$ is $25$ .

19. Which of the following terms does not represent electrical power in a circuit?

(a) ${I^2}R$

(b) $I{R^2}$

(c)  $VI$

(d) ${V^2}/R$

Ans: (b) The term that does not represent electrical power in a circuit is $I{R^2}$ .

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

(a) $220$ W

(b) $75$ W

(c) $50$ W

(d) $25$ W

Ans: (d)The power consumed will be $25$ W.

21. 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 combination would be:

Ans: (c) The ratio of heat produced in series and parallel combination will be $1:4$ .

22. A wire of resistance $R$ is bent in the form of a closed circle, what is the resistance across a diameter of the circle?

Ans: The resistance across a diameter of the circle

23. A charge of $6$ C is moved between two points $P$ and $Q$ having, potential $10$ V and $5$ V respectively. Find the amount of work done.

Ans: The amount of work done, $W = q({V_2} - {V_1})\;$

$\; = {\text{ }}6\left( {10 - 5} \right)$

$= {\text{ }}30$ joule

24. Name the physical quantity whose SI unit is JC

Ans: The physical quantity whose SI unit is JC is Potential.

25. Why are copper wires used as connecting wires?

Ans: Copper wires are used as connecting wires because in case of copper the electrical resistivity for it is low.

26. A wire of resistivity $p$ is stretched to double its length. What is its new resistivity?

Ans: When a wire of resistivity ${\text{p}}$ is stretched to double its length, then the new resistivity remains the same because resistivity depends on the nature of material.

27. What is the resistance of the connecting wire?

Ans: The resistance of a connecting wire made of a good conductor is extremely low.

28. What is the resistance of an ammeter?

Ans: An ammeter's resistance is very minimal, and in an ideal ammeter, it is zero.

29. What is the resistance of a Voltmeter?

Ans: An ideal voltmeter's internal resistance is infinite.

1. How does use of fuse wire protect electrical appliances?

Ans: When a large quantity of current passes through the circuit, the temperature of the wire rises and the fuse wire melts. This prevents current from flowing into the house's other circuits, saving electrical appliances.

2. Calculate the resistance of an electric bulb which allows a $10$A current when connected to a $220$ V power source?

Ans: It is given from the question that an electric bulb allows a $10$A current when connected to a $220$ V power source.

$I = 10$ A ,

$V = 220$V

$R = \dfrac{V}{I}$

$= \dfrac{{220}}{{10}}$

$= 22$ ohm

3. (i) Identify the $V - I$ graphs for ohmic and non-ohmic materials.

Ans: The $V - I$ graphs for ohmic and non-ohmic materials respectively can be represented as shown below:

(ii) Give one example of each.

Ans: (ii) Some examples of ohmic material are Copper, Nichrome and some examples of Non-ohmic material   are Diode, Transistor.

4. What do the following symbols represent in a circuit? Write the name and one function of each?

Ans: (i) It symbolises a battery that maintains a potential difference across the circuit element to allow current to flow.

Ans: It's an ammeter that measures how much current is flowing across a circuit.

5. Define the term “volt”?

Ans: If $1$joule of energy is transferred between two points A and $B$, the potential difference between them is one volt. In an electric circuit, work is done to move one coulomb of charge from one point to another field.

6. Why does the connecting rod of an electric heater not glow while the heating element does?

Ans: As its resistance is lower than that of the heating element, the connecting cord of an electric heater does not glow. As a result, the heating element produces more heat than the connecting cord, and it glows

7. A number of $n$ resistors each of resistance $R$ are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit is series combination and parallel combination?

Ans: Total effective resistance of  the circuit when in series combination ${R_s} = nR$

And for parallel combination is ${R_p} = \dfrac{R}{n}$  and

$\dfrac{{{R_s}}}{{{R_p}}} = \dfrac{{nR}}{{\dfrac{R}{n}}}$

The ratio will be ${n^2}$ .

8. Draw a schematic diagram of a circuit consisting of $3$ V battery, $5$ ohm, $3\Omega$  and $1\Omega$ resistor, an ammeter and a plug key, all connected in series.

Ans: The circuit diagram of a circuit consisting of $3$ V battery, $5$ ohm, $3\Omega$  and $1\Omega$ resistor, an ammeter and a plug key, all connected in series can be represented as show below,

9. A copper wire has diameter $0.5$ mm and Resistivity of $1.6 \times {10^{ - 8}}\Omega m$ What is the length of this wire to make its resistance ? How much does the resistance change if diameter is doubled?

Ans: Diameter of the copper wire, $D = 0.5 \times {10^{ - 3}}$ m

$P = 1.6 \times {10^{ - 8}}$

$R = \dfrac{{\rho l}}{A}$

$= \dfrac{{\rho l}}{{\pi {r^2}}}$

$= \dfrac{{4\rho l}}{{\pi {D^2}}}$

$\Rightarrow l = \dfrac{{\pi R{D^2}}}{{4\rho }}$

$l = \dfrac{{3.14 \times 10 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{4 \times 1.62 \times {{10}^{ - 8}}}}$

$= 121.14$

Length of the wire,  $l = 121.14$ m

New $R' = \dfrac{{4\rho l}}{{\pi {{(D')}^2}}}$

$= \dfrac{1}{4}\dfrac{{4\rho l}}{{\pi {D^2}}}$

$= \dfrac{1}{4}R$

Length of the wire to make its resistance $10\Omega$ is $121.14$m and when the diameter is doubled the new resistance will be one fourth that of the old one.

10. Alloys are used in electrical heating devices rather than pure metals. Give a reason.

Ans: Alloys are utilised in electricity heating devices rather than pure metals because alloys have a higher resistivity and hence produce more heat. Furthermore, alloy is non-combustible (or oxidises easily at higher temperature).

11. On what factor does the resistance of a conductor depend?

Ans: The factors on which Resistance depends are:-

(a) Length of the conductor

(b) Area of cross - section

(c)  Temperature

(d) Nature of material

12. Calculate the number of electrons consisting of one coulomb of charge?

Ans: Let $x =$ no. of electrons

Charge on $1$ electron $= 1.6 \times {10^{ - 19}}$ C , that is

$x = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$x = 6.25 \times {10^{18}}$

The number of electrons consisting of one coulomb of charge is $6.25 \times {10^{18}}$ .

13. What does an electric circuit mean?

Ans: An electric circuit is a current route that is both continuous and closed. Current can flow through an electric circuit if it is complete.

14. Define the unit of current.

Ans: The ampere is the SI unit for electric current. If $1$ coulomb charge flows per second across a conductor cross-section, the current is said to be $1$ ampere.

15. Calculate the number of electrons constituting one coulomb of charge.

Ans: The charge on one electron $= 1.6 \times {10^{ - 19}}$ coulomb.

Number of electrons in one coulomb of charge$= \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$= 6.25 \times {10^{18}}$

16. What is meant by saying that the potential difference between two points is $1$ v?

Ans: If $1$ joule of labour is required to move a charge of $1$ coulomb from one location to another, the potential difference between the two points is said to be $1$ volt.

17. Ammeter burns out when connected in parallel. Give reasons.

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

18. Judge the equivalent resistance when the following are connected in parallel:

(a) Equivalent resistance of $1\Omega$ and ${10^6}\Omega$

Ans: When the resistances are connected in a parallel arrangement, the resultant resistance is given by:

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \cdots$

$\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^6}}}$

$= 1 + {10^{ - 6}}$

$R = 1\Omega$

The equivalent resistance is $1\Omega$ .

(b) Equivalent resistance of  $1\Omega$ , ${10^3}\Omega$ and ${10^6}\Omega$

Ans: $\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^6}}}$

$= 1 + {10^{ - 3}} + {10^{ - 6}}$

19. An electric iron of resistance $20$ takes a current of $5$ A. Calculate the heat developed in $30$ s.

Ans: Resistance of electric iron, $R = 20\Omega$ , current, $I = 5$ A and time $= 30$ s.

Heat generated $H = {I^2}Rt$

$= {5^2} \times 20 \times 30$

$= 15000$ j

The heat developed in $30$ second is $15000$ j .

20. Compute the heat generated while transferring $96000$ coulomb of charge in one hour through a potential difference of $50$ V

Ans: The given information is as shown below,

Charge transferred, $Q = 96000$

Potential Difference, $V = 50$  V.

Heat generated, $H = VQ$

$= 50 \times 96000$

$= 4800000$ j

$= 4.8 \times {10^6}$ j

The heat generated while transferring $96000$  coulomb of charge in one hour

through a potential difference of $50$ V is $4.8 \times {10^6}$ j

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

Ans: Given that current drawn by electric motor $I = 5$ A.

The line voltage $V = 220$ V

Time, $t = 2$ h

Power of motor , $P = VI$

$= 220 \times 5$

$= 1100$ W and

the energy consumed $E = Pt$

$= 2 \times 1100$

$= 2.2$ KWh

The power of the motor and energy consumed in $2$ h are $1100$ W and $2.2$ kWh respectively.

22. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans: A voltmeter is connected in parallel to the resistance across the place where the potential difference is to be determined.

23. 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.

Ans: Given that Voltage of battery, $V = 12$  V,

Current, $I = 2.5$ mA

$= 2.5 \times {10^{ - 3}}$ A

Resistance, $R = V/I$

$= \dfrac{{12}}{{2.5 \times {{10}^{ - 3}}}}$

$= 4800\Omega$

The value of the resistance of the resistor is $4800\Omega$ .

24. 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?

Ans: The given information is as shown below

Each bulb is rated as $10$ W, $220$ V,

It draws current, $I = P/V$

$= \dfrac{{10}}{{220}}$ V

$= 1/22$ A.

The maximum allowable current is $5$ A and all lamps are connected in parallel. Therefore the maximum number of bulbs joined in parallel with each other $= 5 \times 22$whichis$110$.

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

Ans: Current drawn by ${1^{st}}$lamp rated $100$ W at $220$ ,  $V = P/V$

$= 100/220$

$= 5/11$ A.

Current drawn by ${2^{nd}}$lamp rated $60$ W at $220$ , $V = 60/220$$= 3/11\;$A.

In parallel arrangement the total current $= {\text{ }}3/11 + {\text{ }}5/11$

$= {\text{ }}8/11$

$= {\text{ }}0.73$ A.

Current drawn from the line if the supply voltage is $220$ V is $0.73$ A .

26. Which uses more energy, a $250$ W TV set in $1$ hour, or a $1200$ W toaster in $10$ minutes?

Ans: Energy used by a TV set of power $250$ W in $1$ hour $= Pt$

Energy used by toaster of power $1200$ W in $10$ minute $\left( {10/60h} \right) = {\text{ }}200$ Wh.

A $250$ W TV set in $1$ hour uses more energy.

27. An electric heater of resistance $8$ draws $15$ A from the service mains for $2$ hours. Calculate the rate at which heat is developed in the heater.

Ans: Resistance of electric heater, $R{\text{ }} = {\text{ }}8\Omega$,

current, $I{\text{ }} = {\text{ }}15$ A.

Rate at which heat developed in the heater $= \dfrac{{{I^2}Rt}}{t}$

$= 1800$ W.

The rate at which heat is developed in the heater is $1800$ W.

28. In the given figure what is the ratio of current in  $A$

Ans: Observe that it I clearly known to us that $V = IR$

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{R}{{2R}}$

$= \dfrac{1}{2}$

The ratio of current is$1/2$.

29. Two wires of equal cross sectional area, one of copper and other of managing have the same resistance. Which one will be longer?

Ans: Using the equation  $\rho = \dfrac{{RA}}{l}$ , where $\rho$ is the resistivity, $R$ is the resistance and $A$ the area

We can see that copper wire has a lower resistance than manganin, hence copper will last longer.

30. A Rectangular block of iron has dimensions $L \times L \times b$ . What is the resistance of the block measured between the two square ends? Given $p$ resistivity.

Ans: $R = \dfrac{{pb}}{{{L^2}}}$is the resistance of the block measured between the two square ends

31. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances?

Ans: ${R_{series}} = 3R$

${R_{parallel}} = R/3$

The ratio of Resistances is $9$ .

32. Justify for any pair of resistance the equivalent resistance equivalent resistance in parallel.

Ans: As $R = V/I$

From the graph for any pair of resistance the equivalent resistance in series is greater than equivalent resistance in parallel.

$A =$ Series, $B =$ Parallel

33. How many bulbs of $81$ should be joined in parallel to draw a current of $2$ A from a battery of $4$ V?

Ans: $R = V/I$

$= 2\Omega$

let $n$ be the number of bulbs.

$1/R = 1/{R_1} + 1/{R_2} + \cdots + 1/{R_n} = \dfrac{n}{8}$

$\Rightarrow \dfrac{1}{2} = \dfrac{n}{8}$

$\Rightarrow n = 4$

Number of bulbs are $4$ .

34. Two cubes $A$ and $B$ are of the same material. The side of $B$ is thrice as that of $A$ . Find the ratio ${R_A}/{R_B}$ .

Ans: The value of  ${R_A} = \dfrac{{pL}}{A}$ and

${R_B} = \dfrac{{p3L}}{{9A}}$

${R_A}:{R_B} = 3:1$

35. If there are $3 \times {10^{11}}$ electrons are flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron $1.6 \times {10^{19}}$

Ans: Observe as shwn below,

Using the equation

$= 3 \times {10^{11}} \times 1.6 \times {10^{19}}$C

$= 4.8 \times {10^8}$C

$= \dfrac{{4.8 \times {{10}^8}}}{{2 \times 60}}$

$= 4 \times {10^7}$ A

The current flowing through the circuit is $4 \times {10^7}$ A.

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

Ans:   Looking at the equation

$Q = {I^2}RT$

$Q = {I^2}(pL/A)t$

Henceforth, because nichrome wire has a higher resistance than copper wire, it must be heated first.

37. What is represented by joule/coulomb?

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

38. A charge of $2$ C moves between two plates, maintained at a p.d of $IV$ . What is the energy acquired by the charge?

Ans: The energy acquired by the charge, $W = QV$

The energy acquired is $2$ J

39. Which has more resistance: $100$ W bulb or $60$ W bulb?

Ans: As, it is clearly known that$R \propto \dfrac{1}{P}$, thus the resistance of $60$ W bulb is more.

40. What happens to the current in a circuit if its resistance is doubled?

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

41. What happens to the resistance of a circuit if the current through it is doubled?

Ans: Resistance is unaffected since the circuit's resistance is independent of the current flowing through it.

42. How does the resistance of a wire depend upon its radius?

Ans: As $R \propto \dfrac{1}{A}$

$\Rightarrow R \propto V$

Resistance of a wire is directly proportional to its radius.

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

Ans: Since  $R{\text{ }} = {\text{ }}p1/A$ ,

but A and I are the same. It is solely determined by resistivity, hence iron has a higher resistance.

44. Two wires of same material and same length have radii $R$ and $r$ Compare their resistances.

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

1. Two metallic wires A and B are connected, wire A has length I and radius $r$ , while  wire B has  length  $2l$  and  radius $2r$ .  Find  the  ratio  of  total  resistance  of  series combination and the resistance of wire $A$ , if both the wires are of the same material?

Ans: Observe as shown below,

Resistance of metallic wire $A$ , ${R_1} = \dfrac{{\rho l}}{A}$

Resistance of metallic wire $B$ , ${R_2} = \dfrac{{\rho 2l}}{{4\pi {r^2}}}$

Total resistance in series is $R = {R_1} + {R_2}$

$= \dfrac{{\rho l}}{{\pi {r^2}}} + \dfrac{{2\rho l}}{{4\pi {r^2}}}$

$= \dfrac{{3\rho l}}{{2\pi {r^2}}}$

The ratio of the total resistance is series to the resistance of $A$ is

$\dfrac{R}{{{R_1}}} = \dfrac{{\dfrac{{\rho l}}{{\pi {r^2}}}}}{{\dfrac{{3\rho l}}{{2\pi {r^2}}}}}$

The ratio of the total resistance is series to the resistance of $A$ is $2/3$ .

2. Should  the heating  element  of  an  electric  iron be made  of  iron,  silver  or nichrome wire? Justify giving three reasons?

Ans: The following reasons can be found, why the heating element of an electric iron is composed of nichrome wire.

(1) Due to the high resistance, the passage of current generates additional heat.

(2) High melting point.

(3)  At high temperatures, it does not easily oxidised (or burn).

3. (a) Define electric resistance of a conductor?

Ans: A conductor's electric resistance is defined as the resistance it provides to the flow of current.

That is $R = V/I$ and its S.I. unit is ohm , $\Omega$ .

(b) A wire of  length $L$ and resistance $R$  is stretched so that  its  length  is doubled and the area of the cross section is halved. How will its

(i) resistance change

Ans: It is clearly known that resistance,  $R = \dfrac{{\rho l}}{A}$

New length $L' = 2L$ and

$A' = \dfrac{A}{2}$

Therefore ${R^1} = \dfrac{{\rho L'}}{{A'}}$

Therefore, the resistance of a wire becomes 4 times its original resistance.

(ii) resistivity change?

Ans: The size of a wire has no bearing on its resistance. As a result, resistance does not vary.

4. Two  resistors  of  resistance  $R$  and  $2R$  are  connected  in parallel  in  an  electric  circuit. Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: Power consumed by $R$ , ${\rho _1} = \dfrac{{{V^2}}}{R}$

Power consumed by$2R$ , ${\rho _2} = \dfrac{{{V^2}}}{{2R}}$

Ratio $\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{{\dfrac{{{V^2}}}{R}}}{{\dfrac{{{V^2}}}{{2R}}}}$

The ratio of the electric power consumed by $R$ and $2R$ is $2:1$ .

5. The length of different metallic wires but of same area of cross section and made of the same material are given below

(i) Out of these two wires which wire has higher resistance.

Ans: As $R \propto l$ (length of the conductor) and since length of wire $C$ is more than $A$ and $B$ ,  wire $C$ has higher resistance.

Ans: The electrical resistivity of a wire is determined by the nature of the material, not by its dimensions. As a result, the resistivity of all wires is the same as the substance of all wires.

6. Two resistors of resistances $R$ and $2R$ are connected in series with an electrical circuit? Calculate the ratio of the electric power consumed by $R$ and $2R$ ?

Ans: It is clearly known thatElectric power consumed by $R$ , ${P_1} = {I^2}R$

Also, electric power consumed by $2R$ , ${P_2} = {I^2}2R$

$\dfrac{{{P_1}}}{{{P_2}}} = 1/2$

The ratio of the electric power consumed by $R$ and $2R$ is$1:2$.

7. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. the ratio of heat produced in series and parallel combinations would be

Ans: (c)Let resistance of each wire is $R$

In series, resistance is  $= 2R$

Heat produced, ${H_1} = \dfrac{{{V^2}}}{{2R}}t$

In parallel total resistance $= R/2$

Heat produced, ${H_2} = \left( {\dfrac{{{V^2}}}{{\dfrac{R}{2}}}} \right)t$

$= \dfrac{{2{V^2}}}{R}t$

${H_2} = 4{H_1}$

$\dfrac{{{H_1}}}{{{H_2}}} = 1/4$

The ratio of heat produced in series and parallel combinations is $1:4$ .

8. Calculate the following of a circuit shown in the figure.

(i) effective resistance

Ans: Effective resistance, $R = {R_1} + {R_2}$

$= 5 + 10$

The effective resistance is $15\Omega$ .

(ii) current

Ans: Current, $I = V/R$

The current is $0.133$ A.

(iii) Potential difference across $10\Omega$ resistor

Ans: Potential difference across $10\Omega$

$= \dfrac{2}{{15}} \times 10$

The potential difference across $10\Omega$ is $1.33$ volt.

9. A Piece of wire of resistance $20\Omega$ is drawn out so that its length is increased to twice its original length to calculate the resistance of the wire is the new situation?

Ans: Resistance of wire$= 20\Omega$

As $R = \dfrac{{\rho l}}{A}$

And as length of a wire is increased, its area of cross- section decreases, and volume of the wire remains constant.

$R' = \dfrac{{\rho l'}}{{A'}}$

$= \dfrac{{4\rho l}}{{A'}}$

$\dfrac{{R'}}{R} = \dfrac{4}{1}$

$\Rightarrow R' = 80$

The resistance of the wire is the new situation is $80\Omega$ .

10. A  battery  made  of  $5$  cells  each  of  $2$ V  and  have  internal  resistance $0.1\Omega ,0.2\Omega ,$$0.3\Omega ,0.4\Omega$ and $0.5\Omega$  is  connected  across $10\Omega$ resistance.  Draw a circuit diagram and calculate the current flowing through$10\Omega$resistance?

Ans: The internal resistance , $0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5\Omega$

Total resistance $= 1.5 + 10$

$= 11.5$

$I = V/R$

$= 10/11.5$

$I = {\text{ }}0.869$ A

Current flowing through $10\Omega$ resistance is $0.869$ A.

11. In the circuit diagram given here Calculate-

(a) The total effective resistance

Ans: As resistances are in parallel

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$

$= \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{10}}$

$= \dfrac{8}{{10}}$

$R = \dfrac{{10}}{8}$

Total effective resistance is $\dfrac{{10}}{8}\Omega$

(b) The total current

Ans: Total current, $I = V/R$

$= \dfrac{6}{{10/8}}$

The total current is $4.8$ A.

(c) The current through each resistor.

Ans: From the circuit diagram, ${I_1},{I_2}$ and ${I_3}$ be the current through $2\Omega ,5\Omega$ and $10\Omega$  respectively

Therefore, ${I_1} = \dfrac{V}{{{R_1}}}$

${I_2} = \dfrac{V}{{{R_2}}}$

${I_3} = \dfrac{V}{{{R_3}}}$

Current through each$2\Omega ,5\Omega$ and $10\Omega$  is $3$ A ,$1.2$ A, and $0.6$ A respectively.

12. You have two circuits Compare the power used in $2\Omega$ resistor in each case.

(i) a $6V$ battery is series with $1\Omega$ and $2\Omega$ resistor

Ans: Potential difference, $V = 6V$

${R_1} = 1\Omega$

${R_2} = 2\Omega$

Total Resistor $= {R_1} + {R_2}$

$I = \dfrac{V}{R}$

$= \dfrac{6}{3}$

${P_1} = \dfrac{{{V^2}}}{R}$

The power used in $2\Omega$resistor  is $8$ W.

(ii) a $4V$ battery in parallel with  $12\Omega$ and $2\Omega$ resistor

Ans: Potential difference, $V = 4V$

${R_1} = 12\Omega$

${P_2} = \dfrac{{{V^2}}}{R}$

The ratio of both power is $1:1$ .

13. How  much energy  is  given  to  each coulomb  of charge  passing  through  a  $6$  volt battery?

Charge, $Q = 1C$

Energy = total work done

$= {\text{ }}Q{\text{ }}x{\text{ }}V$

$= {\text{ }}1x6$

$= {\text{ }}6$ joule.

Energy  given to each coulomb of charge passing through a $6$ volt battery is $6$ joules.

14. On what factor does the resistance of a conductor depend?

Ans: A conductor's resistance is determined by the following factors:

(i) length of conductor

(ii) Area of cross-section

(iii) Temperature

(iv) Conductors are made from a variety of materials.

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

Ans: When linked to the same source, current flows more freely through a thick wire than via a small wire of the same material. It's because resistance rises as thickness decreases.

16. Let  the  resistance  of  an  electric  component  remain  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?

Ans: T he electrical component's resistance R remains unchanged, but the potential difference across its ends falls to half of its original value. As a result of Ohm's law, new current is reduced to half of its initial value.

17. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans: The following reasons are why the coils of electric toasters and electric irons are built of an alloy rather than a pure metal:

(i) An alloy's resistivity is higher than that of pure metal.

(ii) An alloy does not rust quickly at high temperatures.

18. Draw a schematic diagram of a  circuit  consisting of a battery of  three  cells of $2V$ , each, a $5\Omega$ resistor, $8\Omega$ resistors and a $12\Omega$ and a plug key, all connected in series.

Ans: The diagram of circuit is as follows

19. An electric lamp of $100\Omega$ , a toaster of resistance $50\Omega$ and a water filter of resistance $500\Omega$ are connected in parallel to a $220V$ 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?

Ans: It can be found from the question, voltage, $V = 220V$

${R_1} = 100\Omega$

${R_2} = 50\Omega$

${R_3} = 500\Omega$

$1/R = 1/100 + 1/50 + 1/500$

$R{\text{ }} = {\text{ }}500/16$

$= {\text{ }}31.25\Omega$

The resistance of electric iron, which draws as much current as all three appliances take together is $31.25\Omega$ .

Current passing through electric iron,  $I = V/R$

$= 220/31.25$

$= 7.04$ A.

That is the current passing is $7.04$ A.

20. What  is (a) the  highest total  resistance  that  can  be  secured  by a combination of four resistance of $4\Omega ,8\Omega ,12\Omega$ and $24\Omega$ ?

Ans: When all four resistances must be connected in series, the highest resistance is achieved. In that instance, the outcome

$R = {R_1} + {R_2} + {R_3} + {R_4}$

$= {\text{ }}4 + 8 + 12 + 24$

$= {\text{ }}48\Omega$

The highest resistance is $48\Omega$ .

(b) The  lowest  total  resistance  that  can  be  secured  by a combination of four resistance of $4\Omega ,8\Omega ,12\Omega$ and $24\Omega$ ?

Ans: All four resistances must be connected in parallel to produce the lowest resistance.

$1/R = 1/{R_1} + 1/{R_2} + 1/{R_3} + 1/{R_4}$

$= {\text{ }}1/4{\text{ }} + 1/8{\text{ }} + 1/12{\text{ }} + {\text{ }}1/24$

$= {\text{ }}12/24$

The lowest resistance is $2\Omega$ .

21. Why does the cord of an electric heater not glow while the heating element does?

Ans: When connected to the voltage source, the cord of a heater and the cord of an electric heater are connected in series and carry the same current.

Because the resistance of the cord is so low in comparison to the resistance of the heater element.

As a result, the amount of heat created in the cord is extremely low, but significantly higher in the heater element. As a result, the heating element glows, but the cord does not.

22. A copper wire has diameter $0.5$ mm and resistivity of $1.6 \times {10^{ - 8}}$ m. What will be the length of  this wire  to make  its resistance $10$ ? How much does  the resistance change  if the diameter is doubled?

Ans: The diameter of wire, d = 0.5 mm,

Resistivity, $\rho = 1.6 \times {10^{ - 8}}$

resistance  $R = {\text{ }}10{\text{ }}\Omega$

$R = \rho L/A$

$L = \dfrac{{\pi {D^2}R}}{{4\rho }}$

$= \dfrac{{22 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{7 \times 4 \times 1.6 \times {{10}^{ - 8}}}}$

$= {\text{ }}122.5$ m

As resistance is inversely proportional to the cross-section area of wire, when the diameter is doubled for a given length of material, the resistance reduces.

23. A battery of $9V$ is connected in series with resistance of $0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega$ and $12\Omega$respectively. How much current would flow through the $12$ resistor?

Ans: Potential difference $V = 9V$ .

Total resistance   $= {\text{ }}0.2{\text{ }} + 0.3{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}12{\text{ }}$

$= {\text{ }}13.4{\text{ }}\Omega$

Current in the circuit $I{\text{ }} = {\text{ }}V/R{\text{ }}$

$= {\text{ }}9{\text{ }}V{\text{ }}/{\text{ }}13.{\text{ }}4{\text{ }}$ A.

$= {\text{ }}0.67$

In a series circuit the same current  flows  through all  the resistance, hence current of $0.67$ A will flow through $12{\text{ }}\Omega$ resistor.

24. How many $176\Omega$ resistors (in parallel) are required to carry $5$ A on a $220$ V line?

Ans: Let the resistors of $176\Omega$ be joined in parallel.

Their combined resistance,

$1/R = 1/176 + 1/176 \ldots \ldots$ times

$= n/176\;$ or

$\Rightarrow R = 176/n\Omega$

Given that $V = 220V$ and

$I = 5$ A

$R = V/I$

$= 176/n$

$= 220/5$

$= 44\Omega \;\;\;\;\;\;$

$n{\text{ }} = {\text{ }}176/44{\text{ }}$

$= {\text{ }}4$ ,

A number of $4$ resistors should be joined in parallel.

25. Show  how  you  would  connect  three  resistors,  each  of  resistance  $6\Omega$  so  that  the combination has resistance of

(i) $9\Omega$

Ans: From the question, ${R_1} = {R_2}$

Join three resistors as below to get net resistance of $9\Omega$ :

(ii) $4\Omega$

Ans: Join three resistors as below to get $4\Omega$ net resistance :

26. A hot plate of an electric oven connected to a $220V$ line has two resistance coils $A$ and $B$ . Each of $24\Omega$ resistances, which may be used separately, in series or in parallel. What are the currents in the three cases?

Ans: From the question, potential difference$V = {\text{ }}220{\text{ }}V$ .

Resistance of coil $A =$ Resistance of coil $B$

$= 24{\text{ }}\Omega$

When coil is used separately, the circuit $I = V/R$

$= 220V/24\Omega$

$= {\text{ }}9.2{\text{ }}A$

The current is$9.2{\text{ }}A$.

When coils are used in series total resistance $R = {R_1} + {R_2}$

$= 24 + 24$

$= 48\Omega$

The current flowing, $I = V/R$

$= 220V/48\Omega$

$= 4.6$ A

The current is $4.6$ A.

Two coils are joined in parallel.

Total resistance $R = {\text{ }}1/24{\text{ }} + {\text{ }}1/24$

$= {\text{ }}2/24$

$R = 12{\text{ }}\Omega$ .

Current $I = V/R$

$= 220V/12\Omega$

$= 18.3$ A.

The current is $18.3$ A.

27. Compare the power used in the $2\Omega$ resistor in each of the following circuits:

(i) a $6$ volt battery in series with $1\Omega$ and $2\Omega$ resistors and,

Ans: Suppose a $2\Omega$ resistor is joined to a $6$ V battery in series with $1\Omega$ and $2\Omega$ resistors.

Total resistance $R = 2 + 1 + 2$

$= 5\Omega$

Current $I = 6V/5\Omega$

$= 1.2$ A

Power used in $2A$ resistor $= {I^2}R$

$= 2.88$ W

Power used is $2.88$ W.

(ii) a $4V$ battery in parallel with $12\Omega$ and $2\Omega$ resistors.

Ans: Suppose $2\Omega$ resistor is joined to a $4V$ battery in parallel with $12\Omega$ resistor and $2\Omega$ resistors,

the current flowing in $2\Omega = 4V/2\Omega$

Power used in $2\Omega$ resistor $= {I^2}R$

$= 8$ W

Ratio $= 2.88/8$

$= 0.36:1\;$

The ratio of power used is $0.36:1$ .

28. In  the given  figure what  is  ratio of ammeter  reading when $J$ is connected  to $A$ and then to $B$

Ans: Connect  $J$  to $A$  ,     then

$= 0.6A$

When $J$ is connected to $B$$V = 1 + 2 + 3 + 4$

Ratio of ammeter reading when J is connected to $A$ and then to $B$ is $3:10$ .

29. Given  a  resistor  each  of  resistors  $R$ .  How  will  you  combine  them  to  get  the

(i) maximum effective resistance?

Ans: For maximum resistance $R = nr$  , this is the same as combining a series of numbers.

(ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

Ans: For minimum resistance $R' = r/n$  , this is the same as combining a series of numbers.The ratio of the maximum to minimum resistance is $R/R' = {n^2}$.

30. A wire of  length $L$ and resistance $R$  is stretched so that  its  length  is doubled. How will its

(a) Resistance change

Ans: The resistance of a wire is determined by its length, cross-sectional area, and resistivity as${\text{R = }}\dfrac{{{{\rho l}}}}{{\text{A}}}$

Hence, if the length is doubled and area is halved, then we have

$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_2}}}{{{A_2}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}}$

$= \dfrac{{{l_2}{A_1}}}{{{l_1}{A_2}}}$

Therefore, ${R_2} = 4{R_1}$

Hence, resistance of the wire becomes four times the original value.

(b) Resistively change?

Ans: The substance from which wire is formed has a property called wire resistivity. As a result, changing the wire's size has no effect on its resistivity.

31. Two students perform the experiments on series and parallel combinations of two given resistors ${R_1}$  and ${R_2}$ and plot the following $V - I$ graphs.

Ans: Both students are  correct  because

$AV/A1 = \;$ resistance $R$  and

$A1/AV = l/R\;$

The term "series" refers to high resistance, while "parallel" refers to low resistance.

32. A household uses the following electric appliances. Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. $3.00$

(i) Refrigerator of rating $4$ for ten hours each day.

Ans: Month of June has $30$ days.

Refrigerator of $400$ W is running $2$ hours each day.

Total hours it is run in $30$ days $= 2 \times 30$

$= 60$ h

Energy consumed in kWh is $= 400{\text{ }} \times {\text{ }}60/1000{\text{ }}$

$= {\text{ }}24$ kWh

(ii) Two electric fans of rating $80$each for twelve hours each day.

Ans: Two electric fans of $80$ W are run $12$ hours each day.

Total hours they are run in $30$ days $= 12 \times 30$

Energy consumed in kWh is $= 2 \times 80 \times 360/1000$

$= 57.6$ kWh

(iii) Six electric tubes of rating $18$ W each for 6 hours each day.

Ans: Six electric tubes each of $18$ W are run $6$ hours daily.

Total hours it is run in $30$ days $= 6 \times 30$

$= 180$ h

Energy consumed in kWh is $= 6 \times 18 \times 180/1000$

$= 19.44$  kWh

Net energy consumed in the month of June is $= 24 + 57.6 + 19.44$

$= 101.04$ kWh

Thus, the electric bill is $= 3 \times 101.04$

$= Rs303.12$

1. Two wires  $A$  and  $B$  are  of  equal  length,  different  cross  sectional  areas  and made  of the same metal.

(a) (i) Name the property which is same for both the wires,

Ans: Resistivity - As resistivity is a property of a substance, it is constant for both wires.

(ii) Name the property which is different for both the wires.

Ans: Resistances - As the cross sectional areas of each wires are different, they are treated as separate objects.

(b) If the resistance of wire $A$ is four times the resistance of wire $B$ , calculate

(i) the ratio of the cross sectional areas of the wires and

Ans: Since $R = \dfrac{{\rho l}}{A}$

For wire $A$ , ${R_1} = \dfrac{{\rho l}}{{{A_1}}}$

For wire $B$ , ${R_2} = \dfrac{{\rho l}}{{{A_2}}}$

$\Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{A_1}}}{{{A_2}}}$

Since ${R_1} = 4{R_2}$

$\Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = 1:4$

$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi {r_1}^2}}{{\pi {r_2}^2}}$

$= {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}$

Ratio is ${\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}$ .

(ii) The ratio of the radii of the wire.

Ans: ${\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} = 1/4$

Ratio is $1:2$ .

2. (a) State ohm’s law?

Ans: If the temperature and other physical conditions of the conductor stay constant, the electric current flowing through the conductor is precisely proportional to the potential difference across the conductor's end.

(b) The value of $(I)$ current flowing through a conductor for the corresponding values of $(V)$ potential difference are given below

Plot a graph between $V$ and $I$ and also calculate resistance.

Ans: Along $x$ -axis $IV = 1$ cm

The resistance  is $1.67\Omega$ .

3. (a) Define electrical energy with S.I. unit?

Ans: The effort done by a source of electricity to sustain current in a circuit is known as electrical energy. The joule is its SI unit.

(b) A household uses the following electric appliance; Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. $3.00$ .

(i) Refrigerator of rating $400$ w for ten hour each day.

Ans: Electricity consumed by refrigerator in one day $=$ power $\times$ time

$= 400 \times \;10$

$= {\text{ }}4000$ Wh

$= 4$ kwh

Therefore the electricity consumed is $4$  KWh.

(ii) Two electric fans of rating $80$ w each for twelve hours each day.

Ans: Electricity consumed by $2$ electric fans in $1$ day, power $\times$ time

$= 2 \times 80 \times 12$

$= {\text{ }}1.92$ kwh

Therefore the electricity consumed is $1.92$ KWh.

(iii) Six electric tubes of rating $18$ w each for 6hours each day.

Ans : Electricity consumed by $6$ electric tubes in $1$ day $= 6 \times 18 \times \;6$

$= {\text{ }}0.648$ kwh

Therefore the electricity consumed is $0.648$ KWh.

Total energy consumed in one day $= 4 + 1.92 + 0.648$

$= 6.548$ kwh

Total energy consumed in one month $= 6.568 \times 30$

$= 197.04$ kwh

Cost of $1$ unit (kwh) $=$ Rs $3.00$

Cost of $197.04$ kwh $= 197.04 \times 3$

Electricity bill $= Rs591.12$

The electricity bill of the household for the month of June is Rs. $591.12$.

4. 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 $12\Omega$ resistors. What would be the reading in the ammeter and voltmeter?

Ammeter $A$ has been joined in series with circuit and voltmeter $V$ is joined in parallel to $12$ ohms resistor.

Total voltage of battery $V = 3x2$

$= 6$ V.

Total resistance $R = {R_1} + {R_2} + {R_3}$

$= 5\Omega + 8\Omega + 12\Omega$

$= 25\Omega$

Ammeter reading $= I$

$= 6/25$

$= 0.24$ A.

Voltmeter reading $= {\text{ }}IR$

$= 0.24{\text{ }}x{\text{ }}12$

$= 2.88$ V.

The reading in the ammeter and voltmeter is $0.24$ A and $2.88$ V respectively.

5. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Ans: The following are the benefits of connecting electrical equipment in parallel with the battery rather than in series:

(i) Each connecting electrical device will have the same voltage, and the device will take current according to its resistance.

(ii) It is possible to use separate on/off switches.

(iii) As the total resistance in the parallel circuit falls, a large current can be pulled from the cell.

(iv) Even if one electrical gadget is broken, other devices continue to function normally.

6. How  can  three  resistors  of  resistance  $2\Omega ,3\Omega$and  $6\Omega$  be  connected  to  give  a  total resistance of

(a) $4\Omega$

Ans: If we connect resistance of $3\Omega$ and $6\Omega$ in parallel and resistance of $2\Omega$ is connected in series with the combination, then total resistance of combination is $4\Omega$ .

(b) $9\Omega$

Ans: If all the three resistance are joined in parallel the resulting resistance will be $3\Omega$ .

7. The  value  of  current  $I$  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.

Ans: From the given data the $I - V$ graph is a straight line as shown below:

Resistance of resistor $R{\text{ }} = {\text{ }}{V_A} - {V_B}/{1_A} - {1_B}$

$= {\text{ }}12{\text{ }}V{\text{ }}--{\text{ }}6{\text{ }}V/{\text{ }}3.6{\text{ }}A{\text{ }}--{\text{ }}1.8{\text{ }}A$

$= {\text{ }}6V/{\text{ }}1.8{\text{ }}A{\text{ }}$

$= {\text{ }}3.3{\text{ }}\Omega$

8. Explain the following:

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

Ans: For the filament of electric lamps, we need a robust metal with a high melting point. Because of its high melting point, tungsten is utilised only for electric lamp filament.

(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?

Ans: Electric heating device conductors are composed of an alloy because it has a higher resistance than pure metal and a higher melting point, which prevents it from oxidising at high temperatures.

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

Ans: As the current to all appliances remains constant despite varying resistance, each appliance cannot be turned on or off independently.

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

Ans: As resistance of a wire is inversely proportional to its cross-section area, the resistance will decrease when the area of cross section increases.

(e) Why are copper and aluminum wires usually employed for electric transmission?

Ans: As copper and aluminium wires are good conductors with low resistance, they are commonly utilised for electrical transmission. They can also be drawn into thin wires since they are ductil

## Important Questions of Chapter 12 Class 10 Science - Free PDF Download

Students who are weak in Science and do not have a strong core knowledge might find the chapter electricity quite confusing. This chapter is full of theories and diagrams which confuse students and act as a barrier in their way of achieving good marks. Students must plan to avoid this situation so that they can score the highest possible marks in the final exams. The best way to overcome this problem is a continuous practice. Students can solve some of Class 10 Science Chapter 12 Important Questions regularly. They must induct these practice hours in their preparation schedule. So that every day some time is given to revise and practice. This will make the students efficient and thorough.

Students who are unable to solve the questions must take the help of Important Questions of Electricity Class 10 with Solutions which is available on Vedantu for free. Students can download these questions in PDF format. The downloaded CBSE Class 10 Science Chapter 12 Important Questions act as a guide in preparation for the final exams.

## Class 10 Science Ch 12 Important Questions

Students will learn about the concepts and theories of electricity when they will study Chapter 12 of Class 10. They will gain a detailed knowledge of this chapter after practising Important Questions for Class 10 Science Chapter 12. Some of the knowledge that the students will learn are as follows:

## Electricity

Electricity is considered as a set of physical phenomena which are associated with the presence and motion of electric charge. It is also believed that electricity is somehow related to magnetism which is why both are part of a phenomenon which is called the phenomenon of electromagnetism. This was described in Maxwell's equations. There are many other phenomena which are related to electricity such as lightning, static electricity, electric heating, electric discharges and many others.

An electric field gets produced, when there is some kind of presence of positive or negative charge. The positive and negative charges are considered as electric charges. When the electric charge moves it is known as Electric current, and it produces a magnetic field nearby.

## Attributes of Electricity

There are two primary attributes of electricity: voltage and current. They both have different properties and are quite different from each other, but only in electronic circuits, they are interrelated. Absence of any of these attributes can harm the circuit operation and ruin it. Let's discuss both these attributes:

Voltage is considered as a  force which is responsible for making the current flow in a circuit. Voltage gets measured in volts. In order to get a better understanding of voltage, try to think of a water faucet, in which voltage can get compared to the pressure at which the water is coming out of the faucet. A slow-flowing stream from a faucet resembles a low voltage circuit. In contrast, a high flowing stream resembles a high voltage circuit. A voltage is a mandatory requirement for a current to flow.

The movement of electrical charges is termed as current. When electrons flow through an electronic circuit, it generates a current. Current is measured in terms of amperes (amp). Here also we can take the example of a water faucet to understand the concept of current. It is assumed that the current is at a high level when more water flows in an hour through the faucet, whereas when the flowing water level is low, the current is also low. After the rain, you must have seen that river flows faster than the usual speed because the current remains high at that point of time as more amount of water passes during rains.

## Conductors and Insulators

Another thing that students are going to learn while practising the Electricity Class 10 Important Questions is about the significant classification of elements which is done on the basis of their conductivity of electric charge, i.e. conductors and insulators.

## Conductors:

In simple terms, conductors is the name given to those materials which allow electricity to flow through them easily without any difficulties. The conductors' property that makes them capable of allowing electricity to flow through them is termed as conductivity. When electrons start flowing in the conductors, they start to produce an electric current.  The force which is required for the current to move through the conductor is known as the voltage. Some examples of conductors are copper, gold and iron. Let's discuss some properties of a Conductor.

Materials which are considered as conductors have a minimal resistance because electricity flows through them.

The inductance of the electric conductor occurs when there is a high voltage drop in the conductor.

Inside a perfect conductor, the electric field is considered to be zero. It is zero because it helps to keep the electrons calm so that they don’t accelerate.

There is no electric charge inside the material which is considered as a conductor.

Insulators are considered to be materials within which free flow of electrons from one particle of the element to the other particle is interrupted. If a certain amount of charge is transferred to such an element at a given point of time, then the charge does not get distributed in the surface and remains at the same position. The most common process to charge these elements is by rubbing or charging it through induction. Some examples of insulators are wood, plastic and glass. Let's discuss some properties of insulators:

There are no free electrons in such material because all the electrons are tightly held with each other.

The ability of these materials to stop the electric current from passing through them is known as resistance.

Dielectric length of insulators is vast. Dielectric strength is considered as the maximum electric field that an insulator can handle without suffering an electrical breakdown.

High air permeability is a feature of good insulators as they allow air to pass through their pores.

## Important Questions for Electricity Class 10

To give the students an overview of the Important Questions of Chapter 12 Class 10 Science , we are listing here some of the questions which are most likely to come in the exams:

## What is Electricity?

What are the attributes of electricity?

Write the SI unit of resistivity.

What do you mean by electric current? Mention and define the SI unit of electric current.

What do you understand by electrical resistivity of a material? Describe an experiment which will show the factors on which the resistance of a conducting wire depends.

What do you understand by a conductor? State its properties.

What do you understand by an insulator? State its properties.

Differentiate between indicator and conductor.

State the difference between electric energy and power.

Mention the commercial unit of electric energy.

Convert the unit of electric energy into joules.

## Benefits of Important Questions for Class 10 Science Chapter 12

Class 10 Science Chapter 12 Important Questions offer significant advantages for board exam preparation.

Vedantu's team conducts thorough research to compile a list with a high probability of exam inclusion.

Expert reviews from seasoned professionals with extensive experience ensure accuracy.

Questions adhere to the CBSE board format, providing students insight into exam patterns.

Detailed and explanatory solutions accompany these important questions for comprehensive understanding.

## Important Related Links for CBSE Class 10 Science

These important questions serve as a valuable reference, elucidating key concepts and aiding comprehensive exam preparation. Whether unravelling Ohm's Law or comprehending circuit configurations, these important questions offer understanding. For students grappling with scientific complexities, relying on "Electricity Class 10 Important Questions" proves beneficial, promising improved comprehension and heightened confidence. This resource is not just a tool for academic success but a key to unlocking one's potential in understanding and excelling in the captivating subject of electricity.

## FAQs on Important Questions for CBSE Class 10 Science Chapter 12 - Electricity 2024-25

1. How will Vedantu’s important questions for Class 10 Science Chapter 12 will help you to score well?

Ans: Vedantu’s important questions for Class 10 Science Chapter 12 helps the students by providing an excellent exam strategy to prepare for the board examinations. Science is an essential subject for students who are aspiring for various competitive exams like JEE, NEET, etc in future. Every student aspiring to excel in the board examinations requires deep knowledge and thorough understanding of this subject. Therefore, the important questions for CBSE Class 10 Science help the students in preparing for their exams. All the questions present in these materials are framed under the latest CBSE curriculum and guidelines.

Ans:  Yes, the download option is available on Vedantu website and mobile application for Class 10 Science Chapter 12. Any student can download these study materials from Vedantu website in PDF format at absolutely free of cost. These are high-quality study materials created by our in-house subject matter experts as per the latest CBSE curriculum and guidelines.

3. Define electricity.

Ans: Electricity is one of the most important aspects of our society. Electricity is a set of physical phenomena that has been shaping up our civilization ever since the wake of the industrial revolution all over the entire industries and businesses. Today, life without electricity would be unimaginable and would lead to total chaos if we somehow lose this important source of energy.

4. What are the topics and subtopics covered under this chapter of Class 10 Science?

Resistivity and Resistance

Factors that affect the Resistance of a Conductor

Parallel and Series Combination of Resistors and their applications

Heating Effect of Electric Current and its Applications

Electric Power

The interrelation between P, V, I and R

5. Define the following terms-

Potential difference

Ans: One volt- One volt is defined as the potential difference between two points in a conductor that carries current. Here, one joule of work is done to locomote a charge from one place to the other. It is the SI unit of potential difference.

Potential Difference- Potential difference is defined as the work done to move a charge from one point to another in a current-carrying conductor. The formula of potential difference is 1 volt- 1 joule/ 1coulomb. A voltmeter is used to measure the potential difference. A voltmeter is connected parallelly in a circuit.

6. Define resistance.

Ans: Resistance is defined as the internal property of a substance that offers obstruction to the current flowing through it. The electric resistance of a substance is inversely proportional to its area of cross-section (A) and directly proportional to its length (l). This means when the area will increase the resistance will decrease and vice-versa. And if the length increases, the resistance will also increase.

The formula is: R=⍴*L/A

Here R is the resistance, L is the length, A is the Area and ⍴ is a constant. The SI unit of resistance is the ohm.

7. Define resistivity.

Ans: The resistivity can be defined as the resistance offered by a current-carrying conductor that has an area of cross-section equal to one centimetre square with a length equal to one centimetre. This means it is the resistance of a one-centimetre cube of the current-carrying conductor. The formula for resistivity is ⍴=R*L/A

Here R is the resistance, L is the length and A is the area. The SI unit of resistivity is ohm*metre. To study more about resistivity, students can download the Important Questions free of cost from the Vedantu website or mobile app.

8. Define ohm’s law.

Ans: Ohm’s law states that the potential difference (V) across a current-carrying conductor is directly proportional to the current (I) flowing through it. Mathematically, V ∝ I or V= IR.

Here, V is the potential difference, I is the current and R is a constant called resistance of the conductor. Ohm’s law is followed by both electrolytic conductors as well as metallic objects.

9. Define Joule’s law of heating.

Ans: Joule’s law of heating states that the heat generated in a current-carrying resistor is directly proportional to the square of the current, resisting capability of the resistor and time till which the current flows through the resistor. The formula of Joule’s Law of Heating is given as

H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule.

## CBSE Class 10 Science Important Questions

Cbse study materials.

• CBSE- Electricity
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## Electricity-Sample Questions

• STUDY MATERIAL FOR CBSE CLASS 10 PHYSICS
• Chapter 1 - Electricity
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## Most Important Questions for Class 10 Science Electricity with Solutions

Science Class 10 Electricity Important Questions contain various questions that are useful for students even in their competitive examinations like JEE & NEET. In this chapter, you learn about the various properties and uses of Electricity. You also become aware of the numerous important precautions you need to keep in mind while dealing with electricity and electrical appliances.

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## CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question PDF

CBSEcan never be understated Class 10 Chapter 12 Electricity Science Marks Wise Question with Answers is an extremely useful revision tool. It is very beneficial for students who are aiming to make the most out of their exam preparation. Board exams examine the educational skills of the student. Practising CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question is a must. You can download here Chapter 12 Electricity Science Marks Wise Question with solutions at free of cost.

## CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question

CBSE is one of the most preferred educational board in India. This renowned board was established in the year 1952. CBSE Board is known for its comprehensive syllabus and the quality of education and well-structured question paper. The board aims to give a good education to its students so that they can grow both mentally and physically.

CBSE Board conducts the exam for Class 10. The exams for Class 10 are usually conducted in the month of March every year and the results are announced in the month of May. The questions that are asked in the CBSE board examination are according to the pattern issued by the NCERT. Students who are preparing for their CBSE examination must know the benefits of solving the CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question. The questions are designed by SelfStudys experts according to the CCE guidelines.

## CBSE Chapter 12 Electricity Science Marks Wise Question for Class 10

One of the best way to prepare for the exam is by practising the Chapter 12 Electricity Science Marks Wise Question. Students will get an idea about the examination pattern and the marking scheme. It will also make students familiar with the question paper and the difficulty level of the questions asked in the examination. Some benefits of solving the CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question are given below.

• Helps in brushing up the concepts completely
• Helps to overcome fear before the exam
• Improves time management skills
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• Helps in correcting mistakes

If you are preparing for CBSE board exams, now is the time when you plan efficiently and work more smartly. One way of studying smarter is investing your efforts in the most important questions and make sure you are through with these questions. You must finish the chapters that have high weightage with maximum efficiency so that you can solve any question asked from them. In this page, we will provide you with the CBSE Class 10 marks wise questions.

Ensure that you finish the complete CBSE syllabus of Class 10 based on the CBSE Class 10 marks wise questions of each unit. The chapter with high weightage must be thoroughly completed. At the same time, you must have a basic understanding of all the chapters. Solve marks wise questions of all types.

Class 10 is considered to be the most important part for students aspiring to clear the NEET exam. In CBSE Class 10, several important chapters are presented to the students which are crucial to forming the basic skills required for a medical & engineer career.

CBSE Class 10 syllabus has various important topics, diagrams and definitions that the students require to be thorough with to be able to score well in the Class 10 board exam. To get good marks in Class 10, students require to prepare thoroughly and practice with CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question.

It is not easy to get high marks in CBSE, either. The curriculum of Class 10 has evolved over the years and become more complex. Students will require to put in the effort to remember the topic and work smartly to perform well in their board examinations. This means that it might not be sufficient to just attend the school and blindly mug up the CBSE syllabus. A lot of students go for tuition to be able to clear any doubts they have.

## CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question with Solutions

To help the students to prepare for examination more effectively, we provide CBSE marks wise questions for Class 10 are given here. It is suggested to check these marks wise questions to be able to tackle any question in the examinations.

It is suggested to practice this CBSE Class 10 marks wise questions thoroughly before the exams. Apart from these marks wise questions, students are also suggested to check the below-given links for sample and question papers and prepare more efficiently for the examinations.

As the importance of board examinations can never be understated, CBSE Class 10 mark wise questions Chapter 12 Electricity can prove to be greatly beneficial for students. SelfStudys ensures that students can study better with our marks wise questions with Answers. The carefully crafted questions and answers provide students with a comprehensive understanding of the chapters involved. A lot of these questions are likely to appear in the board examination, making this an ultimate guide for students before their examinations.

## CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question with Answers

Why should students download cbse class 10 chapter 12 electricity science marks wise question from selfstudys.

These CBSE Class 10 Chapter 12 Electricity marks wise questions for exams are framed by our best teachers. They also give solutions to these questions that explain the concepts in a short yet easy to understand manner.

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## CBSE MCQ for Class 10 Science Chapter 12 Electricity Free PDF

The CBSE MCQ for Class 10 Science Chapter 12 Electricity  are provided below, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 10 MCQ PDF for Free, just click ‘ Download pdf ’ button.

## CBSE MCQ for Class 10 Science Chapter 12 Electricity PDF

Checkout our mcqs for other chapters.

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## How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

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1. Case Study Questions 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.

2. CBSE 10th Science Electricity Case Study Questions With Solution 2021

CBSE 10th Standard Science Subject Electricity Case Study Questions With Solution 2021. 10th Standard CBSE. Reg.No. : Science. Time : 00:30:00 Hrs. Total Marks : 16. The rate of flow of charge is called electric current. The SI unit of electric current is Ampere (A). The direction of flow of current is always opposite to the direction of flow ...

3. Case Study Questions Class 10 Science

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks. Case study: 1. We can see that, as the applied voltage is increased the current through the wire also increases.

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 ...

5. CBSE 10th Science Electricity Case Study Questions 2021

CBSE 10th Standard Science Subject Electricity Case Study Questions 2021. If two or more resistances are connected in such a way that the same potential difference gets applied to each of them, then they are said to be connected in parallel. The. current flowing through the two resistances in parallel is, however, not the same.

6. Case Study Questions for Class 10 Science Chapter 12 Electricity

Case Study Questions for Class 10 Science Chapter 12 Electricity. Question 1: The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule (as shown in figure). Actually, Joule represents a very small quantity of energy and ...

7. Class 10 Science Chapter 11 Case Based Questions

Document Description: Case Based Questions: Electricity for Class 10 2024 is part of Class 10 preparation. The notes and questions for Case Based Questions: Electricity have been prepared according to the Class 10 exam syllabus. Information about Case Based Questions: Electricity covers topics like Case Study - 1, Case Study - 2, Case study - 3 and Case Based Questions: Electricity Example ...

8. Electricity Case Study Based Questions Class 10

Students should follow some basic tips to solve Electricity Case Study Based Questions. These tips can help students to score good marks in CBSE Class 10 Science. Generally, the case based questions are in the form of Multiple Choice Questions (MCQs). Students should start solving the case based questions through reading the given passage.

9. Case Study Chapter 12 Electricity

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: R1 > R2.

10. 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.

11. Electricity class 10: CBSE previous question paper problems

The switch if OFF. Q2. A current of 10 A flows through a conductor for two minutes. ( i ) Calculate the amount of charge passed through any area of cross section of the conductor. ( ii ) If the charge of an electron is 1.6 × 10 − 19 C , then calculate the total number of electrons flowing.

12. Case study questions for CBSE 10th

5 Passages. 25 questions. Download case study question pdfs for CBSE Class 10th Maths, CBSE Class 10th English, CBSE Class 10th Sciece, CBSE Class 10th SST. As the CBSE 10th Term-1 Board Exams are approaching fast, you can use these worksheets for FREE for practice by students for the new case study formats for CBSE introduced this year.

13. NCERT Exemplar Class 10 Science Solutions Chapter 12

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 ...

14. Case Study Questions Class 10 Science

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.

15. Electricity Case Study Based Questions Class 10

Electricity Case Study: Here, Students can read Class 10 Electricity Case Based Questions with Answers in PDF File format. Apart from this, You can download Class 10 ...

16. CBSE Class 10 Science Chapter Wise Important Case Study Questions

The Chapter wise Important case study based questions with their solved answers in CBSE Class 10 Science can be accessed from the table below: CBSE Class 10 Science Chapter 1 Chemical Reactions ...

17. Top 15 Case Based Questions

18. Important Questions for CBSE Class 10 Science Chapter 12

H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule. Get chapter-wise important questions for CBSE Class 10 Science Chapter 12 Electricity with answers on Vedantu.

19. PDF CHAPTER12 Electricity

Activity 12.1. Set up a circuit as shown in Fig. 12.2, 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.

20. CBSE 10, Physics, CBSE- Electricity, Sample Questions

Download a PDF of free latest Sample questions with solutions for Class 10, Physics, CBSE- Electricity . All types of questions are solved for all topics. You can also get complete NCERT solutions and Sample papers. ... The voltage is 220V .What is the current and resistant in each case? 15. A piece of wire having a resistance R is cut into ...

21. CBSE Class 10 Science Electricity Important Questions

Science Class 10 Electricity Important Questions contain various questions that are useful for students even in their competitive examinations like JEE & NEET. In this chapter, you learn about the various properties and uses of Electricity. You also become aware of the numerous important precautions you need to keep in mind while dealing with ...

22. CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question PDF

The chapter with high weightage must be thoroughly completed. At the same time, you must have a basic understanding of all the chapters. Solve marks wise questions of all types. CBSE Class 10 Chapter 12 Electricity Science Marks Wise Question PDF. Class 10 is considered to be the most important part for students aspiring to clear the NEET exam.

23. CBSE MCQ for Class 10 Science Chapter 12 Electricity Free PDF

The CBSE MCQ for Class 10 Science Chapter 12 Electricity are provided below, in detailed and free to download PDF format. The solutions are latest, comprehensive, confidenceinspiring, with easy to understandexplanation. To download NCERT Class 10 MCQ PDF for Free, just click ' Download pdf ' button.