NCERT Exemplar Solution for CBSE Class 10 Science Chapter: Electricity (Part-I)

In this article you will get CBSE Class 10 Science chapter 12, Electricity: NCERT Exemplar Problems and Solutions (Part-I). Every question has been provided with a detailed explanation. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Created On: Jun 2, 2017 14:36 IST

class 10 science ncert exemplar, Electricity Class 10 NCERT Exemplar, Class 10 Science NCERT ExemplarHere you get the CBSE Class 10 Science chapter 12, Electricity: NCERT Exemplar Problems and Solutions (Part-I). This part of the chapter includes solutions for Question No. 1 to 18 from the NCERT Exemplar Problems for Class 10 Science Chapter: Electricity. These questions include only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed explanation.

CBSE Class 10 Science Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Science Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Science Chapter, Electricity:

Multiple Choice Questions

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

Circuit Diagram

(a) Maximum in (i)

(b) Maximum in (ii)

(c) Maximum in (iii)

(d) The same in all the cases

Answer. (d)


In series connections the order of elements in the circuit will not affect the amount of current flowing in the circuit.

NCERT Solutions for CBSE Class 10 Science

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

circuit diagram with resistance combinations

(a) Same in all the cases

(b) Maximum in case (i)

(c) Maximum in case (ii)

(d) Maximum in case (iii)

Answer. (d)


numerical pronlems containing resistance

Question 3. Electrical resistivity of a given metallic wire depends upon

(a) Its length

(b) Its thickness

(c) Its shape

(d) Nature of the material

Answer. (d)

Explanation: The resistivity of a material is constant for a particular temperature at a constant temperature.

Resistivity of material does not depend on length, thickness and shape of the material. It only depends on the temperature.

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

(a) 1020

(b) 1016

(c) 1018

(d) 1023

Answer. (a)


Electricity Numericle Problems

Question 5. Identify the circuit in which the electrical components have been properly connected.

Electricity Circuit Diagrams

(a) (i)                          

(b) (ii)                         

(c) (iii)                        

(d) (iv)

Answer. (b)

Explanation: Essential conditions are necessary when electrical components are connected

  • Voltmeter should be connected in parallel.
  • Ammeter is always connected in series.
  • Positive terminals of voltmeter and ammeter should be connected to positive terminal of the cell and their negative terminals should be joined to the negative terminal of the cell.

Thus, the above conditions are satisfied in case (ii).

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

(a) 1/5 Ω

(b) 10 Ω

(c) 5 Ω

(d) 1 Ω

Answer. (d)

Explanation: The maximum resistance is obtained when resistors are connected in series combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, in series = (1/5 + 1/5 + 1/5 + 1/5) = 1 Ω

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

(a) 1/5

(b) 1/25

(c) 1/10

(d) 25

Answer. (b)

Explanation: The minimum resistance is obtained when resistors are connected in parallel combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, parallel to each other =

Resistances in Parallel Numericles

Question 8. The proper representation of series combination of cells obtaining maximum potential is

cells conneced in series combination

(a) (i) 

(b) (ii)                         

(c) (iii)                        

(d) (iv)

Answer. (a)

Explanation: Maximum potential is obtained when cells are connected in series such that, negative terminal of the cell is connected to the positive terminal of the second cell and so on, as shown in the following diagram.

Crrect Combination of Cells in Series

Question 9. Which of the following represents voltage?

Electricity MCQs

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

(a) A/2

(b) 3A/2

(c) 2A

(d) 3A

Answer. (c)


Class 10 Electricity MCQs

Question 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 as shown in figure. Which of the following is true?

VI graph MCQ

(a) R1 = R2 = R3

(b) R1 > R2 > R3

(c) R3 > R2 > R1

(d) R2 > R3 > R1

Answer. (c)


Slope of VI graph is proportional to 1/ Resistance.

It means when slope will be maximum, then resistance will be minimum.

From the figure, we can see that, slope of R1 is maximum; hence its resistance will be minimum.

As, slope of R3 is minimum  so, its resistance will be maximum.

Therefore, R3 > R2 > R1

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

(a) 100%

(b) 200%

(c) 300%

(d) 400%

Answer. (c)

Explanation: If I is current and R is resistance then,

Power, P = I2R

Power in first case, P1 = I2R

100% increase in current means that current becomes 2I

Power in second case, P2 = (2I)2R = 4I2R

Now, increase in dissipated power = P2P1 = 4I2RI2R = 3I2R

Percentage increase in dissipated power = 3P1/ P1 × 100 = 300%

Question 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. (c)

Explanation: The resistivity depends on the nature of the material and the temperature.

It does not depends on dimension of resistor.

Question 14. In an electrical circuit three incandescent bulbs. A, B and C of rating 40W, 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

Answer. (c)

Explanation: We know that power is defined as rate of doing work. A bulb consumes electric energy and produces heat and light. Now, bulb with more power rating will produce more heat and light or we can say that

Power rating of bulb is directly proportional to the brightness produced by bulb.

Therefore, brightness of bulb B with power rating 60 W will be more than the brightness of bulb A having power rating as 40W.

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

(a) 5J

(b) 10J

(c) 20J

(d) 30J

Answer. (c)


Here, firstresistor, R1 = 2Ω

And second resistor, R2 = 4Ω

Voltage of cell, V = 6V

Time taken = t = 5s

Total resistance of the circuit = R = R1 + R2 = 2 + 4 = 6 Ω

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

Heat dissipated by the 4Ω resistor in 5s is given as,

            H = I2Rt

⟹       H = 1 × 4 × 5 = 20J

Question 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?

(a) 1 A

(b) 2 A

(c) 4 A

(d) 5 A

Answer. (d)

Explanation: Here, power = P = 1 KW = 1000 W

Voltage = V = 220 V

Current = I = ?

Now, I = P/V = 1000/220 = 4.5 A

Now rating of fuse wire must be slightly greater than 4.5 A i.e., 5 A.

Question 17. Two resistors of resistance 2 and 4when 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) same potential difference across them when connected in series

(d) different potential difference across them when connected in parallel

Answer. (b)

Explanation: In series combination of resistor, the current through both the resistor are same but potential difference across each will be different.

In parallel combination current across each resistor will be different but the potential difference will be same.

Question 18. Unit of electric power may also be expressed as

(a) volt ampere

(b) kilowatt hour

(c) watt second

(d) joule second

Answer. (a)

Explanation: Electric power = voltage × current

SI Unit of voltage = Volt

SI Unit of current = Ampere

So, unit of electric power is also given by, volt ampere.


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