CBSE Class 10 Physics Electricity MCQs Set H

Question. 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

Question. Unit of electric power may also be expressed as:
a) Volt-ampere
b) Kilowatt-hour
c) Watt-second
d) Joule-second
Answer: a

Question. A cell, a resistor, a key, and ammeter are arranged as shown in the circuit diagrams given below. 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.
Answer: d

Question. 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

Question. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be:
a) 0.4 A
b) 0.6 A
c) 0.8 A
d) 1 A
Answer: d

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.
Answer: c

Question. 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

Question. When a 4 V battery is connected across an unknown resistor there is a current of 100 mA in the circuit. The value of the resistance of the resistor is:
a) 4 Ω
b) 40 Ω
c) 400 Ω
d) 0.4 Ω
Answer: b

Question. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
a) 1/5 Ω
b) 1/25 Ω
c) 1/10 Ω
d) 25 Ω
Answer: b

Question. 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

Question. 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

Question. A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l’ is
a) A/2
b) 3A/2
c) 2A
d) 3A
Answer: c

Question. The maximum resistance which can be made using four resistor each of resistance 1/2 Ω is
a) 2 Ω
b) 1 Ω
c) 2.5 Ω
d) 8 Ω
Answer: a

Assertion and Reason Based MCQs :

Directions : In the following questions, A statement of Assertion (a) is followed by a statement of Reason (R).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is true.

Question. Assertion (A): Two resistance having value R each.
Their equivalent resistance is R/2 .
Reason (R): Resistances are connected in parallel.
Answer: a

Question. Assertion (A): Silver is not used to make electric wires.
Reason (R): Silver is a bad conductor of electricity.
Answer: c

Question. Assertion (A): Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points than their constituent metals.
Answer: c

Question. Assertion (A): The resistivity of conductor increases with the increasing of temperature.
Reason (R): The resistivity is the reciprocal of the conductivity.
Answer: b

Question. Assertion (A): Electric appliances with metallic body have three connections, whereas an electric bulb has a two pin connection.
Reason (R): Three pin connections reduce heating of connecting wires.
Answer: c

Question. Assertion (A): Copper is used to make electric wires.
Reason (R): Copper has very low electrical resistance.
Answer: a

Question. Assertion (A): Bending a wire does not affect electrical resistance.
Reason (R): Resistance of wire is proportional to resistivity of material.
Answer: a

Very Short Answer Type Questions :

Question. What does the cord of an electric oven not glow while its heating element does ?
Answer: Cord is made up of copper wire whereas heating element is made up of alloy.

Question. Two unequal resistances are connected in parallel.
If you are not provided with any other parameters (eg. numerical values of I and R), what can be said about the voltage drop across the two resistors?
Answer: Voltage-drop is same across both.

Question. Should the resistance of a voltmeter be low or high? Give reason.
Answer: High. In parallel connection, less current passes through high resistance.

Question. What is meant by the statement. “The resistance of a conductor is one ohm” ?
Answer: The resistance of a conductor is said to be 1 ohm if a current of 1 ampere flows through it when the potential difference across it is 1 volt.

Question. Write SI unit of resistivity.
Answer: Ohm metre (ohm m).

Question. Why are the heating elements of electric toasters and electric irons made of an alloy rather than a pure metal ? 
Answer: Due to high resistivity of alloys rather than its constituting metals.

Question. Write the mathematical expression for Joule’s law of heating.
Answer: Mathematical expression of Joule’s law of heating is : H = I2Rt
Where, H = Produced Heat
I = Current flowing through the device
t = Time of current flow
R = Resistance of the appliance

Short Answer Type Questions :

 Question. Give reason for the following:
(i) Why are copper and aluminium wires used as connecting wires ?
(ii) Why is tungsten used for filament of electric lamps?
(iii) Why is lead-tin alloy used for fuse wires ?
Answer: (i) These are good conductors of electricity/low resistance, low resistivity.
(ii) Very high melting point and high resistivity.
(iii) Low melting point.

Question. Define electric power. Write an expression relating electric power, potential difference and resistance.
Answer: Electric power : It is the amount of electric energy consumed in a circuit per unit time.
Expression : P = V2/R

Question. How many 132 Ω resistors in parallel are required to carry 5 A on a 220 V line ?
Answer: Given V = 220 V, I = 5 A
V = IR
Or R = V/I
In parallel combination, let the no. of resistors = x
132/x = 220/5
or, 132/x = 44
or, x = 132/44
∴ x = 3
The number of resistors = 3

Question. (a) State the relation correlating the electric current flowing in a conductor and the voltage applied across it. Also, draw a graph to show this relationship.
(b) Find the resistance of a conductor if the electric current flowing through it is 0.35 A when the potential difference across it is 1.4 V.
Answer: (a) The flow of current (I) in the conductor is directly proportional to the potential difference (V) established across it provided the physical
conditions remain same.
Or V = IR
Graph: 2
(b) Given :
Potential Difference (V)= 1.4 V
Current (I) = 0.35 A
As per formula, V = IR
So, V/I = 1.4/0.35 = 4 ohm

 Question. (a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.
(c) Why are alloys commonly used in electrical heating devices ? Give reason.
Answer: (a) Factors on which resistance of a conductor depends :
(i) Length of conductor [or R ∝ l]
(ii) Area of cross-section of the conductor
[or R ∝ 1/A]
(b) Metals are good conductor of electricity – as they have low resistivity/have free electrons. Glass is a bad conductor of electricity – as it has high resistivity/have no free electrons. 
(c) Reason: Alloys have high resistivity/high melting point/alloys do not oxidize (or burn) readily at high temperatures.

 Question. Compute the heat generated while transferring 96,000 coulomb of charge in two hours through a potential difference of 40 V. A
Answer: Given, Charge (Q)= 96000 C, Time (t) = 2 h,
Potential difference (V) = 40 V
Heat generated, H = VIt (where I = Q/t)
H = d
H = V × Q
H = 40 ×96000
H = 384000 J

Question. Calculate the resistance of a 1 km long copper wire of area of cross section 2 × 10–2 cm2. The resistivity of copper is 1.623 × 10–8 ohm-meter.
Answer: R = ρ l/A
= 1 x 623 x 10-8 x 100/2 x 10-2 x 10-4 m2
= 0.81 × 10 Ω = 8.1 Ω.

Question. Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ` 6.00.
(i) Electric heater of 1000 W for 5 hours daily.
(ii) Electric refrigerator of 400 W for 10 hours daily.
Answer: P1 = 1000 W = 1000/1000 kW, t1 = 5h
P2 = 400 W = 400/1000 1000 kW, t2 = 10h
No. of days, n = 30
E1 =P1 × t1 × n 
= 1 kW × 5h × 30 = 150 kWh 
E2 = P2 × t2 × n
= 400/1000 kW × 10 h × 30
= 120 kWh
∴ Total energy = (150 + 120) kWh = 270 kWh ½
∴ Total cost = 270 × 6 = ` 1620

Question. Three resistors of 10 Ω, 15 Ω and 20 Ω are connected in series in a circuit. If the potential drop across the 15 Ω resistor is 3 V, find the current in the circuit and potential drop across the 10 Ω resistor.
Answer: In series circuit same current flows through all the
resistors. Current through 15 Ω resistor,
I = V/R = 3 V/15Ω = 1/5
= 0.2 A
∴ Current in the circuit = 0.2 A
∴ Potential drop across 10 Ω resistor is
V = IR
= 0.2 A × 10 Ω
= 2 V

Question. (a) Write Joule’s law of heating.
(b) Two lamps one rated 100 W 220 V, and the other 60 W 220 V, are connected in parallel to electric mains supply. Find the currents drawn by two bulbs from the line, if the supply voltage is 220 V.
Answer: (a) Joule’s law of heating: Heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to the resistance for a given current and (iii) directly proportional to the time for which the current flows through the resistor.
H = I2Rt where, H = Heat produced, I = current,
R = Resistance of the conductor and t = Time for
which the current flows through the resistor.
(b) Current in 1st bulb,
I1 = p1/V = 100/220 = 5/11 A or 0.45 A
Current in 2nd bulb,
I2 = p2/V = 60/220 = 3/11 A or 0.27 A 

Question. A bulb is rated at 200V – 40W. What is its resistance ? 5 such bulbs are lighted for 5 hours. Calculate the electrical energy consumed ? Find the cost if the rate is 5.10 per kWh.
Answer: V = 200 V, P = 40 W
P = VI
I = V/I = 40/200 = 1/5 A
R = V/I = 200/1/5 
= 200 × 5 = 1000 Ω
Total Power = 40 W × 5 = 200 W
Time = 5 hrs
Electrical energy = 200 W × 5 hrs.
= 1000 Wh
= 1 kWh.
Cost of 1 kWh = ₹ 5.10 
∴ Total cost = ₹ 1 × 5.10 = ₹ 5.10

Question. What is electrical resistivity ? Derive its SI unit.
In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change ? Justify your answer.
Answer: Electrical resistivity of the material of a conductor is the resistance offered by the conductor of length 1 m and area of cross-section 1 m2.
ρ = RA/l
Unit of ρ = ohm meter2 / metre = ohm meter
Resistance of wire is doubled if its length is doubled.
Hence current is reduced to half.
∴ Ammeter reading = 100 mA/2 = 50 mA

Question. A circuit has a line of 5 A. How many lamps of rating 40 W, 200 V can simultaneously run on this line safely ?
Answer: Given, V = 200 V, P = 40 W, I = 5A, n = ?
nP = VI
n = VI/P = 200 X 5/40
= 1000/4 = 25 lamps

Question. (i) Consider a conductor of resistance ‘R’, length ‘L’, thickness ‘d’ and resistivity ‘r’.
Now this conductor is cut into four equal parts.
What will be the new resistivity of each of these parts? Why?
(ii) Find the resistance if all of these parts are connected in:
(a) Parallel
(b) Series
(iii) Out of the combinations of resistors mentioned above in the previous part, for a given voltage which combination will consume more power and why?

Long Answer Type Questions :

Question. (a) An electric bulb is rated at 200 V-100 W. What is its resistance ? 
(b) Calculate the energy consumed by 3 such bulbs if they glow continuously for 10 hours for complete month of November. 
(c) Calculate the total cost if the rate is Rs 6.50 per unit.
Answer: (a) Given, V = 200 volts and P = 100 watt
As P = V2/R or R = V2/R = (200)2/100W = 40000/100Ω = 400 Ω
(b) Electrical energy consumed, E = number of units ×
Power of each unit × time × total days
Here, n = 3, P = 100 W, t = 10 hours, Days = 30
So, E = 3 × 100 W × 10 h × 30 = 90,000 Wh
= 90 kWh 2
(c) Total cost of electricity = Total unit of energy
consumed × Cost per unit
= 90 kWh × 6.50 = ₹ 585

Question. A bulb is rated 40W, 220V. Find the current drawn by it, when it is connected to a 220V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25W, 220V,will be there be any change in the value of current and resistance ?
Justify your answer and determine the change.
Answer: P = 40 W, V = 220V
P = VI
∴ I = P/V = 40/220 = 2/11 A = 0.18 A
From Ohm’s law,
V = IR
R = V/I = 220/2 = 1210 W
When replaced by 25 W, 220 V lamp:
I = P/V = 25/220 = 5/44 A = 0.113 A
R = V2/P = 2202/25 = 1936 Ω
Yes there is change in current and resistance.
Change in current = 0.18 – 0.1136 = 0.0664 A Change in resistance = 1936 – 1210 = 726 ohm.
Hence, from the above justification, we can see that current decreases and resistance increases when we use a 25 W bulb in place of a 40 W.

Question. (a) Define Power and state its SI unit.
(b) A torch bulb is rated 5 V and 500 mA. Calculate:
(i) Power (ii) Resistances (iii) Energy consumed when it is lighted for 2½ hours.
Answer: (a) Power: It is the amount of electric energy
consumed in a circuit per unit time.
P = W/t
Its S.I unit is Watt (W).
(b) V = 5V
I = 500 mA = 0.5 A 2
(i) P = V × I = 5 × 0.5 = 2.5 W
(ii) Resistance R = V/I = 5/0.5 = 10 ohms
(iii) Energy consumed = P × t
= 2.5 × 2.5
= 6.25 Wh