CBSE Class 10 Science: Chapter 12 Electricity MCQ Questions (with Answers)
Welcome to Part 1 of our 8-part series on Chapter 12, Electricity. This post is dedicated to Multiple Choice Questions (MCQs) designed as per the latest CBSE Class 10 board pattern. Practice these questions, including highlighted Previous Year Questions (PYQs), to strengthen your concepts.
Part 1: Multiple Choice Questions (MCQ)
The unit of electric current is:
Correct Option: (b) Ampere
Electric current is the rate of flow of electric charge. Its SI unit is Ampere (A). Volt is the unit of potential difference, Ohm is the unit of resistance, and Watt is the unit of power.
A device used to measure potential difference is known as a:
Correct Option: (b) Voltmeter
A voltmeter is used to measure the potential difference (voltage) between two points in a circuit. It is always connected in parallel across the points.
Ohm's law gives a relationship between:
Correct Option: (c) Potential difference and current
Ohm's Law states that the potential difference (V) across the ends of a conductor is directly proportional to the electric current (I) flowing through it, provided the temperature remains constant. (V ∝ I)
The resistivity of a conductor depends on:
Correct Option: (c) Its temperature and nature of material
Resistivity (ρ) is an intrinsic property of a material. It depends only on the nature of the material and its temperature. It does *not* depend on the dimensions (length or area) of the conductor.
A wire of resistance R is cut into five equal parts. These parts are then connected in parallel. The equivalent resistance of this combination is R'. The ratio R/R' is:
Correct Option: (d) 25
Let the original resistance be R.
When cut into 5 equal parts, the resistance of each new part is R/5 (since R ∝ L).
When these 5 parts are connected in parallel, the equivalent resistance R' is given by:
1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
1/R' = 5/(R/5)
1/R' = 25/R
So, R' = R/25.
Therefore, the ratio R/R' = R / (R/25) = 25.
The commercial unit of electrical energy is:
Correct Option: (c) Kilowatt-hour (kWh)
The commercial unit of electrical energy is the kilowatt-hour (kWh), commonly known as a 'unit' in electricity bills. 1 kWh = 3.6 x 10^6 Joules.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
Correct Option: (d) 25 W
First, find the resistance (R) of the bulb, which is constant.
P = V²/R => R = V²/P = (220 * 220) / 100 = 484 Ω.
Now, calculate the new power (P') when operated at 110 V:
P' = (V')²/R = (110 * 110) / 484 = 12100 / 484 = 25 W.
Joule's law of heating states that the heat produced in a resistor is proportional to:
Correct Option: (d) All of the above
Joule's Law of Heating is given by the formula H = I²Rt. This means the heat (H) produced is directly proportional to the square of the current (I²), the resistance (R), and the time (t) for which the current flows.
What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
Correct Option: (d) 1 Ω
To get the maximum equivalent resistance, the resistors must be connected in series.
R_series = R1 + R2 + R3 + R4 + R5
R_series = (1/5) + (1/5) + (1/5) + (1/5) + (1/5) = 5/5 = 1 Ω.
What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
Correct Option: (a) 1/25 Ω
To get the minimum equivalent resistance, the resistors must be connected in parallel.
1/R_parallel = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5
1/R_parallel = 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5) + 1/(1/5)
1/R_parallel = 5 + 5 + 5 + 5 + 5 = 25
R_parallel = 1/25 Ω.
Alloys are commonly used in electrical heating devices, like toasters and irons, because:
Correct Option: (b) They have high resistivity and high melting point
Heating elements require high resistance to produce a large amount of heat (H ∝ R). They also need a high melting point so they do not melt or oxidize at high temperatures. Alloys (like Nichrome) have both these properties.
1 Volt is equivalent to:
Correct Option: (a) 1 Joule / 1 Coulomb
Potential difference (Voltage) is defined as the work done (W) to move a unit charge (Q) from one point to another.
V = W / Q.
So, 1 Volt = 1 Joule / 1 Coulomb.
How is an ammeter connected in a circuit to measure current?
Correct Option: (b) In series
An ammeter measures the current flowing *through* a component. To measure the total current, it must be placed in series in the circuit so that all the current flows through it. It has very low resistance.
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The SI unit of resistivity is:
Correct Option: (b) Ohm-meter (Ω m)
From the formula R = ρ(L/A), we can rearrange for resistivity (ρ):
ρ = R(A/L).
Unit of ρ = (Ohm * m²) / m = Ohm-meter (Ω m).
Two resistors of 2 Ω and 4 Ω are connected in parallel. The equivalent resistance is:
Correct Option: (b) 4/3 Ω
For parallel combination: 1/R_p = 1/R1 + 1/R2
1/R_p = 1/2 + 1/4
1/R_p = (2 + 1) / 4 = 3/4
R_p = 4/3 Ω.
The electrical resistivity of a given metallic wire depends upon:
Correct Option: (d) the nature of the material
Resistivity (ρ) is an intrinsic property of a material and depends only on the nature of the material and its temperature. It does not depend on the physical dimensions (length, thickness, shape).
When 1 Ampere of current flows for 1 second, the charge that has flowed is:
Correct Option: (c) 1 Coulomb
Current is defined as the rate of flow of charge (I = Q/t).
Therefore, Charge (Q) = Current (I) × Time (t).
Q = 1 A × 1 s = 1 Coulomb (C).
What happens to the resistance of a wire if its length is doubled?
Correct Option: (b) It becomes double
Resistance (R) is directly proportional to the length (L) of the wire (R ∝ L), given R = ρ(L/A). If the length (L) is doubled, the resistance (R) also gets doubled.
What happens to the resistance of a wire if its radius is doubled?
Correct Option: (c) It becomes one-fourth
Resistance (R) is inversely proportional to the area of cross-section (A). Area A = πr².
So, R is inversely proportional to r² (R ∝ 1/r²).
If the radius (r) is doubled (r' = 2r), the new area A' = π(2r)² = 4πr² = 4A.
The new resistance R' = ρ(L/A') = ρ(L/4A) = (1/4) * R.
It becomes one-fourth of the original.
The heating element of an electric iron is made of:
Correct Option: (c) Nichrome
Nichrome (an alloy of nickel and chromium) is used for heating elements because it has high resistivity (produces a lot of heat) and a high melting point (does not melt or oxidize easily at high temperatures).
Which of the following is not a unit of electric power?
Correct Option: (c) Kilowatt-hour (kWh)
Kilowatt-hour (kWh) is the unit of electrical *energy* (Energy = Power × Time). Watt, Volt-Ampere (P=VI), and I²R are all expressions or units for electric *power* (rate of energy consumption).
In a parallel circuit, which quantity remains the same across all components?
Correct Option: (b) Potential Difference
In a parallel combination, the potential difference (voltage) across each resistor is the same and is equal to the voltage of the source. The current, however, gets divided among the branches.
In a series circuit, which quantity remains the same through all components?
Correct Option: (a) Current
In a series combination, the electric current is the same through every component in the circuit. The potential difference (voltage) gets divided among the resistors.
The filament of an electric bulb is made of tungsten because:
Correct Option: (b) It has a very high melting point
Tungsten is used for filaments because it has an extremely high melting point (approx 3380 °C). This allows it to be heated to a very high temperature (to glow and produce light) without melting.
1 kWh is equal to:
Correct Option: (b) 3.6 × 10⁶ J
1 kWh = 1 kilowatt × 1 hour
= 1000 Watt × 3600 seconds
= 3,600,000 Watt-seconds
= 3.6 × 10⁶ J (Since 1 Watt-second = 1 Joule)
