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CUET Humanities Exam  >  CUET Humanities Tests  >  Test: Current Electricity - 2 - CUET Humanities MCQ

Test: Current Electricity - 2 - CUET Humanities MCQ


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15 Questions MCQ Test - Test: Current Electricity - 2

Test: Current Electricity - 2 for CUET Humanities 2024 is part of CUET Humanities preparation. The Test: Current Electricity - 2 questions and answers have been prepared according to the CUET Humanities exam syllabus.The Test: Current Electricity - 2 MCQs are made for CUET Humanities 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Current Electricity - 2 below.
Solutions of Test: Current Electricity - 2 questions in English are available as part of our course for CUET Humanities & Test: Current Electricity - 2 solutions in Hindi for CUET Humanities course. Download more important topics, notes, lectures and mock test series for CUET Humanities Exam by signing up for free. Attempt Test: Current Electricity - 2 | 15 questions in 15 minutes | Mock test for CUET Humanities preparation | Free important questions MCQ to study for CUET Humanities Exam | Download free PDF with solutions
Test: Current Electricity - 2 - Question 1
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Resistance wires are not made of

Detailed Solution for Test: Current Electricity - 2 - Question 1

Resistance wires are generally made of nichrome, manganin and constantan.

Test: Current Electricity - 2 - Question 2
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When three identical bulbs of 60 W, 200 V rating are connected in series to a 200 V supply, the power drawn by them will be

Detailed Solution for Test: Current Electricity - 2 - Question 2

Power of each bulb = 60 W
When three bulbs are connected in series, power drawn will be

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Test: Current Electricity - 2 - Question 3
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In the network given, P = 10Ω, Q = 20Ω, R = 15Ω, S = 30Ω and voltage = 6V. The current passing through the battery (of negligible internal resistance) is

Detailed Solution for Test: Current Electricity - 2 - Question 3


Or equivalent resistance is

Test: Current Electricity - 2 - Question 4
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For the wheatstone bridge shown in the figure, total current is 2 A. What will be the current through BD?
Detailed Solution for Test: Current Electricity - 2 - Question 4

As 5/10 = 15/30
Since the bridge is balanced, so no current flows through the galvanometer as potential at point B is equal to potential at point D.

Test: Current Electricity - 2 - Question 5
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Which of the following ways is/are correct to increase current in the circuit?

  1. Decrease the voltage while decreasing the resistance.
  2. Increase the voltage while holding the resistance constant.
  3. Decrease the resistance while holding the voltage constant.
  4. Increase the voltage while increasing the resistance.
  5. Raise the voltage, but only if you raised the resistance by same factor.
Detailed Solution for Test: Current Electricity - 2 - Question 5

Current in the circuit can be increased by increasing the voltage while holding resistance constant.
It can also be increased by decreasing resistance while holding the voltage constant.

Test: Current Electricity - 2 - Question 6
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An electric bulb is rated 220 volt - 100 watt. What will be the power consumed by it when operated on 110 volt?
Detailed Solution for Test: Current Electricity - 2 - Question 6

Test: Current Electricity - 2 - Question 7
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The resistance of wire of uniform diameter d and length l is R. The resistance of another wire of the same material, but diameter 2d and length 4l will be
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Test: Current Electricity - 2 - Question 8
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An aluminium rod and a copper rod have the same length and resistance. The specific resistance of copper is half of aluminium, but its density is thrice of aluminium. What will be the ratio of the mass of aluminium rod to the mass of copper rod?
Detailed Solution for Test: Current Electricity - 2 - Question 8

Resistance of wire is given by,
R = ρ (l/A) ...(i)
Where ρ is specific resistance of wire, A is area of cross-section of wire, and l is length of wire.
If V is volume of wire, then
R = ρ (l2/V) ...(ii)
If d is density and m is mass of wire, then

Test: Current Electricity - 2 - Question 9
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The resistance of a bulb filament is 100 Ω at a temperature of 100°C. If its temperature coefficient of resistance is 0.005 per °C, then its resistance will become 115 Ω at a temperature of
Detailed Solution for Test: Current Electricity - 2 - Question 9

R = R0(1 + αΔT)
115 = 100[1 + 0.005(T - 100)]
T = 130°C

Test: Current Electricity - 2 - Question 10
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A potentiometer has uniform potential gradient. The specific resistance of the material of the potentiometer wire is 10-7 ohm-metre, and the current passing through it is 0.1 ampere, cross-section of the wire is 10-6m2. The potential gradient along the potentiometer wire is
Detailed Solution for Test: Current Electricity - 2 - Question 10

Test: Current Electricity - 2 - Question 11
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Two batteries, one of emf 18 V and internal resistance 2Ω and the other of emf 12 V and internal resistance 1Ω, are connected as shown. The voltmeter V will record a reading of
Detailed Solution for Test: Current Electricity - 2 - Question 11

Here E1 = 18 V and r1 = 2Ω; E2 = 12 V and r2 = 1Ω
Equivalent e.m.f in the circuit is,
E = E1 - E2 = 18 - 12 = 6 V
Equivalent resistance of the circuit is,
r = r1 + r2 = 2 + 1 = 3Ω
Current in the circuit is,
I = E/r = 6/3 = 2 A
The reading of the voltmeter = Potential difference across either of two cells,
V = E1 - Ir1 = 18 - 2 × 2 = 14 V

Test: Current Electricity - 2 - Question 12
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If the current through 3 Ω resistor is 0.8 A, then the potential drop through 4 Ω resistor is
Detailed Solution for Test: Current Electricity - 2 - Question 12

Current through 3 Ω resistor = 0.8 A
Potential across PQ or 3 Ω resistor = 0.8 × 3 = 2.4 V
Potential across PQ or 6 Ω resistor = 2.4 V (Potential is same in parallel)
Current across 6 Ω resistor = 2.4/6 = 0.4 V
Current flowing through 4 Ω resistor = 0.8 A + 0.4 A = 1.2 A
Potential across 4 Ω resistor = 1.2 × 4 = 4.8 V

Test: Current Electricity - 2 - Question 13
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Two identical cells, whether connected in parallel or in series, give the same current when connected to an external resistance of 1.5Ω. Find the value of internal resistance of each cell.
Detailed Solution for Test: Current Electricity - 2 - Question 13

Let n cells be in series and m in parallel, then

⇒ nRm + nr = Rm + mnr
⇒ 6 + 2r = 3 + 4r
⇒ 2r = 3
⇒ r = 1.5Ω

Test: Current Electricity - 2 - Question 14
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The resistance of a galvanometer is 50 Ω. When 0.01 A current flows in it, full scale deflection is obtained in the galvanometer. The resistance of shunt connected to convert the galvanometer into an ammeter of range 5 A will be
Detailed Solution for Test: Current Electricity - 2 - Question 14

Shunt resistance, 
Rs = 

Test: Current Electricity - 2 - Question 15
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In a Wheatstone bridge, all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is
Detailed Solution for Test: Current Electricity - 2 - Question 15

This is a case of balanced Wheatstones bridge. The resistance R  in galvanometer arm BD is ineffective as VB = VD.

Resistance of arm ABC = R + R = 2R
Resistance of arm ADC = R + R = 2R
These are connected in parallel. Hence, the equivalent resistance of the combination = 

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