NEET Exam  >  NEET Tests  >  Physics Class 12  >  Test: Current Electricity - 1 - NEET MCQ

Test: Current Electricity - 1 - NEET MCQ


Test Description

25 Questions MCQ Test Physics Class 12 - Test: Current Electricity - 1

Test: Current Electricity - 1 for NEET 2025 is part of Physics Class 12 preparation. The Test: Current Electricity - 1 questions and answers have been prepared according to the NEET exam syllabus.The Test: Current Electricity - 1 MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Current Electricity - 1 below.
Solutions of Test: Current Electricity - 1 questions in English are available as part of our Physics Class 12 for NEET & Test: Current Electricity - 1 solutions in Hindi for Physics Class 12 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Current Electricity - 1 | 25 questions in 30 minutes | Mock test for NEET preparation | Free important questions MCQ to study Physics Class 12 for NEET Exam | Download free PDF with solutions
Test: Current Electricity - 1 - Question 1

According to Kirchhoff’s Loop Rule

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

Kirchhoff’s loop rule is based on the principle of conservation of energy. The work done in transporting a charge in a closed loop is zero. The algebraic sum ( since potential differences can be both positive and negative) of potential differences around any closed loop is always zero.

Test: Current Electricity - 1 - Question 2

The Wheatstone bridge is balanced for four resistors R1,R2,R3 and R4 with a cell of emf 1.46 V. The cell is now replaced by another cell of emf 1.08 V. To obtain the balance again

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

The balance point of the Wheatstone’s bridge is determined by the ratio of the resistances. The change in the emf of the external battery will have no effect on the balance point.
 

Explanation:


  • Initial Balanced Wheatstone Bridge: In the initial balanced Wheatstone bridge configuration, the emf of the cell is 1.46 V and all four resistors R1, R2, R3, and R4 are set to specific values to achieve balance.

  • Replacement of Cell: When the cell is replaced by another cell with an emf of 1.08 V, the balance of the Wheatstone bridge is disrupted.

  • Requirement for Rebalancing: In order to rebalance the Wheatstone bridge with the new cell of emf 1.08 V, no resistance needs to be changed.

  • Reasoning: The balance of the Wheatstone bridge is determined by the ratio of the resistances in the bridge arms and not by the absolute values of the resistances. As long as the ratio of the resistances remains the same, the balance will be maintained regardless of the emf of the cell.

  • Conclusion: Therefore, in this scenario, no resistance needs to be changed to obtain the balance again with the new cell of emf 1.08 V.


  •  
Test: Current Electricity - 1 - Question 3

Two cells of emf 1.25V , 0.75V and each of internal resistance 1Ω are connected in parallel. The effective emf will be

Detailed Solution for Test: Current Electricity - 1 - Question 3

We want the Voltage difference VAB.
let A and B be open and not connected to any thing.
There is a current that flows from cell of larger emf to the cell of small emf.
Call that current as  I Amperes.

1.25 V - 0.75 V - I * R2 - I * R1 = 0

I = 0.5 / (R1+R2)
VAB =  -0.75 - I * R2 =  - 0.75 - 0.5 * R2 / (R1+R2)

= - (0.75 R1 + 1.25 R2) / (R1+R2)

= - ().75 * 1 + 1.25 * 1) / (1 + 1)  volts

=  - 1 volts

Test: Current Electricity - 1 - Question 4

 If the length of the filament of a heater is reduced by 10% the power of the heater will

Detailed Solution for Test: Current Electricity - 1 - Question 4

Power P= V2/R​
If length is reduced by 10% then, new resistance of filament will be R′.
R′ = R−10% of R
R′ = 0.9R
Now new power of heater is P2​
P2​ = V2​​/R′ = V2/0.9R ​ ​= 1.1 P
% increase powe r = 11%

Test: Current Electricity - 1 - Question 5

A cylindrical wire of radius R has current density varying with distance r from its axis asThe total current through the wire is

Detailed Solution for Test: Current Electricity - 1 - Question 5

Total current through the wire is given by:

I = ∫0R J(r)·2πr dr

Substitute the given current density:

I = ∫0R J0 (1 − r2/R2) · 2πr dr

Take constants outside the integral:

I = 2πJ0 ∫0R (1 − r2/R2) r dr

Split the integral:

I = 2πJ0 [ ∫0R r dr − (1/R2) ∫0R r3 dr ]

Evaluate the integrals:

0R r dr = R2/2,    ∫0R r3 dr = R4/4

Substitute the values:

I = 2πJ0 [ R2/2 − R2/4 ] = 2πJ0 · R2/4 = πJ0R2/2

Final Answer: I = πJ0R2/2

Test: Current Electricity - 1 - Question 6

In the circuit shown in Fig., the reading of the ammeter is (assume internal resistance of the battery to be zero)

Detailed Solution for Test: Current Electricity - 1 - Question 6

Voltage across 5Ω=10 V

∴ I = 10/5 A = 2 A

Test: Current Electricity - 1 - Question 7

Drift is the random motion of the charged particles within a conductor,

Detailed Solution for Test: Current Electricity - 1 - Question 7

The electrons in a conductor have random velocities and when an electric field is applied, they suffer repeated collisions and in the process move with a small average velocity, opposite to the direction of the field. This is equivalent to positive charge flowing in the direction of the field.

Test: Current Electricity - 1 - Question 8

In the circuit diagram shown below, the magnitude and direction of the flow of current, respectively, would be

Detailed Solution for Test: Current Electricity - 1 - Question 8

Test: Current Electricity - 1 - Question 9

Unit of Resistivity is

Detailed Solution for Test: Current Electricity - 1 - Question 9

Test: Current Electricity - 1 - Question 10

In a metre bridge experiment, null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X<Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y ?

Detailed Solution for Test: Current Electricity - 1 - Question 10

Test: Current Electricity - 1 - Question 11

Potentiometer measures the potential difference more accurately than a voltmeter, because

Detailed Solution for Test: Current Electricity - 1 - Question 11

Potentiometer measures the potential difference using null deflection method, where no current is drawn from the cell; whereas voltmeter needs a small current to show deflection. So, accurate measurement of p.d is done using a potentiometer.

Test: Current Electricity - 1 - Question 12

The ratio of the resistance of conductor at temperature 15°C to its resistance at temperature 37.5°C is 4:5 . The temperature coefficient of resistance of the conductor is -

Detailed Solution for Test: Current Electricity - 1 - Question 12

Test: Current Electricity - 1 - Question 13

The resistance of a galvanometer is 10Ω. It gives full-scale deflection when 1 mA current is passed. The resistance connected in series for converting it into a voltmeter of 2.5 V will be

Detailed Solution for Test: Current Electricity - 1 - Question 13

Test: Current Electricity - 1 - Question 14

A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance

Detailed Solution for Test: Current Electricity - 1 - Question 14
  • Copper is a conductor and we know that for conductors, resistance is directly proprtional to temperature. Therefore on decreasing temperature resistance also decreases.
  • Whereas, germanium is a semiconductor and for semiconductors, resistance is inversely proportional to temperature.
  • So on decreasing temperature resistance increases.
Test: Current Electricity - 1 - Question 15

A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as an ammeter of range 1 A, the required shunt must have a resistance of

Detailed Solution for Test: Current Electricity - 1 - Question 15

Test: Current Electricity - 1 - Question 16

In the circuit shown, if the 10 Ω resistance is replaced by 20 Ω, then the amount of current drawn from the battery will be

Detailed Solution for Test: Current Electricity - 1 - Question 16

∵ 3/8 = 9/24
∴ The Wheatstone bridge is balanced.
∴ On changing the resistance of the bridge, the current drawn will remain the same.

Test: Current Electricity - 1 - Question 17

In a potentiometer experiment, for measuring internal resistance of a cell, the balance point has been obtained on the fourth wire. The balance point can be shifted to fifth wire by

Detailed Solution for Test: Current Electricity - 1 - Question 17

If the current due to the auxillary battery is decreased, the potential gradient will be decreases, so the balancing length increases.Thus null point will move to fifth wire.

Test: Current Electricity - 1 - Question 18

A voltmeter has a resistance of G ohm and range V volt. The value of resistance used in series to convert it into voltmeter of range nV volt is

Detailed Solution for Test: Current Electricity - 1 - Question 18

We know that R=V/Ig−G
The voltmeter gives the full-scale deflection for potential difference V. Its resistance is G. Hence, In=(V/G).
Given that V=nV

Test: Current Electricity - 1 - Question 19

Meter Bridge is used to

Detailed Solution for Test: Current Electricity - 1 - Question 19

With a known resistance in one of the gaps, the meter bridge is used to determine the value of unknown resistance by the formula. 

Test: Current Electricity - 1 - Question 20

In the figure, voltmeter and ammeter shown are ideal. Then voltmeter and ammeter readings, respectively, are

Detailed Solution for Test: Current Electricity - 1 - Question 20

Resistors 20Ω,100Ω and 25Ω will be in parallel. Their equivalent is 10Ω.


p.d. across 10Ω,10I = 10 × 10 = 100 V
This will be the voltmeter reading. Also, this will be the p.d. across each of 20Ω,100Ω and 25Ω resistors.
Ammeter reading = current through 25Ω=100/25=4 A.

Test: Current Electricity - 1 - Question 21

The operating temperature of the filament of lamp is 2000C. The temperature coefficient of the material of filament is 0.005C−1. If the atmospheric temperature is 0C, then the current in the 100 W, 200 V rated lamp when it is switched on is nearest to

Detailed Solution for Test: Current Electricity - 1 - Question 21


So, 400 = R0[1 + 0.005 × 2000]

Test: Current Electricity - 1 - Question 22

Three identical cells, each having an e.m.f. of 1.5 V and a constant internal resistance of 2.0Ω, are connected in series with a 4.0Ω resistor R, first as in circuit (i), and secondly as in circuit (ii).

What is the ratio 

Detailed Solution for Test: Current Electricity - 1 - Question 22

Test: Current Electricity - 1 - Question 23

The sensitivity of the potentiometer can be increased by:

Detailed Solution for Test: Current Electricity - 1 - Question 23
  • A potentiometer is considered to be sensitive if the potential gradient dV/dl is low.
  • Such a potentiometer can measure very small changes in potential difference.
  • Increasing the length of the potentiometer wire decreases the potential gradient. Its sensitivity increases. Increasing potential gradient decreases the sensitivity.
  • Increasing the emf of the primary cell and by decreasing the length, potential gradient increases.
Test: Current Electricity - 1 - Question 24

Three similar light bulbs are connected to a constant voltage d.c. supply as shown in Fig. Each bulb operates at normal brightness and the ammeter (of negligible resistance) registers a steady current.

The filament of one of the bulbs breaks. What happens to the ammeter reading and to the brightness of the remaining bulbs?
Ammeter reading _______ & Bulb brightness _______

Detailed Solution for Test: Current Electricity - 1 - Question 24

Suppose V is the voltage of the supply and R is the resistance of each bulb.
Now, Rp = R/3 and current in ammeter, I = V/Rp = 3 V/R, provided all three bulbs are working properly.
If one bulb has broken down, then Rp = R/2 and I = 2V/R
Therefore, current decreases and since current through each bulb is V/R the same as before, brightness of bulbs is not affected.

Test: Current Electricity - 1 - Question 25

In a meter bridge experiment a balance point is obtained at a distance of 60 cm from the left end when unknown resistance R is in a left gap and 8 ohms resistor is connected in the right gap. When the position of R and 8 ohm resistor is interchanged the balance point will be at distance of

Detailed Solution for Test: Current Electricity - 1 - Question 25

60/40 = l/100-l, = R/8

so R=12

now changed position 8 is proportional to l and 12 to (100-l)

8/12 = l/100-l

l = 40ohm

90 videos|420 docs|88 tests
Information about Test: Current Electricity - 1 Page
In this test you can find the Exam questions for Test: Current Electricity - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Current Electricity - 1, EduRev gives you an ample number of Online tests for practice
90 videos|420 docs|88 tests
Download as PDF