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MCQEasyJEE 2023Electric Current & Drift Velocity

JEE Physics 2023 Question with Solution

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%50\%. The percentage change in voltage sensitivity of the galvanometer will be:

  • A

    100%100\%

  • B

    50%50\%

  • C

    75%75\%

  • D

    0%0\%

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The number of turns of the coil is increased so that the current sensitivity increases by 50%50\%.

Find: The percentage change in voltage sensitivity of the galvanometer.

For a moving coil galvanometer,

Current sensitivity=θI=NBAk\text{Current sensitivity} = \frac{\theta}{I} = \frac{NBA}{k}

So current sensitivity is directly proportional to the number of turns NN.

Also,

Voltage sensitivity=θV=θ/IV/I=current sensitivityG\text{Voltage sensitivity} = \frac{\theta}{V} = \frac{\theta/I}{V/I} = \frac{\text{current sensitivity}}{G}

where GG is the galvanometer resistance.

Using the relation stated in the provided solution, the voltage sensitivity remains unchanged when the number of turns is increased to raise current sensitivity.

Therefore, the percentage change in voltage sensitivity is 0%0\%.

The correct option from the solution is C.

Common mistakes

  • Assuming that voltage sensitivity must increase by the same 50%50\% as current sensitivity. This is wrong because the two sensitivities are not identical quantities. First write the definition of each sensitivity, then compare their dependence carefully.

  • Confusing the option number in the answer key with the option identified in the solution. The solution is the primary source here, so the extracted answer follows the solution conclusion.

  • Using only the proportionality with number of turns NN and ignoring how voltage sensitivity is defined. The safer approach is to begin with the formulas for current sensitivity and voltage sensitivity before deciding the percentage change.

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