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Physics · Current Electricity

JEE Main 2024 — 27 January, Shift 2 — Question 32

Wheatstone bridge principle is used to measure the specific resistance (S1)\left(\mathrm{S}_{1}\right) of given wire, having length LL, radius rr. If XX is the resistance of wire, then specific resistance is : S1=X(πr2 L)\mathrm{S}_{1}=\mathrm{X}\left(\frac{\pi \mathrm{r}^{2}}{\mathrm{~L}}\right). If the length of the wire gets doubled then the value of specific resistance will be :

  1. Option A:

    S14\frac{S_{1}}{4}

  2. Option B:

    2 S12 \mathrm{~S}_{1}

  3. Option C:

    S12\frac{\mathrm{S}_{1}}{2}

  4. Option D:

    S1\mathrm{S}_{1}

    Correct

Answer: D

Step-by-step solution

As specific resistance does not depends on dimension of wire so, it will not change.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 2
Subject
Physics
Chapter
Current Electricity
Topic
Ohm's Law and Calculation of Resistance

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