Physics · Current Electricity

JEE Main 2024 — 27 January, Shift 1 — Question 45

A wire of length 10 cm and radius $\sqrt{7} \times 10^{-4} \mathrm{~m}$ connected across the right gap of a meter bridge. When a resistance of $4.5 \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is $\mathrm{R} \times 10^{-7} \Omega \mathrm{~m}$ , then value of R is :

Option A:
63
Option B:
70
Option C:
66
Correct
Option D:
35

Answer: C

Step-by-step solution

For null point, $\frac{4.5}{60}=\frac{R}{40}$

Also, $R=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \ell}{\pi \mathrm{r}^{2}}$

$4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60$

$\rho=66 \times 10^{-7} \Omega \times \mathrm{m}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Current Electricity
Topic: Electrical Measuring Instruments

A wire of resistance R and length L is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :

Answer: C

Step-by-step solution

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