Physics · Current Electricity

NEET (UG) 2024 — Question 32

A wire of length ' $r$ and resistance $100 \Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Option A:
$26 \Omega$
Option B:
$52 \Omega$
Correct
Option C:
$55 \Omega$
Option D:
$60 \Omega$

Answer: B

Step-by-step solution

$R=\frac{\rho l}{A}$

$R^{\prime}=\frac{\rho l}{10 A}=\frac{R}{10}$

$R_{S}=5 \times \frac{R}{10} \quad$ [series]

$R_{S}=50$

$R_{P}=\frac{R}{50} \quad$ [parallel]

$R_{\text {eq }}=R_{S}+R_{P}$

=52 \Omega

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2024
Paper: 2024 paper
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge

Answer: B

Step-by-step solution

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