A bob is whirled in a horizontal plane by means of a string with an initial speed of ω \;rpm . The tension in the string is T . If speed becomes 2 ω while keeping the same radius, the tension in the string becomes:

Physics · Horizontal Circular Motion

NEET (UG) 2024 — Question 35

A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega \mathrm{\;rpm}$ . The tension in the string is $T$ . If speed becomes $2 \omega$ while keeping the same radius, the tension in the string becomes:

Option A:
$T$
Option B:
$4 T$
Correct
Option C:
$\frac{T}{4}$
Option D:
$\sqrt{2} T$

Answer: B

Step-by-step solution

F_{cp} = m a_{cp}

F_{cp} = m \omega^2 r

T = m \omega^2 r

\text{Now\; speed\; becomes\; } 2\omega

T' = m (2\omega)^2 r

T' = 4 m \omega^2 r

T' = 4T

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2024
Paper: 2024 paper
Subject: Physics
Chapter: Horizontal Circular Motion
Topic: Problems involving application of circular motion

A particle moving with uniform speed in a circular path maintains:

Answer: B

Step-by-step solution

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