Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

T/4

\sqrt{2}T

Step-by-Step Solution

Identify the Principle: For a bob whirled in a horizontal circle, the tension ( $T$ ) in the string provides the necessary centripetal force ( $F_c$ ).
Formula: The centripetal force is given by $F_c = m \omega^2 r$ , where $m$ is the mass, $\omega$ is the angular velocity, and $r$ is the radius [Source 51, 52]. Therefore, $T = m \omega^2 r$ .
Analyze the Change:

Initial Tension: $T_1 = T = m \omega^2 r$ .
Final Angular Velocity: $\omega' = 2\omega$ .
Radius remains constant.
Final Tension: $T_2 = m (\omega')^2 r = m (2\omega)^2 r = m (4\omega^2) r = 4 (m \omega^2 r)$ .

Conclusion: Since $m \omega^2 r = T$ , the new tension $T_2 = 4T$ .

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