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NEET
Exam
NEET
Subject
Physics
Chapter
Motion in Two Dimensions

NEET · Physics · Motion in Two Dimensions

A small block of mass m is tied to a string of length L and whirled...

A small block of mass mm is tied to a string of length LL and whirled in a circle on a smooth horizontal table. The string will break if the tension exceeds a certain maximum value. The maximum angular velocity with which the block can be whirled without breaking the string is ω0\omega_0. If the string is shortened to a length of L2\dfrac{L}{2}, what will be the new maximum angular velocity before the string breaks?

Options

  1. A2ω02\omega_0
  2. Bω02\dfrac{\omega_0}{2}
  3. Cω02\dfrac{\omega_0}{\sqrt{2}}
  4. D2ω0\sqrt{2}\omega_0

Correct answer

D. 2ω0\sqrt{2}\omega_0

Step-by-step solution

The tension in the string provides the necessary centripetal force for the circular motion of the block.

The maximum tension T_\max the string can withstand before breaking is given by:
T_\max = m L \omega_0^2

When the length of the string is halved to L2\dfrac{L}{2}, the maximum tension the string can bear remains the same. Let the new maximum angular velocity be ω′\omega'.

T_\max = m \left(\dfrac{L}{2}\right) (\omega')^2

Equating the two expressions for T_\max:
mLω02=m(L2)(ω′)2m L \omega_0^2 = m \left(\dfrac{L}{2}\right) (\omega')^2

ω02=(ω′)22\omega_0^2 = \dfrac{(\omega')^2}{2}

(ω′)2=2ω02(\omega')^2 = 2\omega_0^2

ω′=2ω0\omega' = \sqrt{2}\omega_0.

Answer: 2ω0\sqrt{2}\omega_0
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