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CGP EDU Academic Team
Published on: September 13, 2026
A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/ p π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is
Text Solution
Verified by ExpertsThe correct answer is:
D
\(T \tan \theta = M \omega^{2} R\) …(i)
\(\tau \sin \theta = M \omega^{2} L \sin \alpha\) … (ii)
From (i) and (ii)
\tau = M r^2 g
\(= M 4 \pi^{2} n^{2} L\)
\(= M_{4,7}! \left| 2_{,7} \right|^{2} t\)
16 mL
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