MHT CET · Physics · Work Power Energy
A string of \(L\) is fixed at one end and carries a mass \(M\) at the other end. The string makes \(\frac{\pi}{2}\) revolutions per second around the vertical axis through the fixed end as shown in the figure, then the tension in the string is
- A \(2 M L\)
- B \(M L\)
- C \(16 M L\)
- D \(4 M L\)
Answer & Solution
Correct Answer
(C) \(16 M L\)
Step-by-step Solution
Detailed explanation
Given: \(\omega=\frac{2}{\pi}\) rev.per sec
\(\Rightarrow \omega=2 \pi \frac{2}{\pi} \mathrm{rad} / \mathrm{s}=4 \mathrm{rad} / \mathrm{s}\)
The horizontal force balance reads:
\(T \sin \theta=M R \omega^2\)
From figure \(R=L \sin \theta\)
\(\Rightarrow T=M L \omega^2\)
On introducing, \(\omega=4 \mathrm{rad} / \mathrm{s}\)
\(\Rightarrow T=M L(4)^2=16 \mathrm{ML}\)
\(\Rightarrow \omega=2 \pi \frac{2}{\pi} \mathrm{rad} / \mathrm{s}=4 \mathrm{rad} / \mathrm{s}\)
The horizontal force balance reads:
\(T \sin \theta=M R \omega^2\)
From figure \(R=L \sin \theta\)
\(\Rightarrow T=M L \omega^2\)
On introducing, \(\omega=4 \mathrm{rad} / \mathrm{s}\)
\(\Rightarrow T=M L(4)^2=16 \mathrm{ML}\)
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