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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
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The correct answer is:
$16 M L$
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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