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A string of length ' $L$ ' fixed at one end carries a mass ' $m$ ' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through fixed end. The tension in the string is
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36 mL
Consider the free body diagram
On considering horizontal force balance:
$\mathrm{T} \sin \theta=\mathrm{m} \omega^2 \mathrm{R}$
Using the geometry, $\mathrm{R}=\mathrm{L} \sin \theta$
$\therefore \mathrm{T}=\left(\mathrm{m} \omega^2 \mathrm{~L}\right)$
Given, $\omega=\left(\frac{3}{\pi}\right)$ r.p.s. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$
$\therefore \mathrm{T}=36 \mathrm{~mL}$
On considering horizontal force balance:
$\mathrm{T} \sin \theta=\mathrm{m} \omega^2 \mathrm{R}$
Using the geometry, $\mathrm{R}=\mathrm{L} \sin \theta$
$\therefore \mathrm{T}=\left(\mathrm{m} \omega^2 \mathrm{~L}\right)$
Given, $\omega=\left(\frac{3}{\pi}\right)$ r.p.s. $=\left(\frac{3}{\pi}\right)(2 \pi) \mathrm{rad} / \mathrm{s}=6 \mathrm{rad} / \mathrm{s}$
$\therefore \mathrm{T}=36 \mathrm{~mL}$
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