A string of length L is fixed at one end and | vedclass

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Question

(d) $T\sin 	heta = M{\omega ^2}R$&hellip;(i)

$T\sin 	heta = M{\omega ^2}L\sin 	heta $&hellip;(ii)

From (i) and (ii)

$T = M{\omega ^2}L$

Vedclass · Accepted Answer

$16 \,ML$

Vedclass · Answer

(d) $T\sin 	heta = M{\omega ^2}R$&hellip;(i)

$T\sin 	heta = M{\omega ^2}L\sin 	heta $&hellip;(ii)

From (i) and (ii)

$T = M{\omega ^2}L$

$ = M\,4{\pi ^2}{n^2}L$

$ = M\,4{\pi ^2}{\left( {\frac{2}{\pi }} ight)^2}L$

$ = 16\,ML$

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2/\pi$ revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is

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