A solid cylinder length is suspended symmetrically through two massless strings, as shown in figure.

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6.System of Particles and Rotational Motion

hard

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of $4\,ms ^{-1}$, is$........cm$. $\left(\right.$ take $\left.g=10\,ms ^{-2}\right)$

A

$60$

B

$30$

C

$120$

D

$150$

(JEE MAIN-2022)

Solution

From energy conservation

$mgh =\frac{1}{2} m v^{2}+\frac{1}{2} I ^{2}$

$mgh =\frac{1}{2} mv ^{2}+\frac{1}{2} \frac{ mR ^{2}}{2} \omega^{2}$

$10\,h =\frac{16}{2}+\frac{16}{4} \Rightarrow h =1.2\,m =120\,cm$

Standard 11

Physics

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