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In the arrangement shown in the figure, the ends P and Q of an un stretchable string move downwards uniform speed U. Pulleys A and $B$ are fixed. Mass M moves upwards with a speed.

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$2 U \cos θ$

$U \cos θ$

$\frac{2 U}{\cos θ}$

$\frac{U}{\cos θ}$

answer is D.

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Detailed Solution

In the right angle $∆ P Q R$

$l^{2} = c^{2} + y^{2}$

Differentiating this equation with respect to time, we get

$2 l \frac{d l}{d t} = 0 + 2 y \frac{d y}{d t} or (- \frac{d y}{d t}) = \frac{l}{y} (- \frac{d l}{d t})$

In the arrangement shown in the figure the ends P and class 11 physics CBSE

Here,

$- \frac{d y}{d t} = v_{M}, \frac{l}{y} = \frac{1}{\cos θ}$ and $- d l / d t = U$

Hence,

$v_{M} = \frac{U}{\cos θ}$

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