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(D) Let $l$ be the length of the string from the pulley to the mass $M$. From the geometry of the figure,we have the relation $l^2 = b^2 + y^2$,where

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$\frac{U}{\cos 	heta}$

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(D) Let $l$ be the length of the string from the pulley to the mass $M$. From the geometry of the figure,we have the relation $l^2 = b^2 + y^2$,where $b$ is the horizontal distance from the pulley to the vertical axis of the mass,and $y$ is the vertical distance of the mass from the line joining the pulleys.Since the string is unstretchable and the ends $P$ and $Q$ move downwards with speed $U$,the length $l$ of the string segment decreases at a rate of $U$. Thus,$\frac{dl}{dt} = -U$.Differentiating the relation $l^2 = b^2 + y^2$ with respect to time $t$,we get:$2l \frac{dl}{dt} = 2b \frac{db}{dt} + 2y \frac{dy}{dt}$.Since the pulleys are fixed,the horizontal distance $b$ remains constant,so $\frac{db}{dt} = 0$.Substituting $\frac{dl}{dt} = -U$ and $\frac{db}{dt} = 0$ into the equation:$2l(-U) = 2y \frac{dy}{dt}$.Solving for the vertical speed of the mass $v_y = \frac{dy}{dt}$:$\frac{dy}{dt} = -\frac{l}{y} U$.From the triangle formed,$\cos 	heta = \frac{y}{l}$,so $\frac{l}{y} = \frac{1}{\cos 	heta}$.Therefore,the speed of the mass $M$ is $v = |\frac{dy}{dt}| = \frac{U}{\cos 	heta}$.

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

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