Question

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

• A. $2Ucos\theta$
• B. $Ucos\theta$
• C. $\frac{2U}{cos\theta }$
• D. $\frac{U}{cos\theta }$

Answer 1

Answer: (D)

As $P$ and $Q$ fall down, the length $l$ decreases at the rate of U m/s

From the figure, $l^2=b^2+v^2$

Differentiating with respect to time

$2l\times \frac{dl}{dt}=2b\times \frac{db}{dt}+2y\times \frac{dy}{dt}$

(As $\frac{db}{dt}=0,\frac{dl}{dt}=U)$

$\Rightarrow \frac{dy}{dt}=\left(\frac{l}{y}\right)\times \frac{dl}{dt}$

$\Rightarrow \frac{dy}{dt}=\left(\frac{1}{cos\theta }\right)\times U$

$=\frac{U}{cos\theta }$