Question

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

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

Answer 1

The Correct Option is B

In the right angle $\Delta$PQR
$ \, \, \, \, \, t^2=c^2+y^2$
Differentiating this
equation with respect to
time, we ge
$2t\frac{dl}{dt}=0+2y\frac{dy}{dt}$
or $ \, \, \, \, \, \, \, \big(-\frac{dy}{dt}\bigg)=\frac{l}{y}\bigg(-\frac{dl}{dt}\bigg)$
$Here, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -\frac{dy}{dt}=v_{M} \, \, \, \Rightarrow \, \, \, \frac{1}{y}=\frac{1}{cos\theta}$
and -dl / dt =U
Hence,$ \, \, \, \, \, \, \, \, \, \, v_M=\frac{U}{cos\theta}$
$\therefore \, \, $ Correct option is (b).
Hence ,$ \, \, \, \, \, \, \, \, \, \, v_M=\frac{U}{cos\theta}$
$\therefore \, \, $ Correct option is (b).