question_answer

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

A) \[2u\cos \theta \]

B) \[u\cos \theta \]

C) \[\frac{2u}{\cos \theta }\]

D) \[\frac{u}{\cos \theta }\]  

Correct Answer: D

Solution :

 Referring figure,             \[{{l}^{2}}={{b}^{2}}+{{y}^{2}}\] Differentiating with respect to time, we get             \[2l\frac{dl}{dt}=2y\frac{dy}{dt}\] \[\Rightarrow \]   \[\frac{dl}{dt}=\frac{ldy}{ydt}=\frac{1}{\cos \theta }\cdot \frac{dl}{dt}=\frac{u}{\cos \theta }\] (since as \[P\] and \[Q\] move down, the length\[l\]decreases at the rate of \[U\,\,m/s\])