Question

In the arrangement shown in figure, pulleys are fixed and the ends $P$ and $Q$ of an unstretchable string move downwards with a uniform speed $u$. Mass $m$ moves upwards with a speed of

• A. $\text{u cos}\theta$
• B. $\frac{u}{\text{cos}\theta }$
• C. $\frac{2u}{\text{cos}\theta }$
• D. $\text{u sin}\theta$

Answer 1

The correct option is B $\frac{u}{\text{cos}\theta }$

Let $v$ be the upward velocity of block of mass $m$.
The given condition is shown in the below figure.


Let $c_1$ and $c_2$ be the parts of string on pulleys and $L$ be the length of the string connecting pulley and mass $m$.
So, from method of intercept we have,
$x_2+c_1+2L+c_2+x_1=\text{constant}$
$\Rightarrow x_2+c_1+2\sqrt{y^2+d^2}+c_2+x_1=\text{constant}$
$\Rightarrow \frac{d{x}_2}{dt}+0+2\frac{1}{2\sqrt{y^2+d^2}}2y\frac{dy}{dt}+0+\frac{d{x}_1}{dt}=0$
$\Rightarrow \frac{d{x}_2}{dt}+\frac{d{x}_1}{dt}+2\left(\frac{y}{\sqrt{y^2+d^2}}\right)\frac{dy}{dt}=0$
$\Rightarrow u+u+2\left(\cos \theta \right)v=0$
$\Rightarrow 2u+\left(2\cos \theta \right)v=0\Rightarrow u+vcos\theta =0$
$\Rightarrow v=\frac{-u}{\cos \theta }$
This $-ve$ sign shows that the mass $m$ will move up as opposite to the string. Thus, magnitude of $v$ will be $\frac{u}{\cos \theta }$