Question

In the arrangement shown in the figure, if $v_1$ and $v_2$ are instantaneous velocities of masses $M_1$ and $M_2$ respectively, and angle $ACB=2\theta$ at the instant, then

• A. $\theta ={\cos }^{-1}\left(\frac{v_2}{2v_1}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{v_1}{2v_2}\right)$
• C. $\theta ={\tan }^{-1}\left(\frac{v_1}{2v_2}\right)$
• D. $\theta ={\sin }^{-1}\left(\frac{v_1}{v_2}\right)$

Answer 1

The correct option is A $\theta ={\cos }^{-1}\left(\frac{v_2}{2v_1}\right)$

Let us consider $M_2$ moves down with velocity $v_2$ and $M_1$ and pulley 1 move up with velocity $v_1$.

Let the distance b/w $O$ and $C$ be $h$.
Also let $AO=OB=x=\text{constant}$
$\therefore AC=CB=\sqrt{x^2+h^2}$

Here, $AC,CB$ and the string above $M_2$ are the intercepts of the same string.
$\Rightarrow AC+CB+l_1=\text{constant}$

Differentiating w.r.t time, we get
$\Rightarrow \frac{d}{dt}\left(2\sqrt{x^2+h^2}\right)+\frac{d}{dt}\left(l_2\right)=0$
$\Rightarrow \frac{2}{2\sqrt{x^2+h^2}}2h\frac{dh}{dt}+\frac{d{l}_2}{dt}=0$

$\Rightarrow 2\cos \theta \frac{dh}{dt}-v_2=0$
(taking downward velocity negative and $\frac{h}{\sqrt{x^2+h^2}}=\cos \theta$)

$\Rightarrow v_2=2\cos \theta \frac{dh}{dt}$
Since $\frac{dh}{dt}$ is the rate with which the pulley (1) rises up, $\frac{dh}{dt}=v_1$
$\Rightarrow v_2=2\cos \theta v_1$
$\therefore \theta ={\cos }^{-1}\left(\frac{v_2}{2v_1}\right)$