Question

The ring $M_1$ and block $M_2$ are held in the position shown in figure. Now the system is released. If $M_1$ $>$ $M_2$ find $\frac{v_1}{v_2}$ when the ring $M_1$ has descended along the smooth fixed vertical rod by the distance $y=h$ .

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let ring has moved down a distance $y$. From figure (b), we have $y^2+l^2=s^2$ ....(i)

Here $l$ is the length of string which is constant differentiating equation (i) with respect to, time, we get

$2y\frac{dy}{dt}+0=2s\frac{ds}{dt}$ or $\frac{dy}{dt}=\frac{s}{y}.\frac{ds}{dt}$ .........(ii)

Here $\frac{dy}{dt}=v_1$ and $\frac{ds}{dt}=v_2$ For $y$ = $h,s$ = $\sqrt{h^2+l^2}$

$\therefore$ equation (ii) takes the form $v_1$ = $\frac{\sqrt{h^2+l^2}}{h}.v_2$ or $\frac{v_1}{v_2}=\frac{\sqrt{h^2+l^2}}{h}$