JEE Main · 2010 · 11 Apr 2010 · Offline · Laws of Motion

A block sliding on a smooth plane inclined at angle $\varphi$ to the vertical accelerates down the slope at $g\cos \varphi$, since the slope makes $90^{\circ }-\varphi$ with the horizontal. The vertical part of that acceleration is $g\cos \varphi$ multiplied by $\cos \varphi$ again, namely $g{\cos }^2\varphi$. For the block on the plane at $30^{\circ }$ to the vertical this is $g{\cos }^{2}30^{\circ }=\frac{3}{4}g$, and for the one at $60^{\circ }$ it is $g{\cos }^{2}60^{\circ }=\frac{1}{4}g$. Their vertical accelerations differ by $g\left(\frac{3}{4}-\frac{1}{4}\right)=\frac{g}{2}=4.9$ m/s${}^2$, and since both parts are vertical, the relative acceleration of one block with respect to the other is $4.9$ m/s${}^2$ directed vertically.