Question 141 of 210

In the given arrangement of a doubly inclined plane two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$, for which $M=10kg$ will move down with an acceleration of $2m{s}^{-2}$, is: (take $g=10m{s}^{-2}$ and $\tan 37^{\circ}=\dfrac{3}{4}$)

A$9kg$B$4.5kg$C$6.5kg$D$2.25kg$

Show answer and solution▾

Answer: (B) $4.5kg$

For $M$ block:
$10g\sin 53^{\circ}-\mu (10g)\cos 53^{\circ}-T=10\times 2$
$\Rightarrow T=80-15-20$
$\Rightarrow T=45N$
For $m$ block:
$T–mg\sin 37^{\circ}-\mu mg\cos 37^{\circ}=m\times 2$
$\Rightarrow 45=10m$
$\Rightarrow m=4.5kg$