### Laws Of Motion Question 13

#### Question 13 - 2024 (31 Jan Shift 1)

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 $M=10 kg$ will move down with an acceleration of $2 m / s^{2}$, is : $\left(\right.$ take $g=10 m / s^{2}$ and $\left.\tan 37^{\circ}=3 / 4\right)$

(1) $9 kg$

(2) $4.5 kg$

(3) $6.5 kg$

(4) $2.25 kg$

## Show Answer

#### Answer: (2)

#### Solution:

For M block

$10 g \sin 53^{\circ}-\mu(10 g) \cos 53^{\circ}-T=10 \times 2$

$T=80-15-20$

$T=45 N$

For $m$ block

$T-mg \sin 37^{\circ}-\mu mg \cos 37^{\circ}=m \times 2$

$45=10 m$

$m=4.5 kg$