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 $\mathrm{M}$ and $\mathrm{m}$ 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 $\mathrm{m}$, for which $\mathrm{M}=10 \mathrm{~kg}$ will move down with an acceleration of $2 \mathrm{~m} / \mathrm{s}^{2}$, is : $\left(\right.$ take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ and $\left.\tan 37^{\circ}=3 / 4\right)$
(1) $9 \mathrm{~kg}$
(2) $4.5 \mathrm{~kg}$
(3) $6.5 \mathrm{~kg}$
(4) $2.25 \mathrm{~kg}$
Show Answer
Answer: (2)
Solution:
For M block
$10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 \times 2$
$\mathrm{T}=80-15-20$
$\mathrm{T}=45 \mathrm{~N}$
For $\mathrm{m}$ block
$\mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} \times 2$
$45=10 \mathrm{~m}$
$\mathrm{m}=4.5 \mathrm{~kg}$
