JEE MAIN - Physics (2024 - 31st January Morning Shift - No. 4)
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 $$M=10 \mathrm{~kg}$$ will move down with an acceleration of $$2 \mathrm{~m} / \mathrm{s}^2$$, is: (take $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$ and $$\left.\tan 37^{\circ}=3 / 4\right)$$
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Explanation
_31st_January_Morning_Shift_en_4_2.png)
For M block
$$\begin{aligned} & 10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 \times 2 \\ & \mathrm{~T}=80-15-20 \\ & \mathrm{~T}=45 \mathrm{~N} \end{aligned}$$
For $$\mathrm{m}$$ block
$$\begin{aligned} & \mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} \times 2 \\ & 45=10 \mathrm{~m} \\ & \mathrm{~m}=4.5 \mathrm{~kg} \end{aligned}$$
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