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QJEEM24Ph5d6f: JEE Main 2024 Physics - Class 11 JEEM24Ph5d6f Solution: In the given arrangement of a doubly inclined plane two blocks…

In the given arrangement of a doubly inclined plane two blocks of masses MM and mm 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 mm , for which M=10 kgM=10 \mathrm{~kg} will move down with an acceleration of 2 m/s22 \mathrm{~m} / \mathrm{s}^2 , is: (take g=10 m/s2\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 and tan⁡37∘=3/4)\left.\tan 37^{\circ}=3 / 4\right)

JEE Main 2024 (Online) 31st January Morning Shift Physics - Laws of Motion Question 19 English

Multiple Choice Options

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

Step-by-Step Verified Solution

JEE Main 2024 (Online) 31st January Morning Shift Physics - Laws of Motion Question 19 English Explanation

For M block

amp;10 gsin⁡53∘−μ(10 g)cos⁡53∘−T=10×2amp; T=80−15−20amp; T=45 N\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 m\mathrm{m} block

amp;T−mgsin⁡37∘−μmgcos⁡37∘=m×2amp;45=10 mamp; m=4.5 kg\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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