For M block:
10gsin53∘−μ(10g)cos53∘−T=10×2
⇒T=80−15−20
⇒T=45N
For m block:
T–mgsin37∘−μmgcos37∘=m×2
⇒45=10m
⇒m=4.5kg
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…
JEE Main 2024 — Physics Mechanics
2024mcqmedium
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 2ms−2, is: (take g=10ms−2 and tan37∘=43)
Official previous-year question
Held on 31 Jan 2024 · Verified 6 Jul 2026.
Options
- A
9kg
- B
4.5kg
- C
6.5kg
- D
2.25kg
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