JEE Advanced
Physics
Laws of Motion
2021
JEE Advanced 2021 (Paper 2)
JEE Advanced Physics Question (2021) — Solution
Question
One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q , at a height L above the hinge at point O . A block of weight α W is attached at the point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of 2 2 W . Which of the following statement(s) is(are) correct?
Options
- A. The vertical component of reaction force at O does not depend on α .
- B. The horizontal component of reaction force at O is equal to W for α = 0 . 5 .
- C. The tension in the rope is 2 W for α = 0 . 5 .
- D. The rope breaks if α > 1 . 5 .
Answer
D. The rope breaks if α > 1 . 5 .
Step-by-step solution
∵ Rod is in equilibrium, net torque on the rod about O will be zero. W l 2 + α W l = T 2 l ⇒ W 2 + α W = T 2 ⇒ T = 2 W α + 0 . 5       . . . 1 Now checking options- (A) Force balance on rod, N v + T 2 = W + α W ⇒ N v = W + α W - W 2 + α W ⇒ N v = W 2 Hence, option (A) is correct. (B) Write horizontal force equation, N H = T 2 ⇒ N H = W α + 0 . 5 If α = 0 . 5 , then, N H = W and T = 2 W Hence, option (B) is correct. (D) ∴ Maximum Tension of string = 2 2 W So, to break. T > 2 2 W ⇒ 2 W α + 0 . 5 > 2 2 W ⇒ α > 1 . 5 So, string will break. option (D) correct.
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