JEE Main Physics question

Physics Laws of Motion JEE Main

A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction between them is √3. The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0°. Then [IIT JEE 2009]

(A)At θ = 30°, the block will start sliding down the plane
(B)The block will remain at rest on the plane up to certain θ and then it will toppleCorrect
(C)At θ = 60°, the block will start sliding down the plane and continue to do so at higher angles
(D)At θ = 60°, the block will start sliding down the plane and on further increasing θ, it will topple

Answer: (B)

Solution

Sliding requires tanθ > μ = √3, i.e. θ > 60°. Toppling about the lower edge occurs when tanθ exceeds (base/2)/(height/2) = 10/15 = 2/3, i.e. θ > tan⁻¹(2/3) ≈ 33.7°. Since

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Question ID VK-152957