IIT-JEE 2009 · Paper 1 · Physics Q42

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PhysicsSingle correct+3 / -1Question 42 of 60

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 $\sqrt3$ . The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from 0°. Then

\theta=30°

, the block will start sliding down the plane

the block will remain at rest on the plane up to certain

\theta

and then it will topple

\theta=60°

, the block will start sliding down the plane and continue to do so at higher angles

\theta=60°

, the block will start sliding down the plane and on further increasing

\theta

, it will topple at certain

\theta

Answer and worked solution

Official answer: B

Sliding starts when $\tan\theta=\mu=\sqrt3$ , i.e. $\theta=60°$ .

Toppling starts when the weight's line of action passes the lower edge: $\tan\theta=\dfrac{\text{base}}{\text{height}}=\dfrac{10}{15}=\dfrac23$ , i.e. $\theta\approx33.7°$ .

So the block stays at rest until about 34° and then topples, long before it could slide.

(B)

Topics:Toppling Friction

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IIT-JEE 2009 · Paper 1, Physics question 42