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 $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then

(B) the block will remain at rest on the plane up to certain $\theta$ and then it will topple

(A) at $\theta=30^{\circ}$, the block will start sliding down the plane

(C) at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles

(D) at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Laws of Motion

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^{\circ}

. Then

(A)at $θ = 3 0^{\circ}$ , 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 topple✓ Correct
(C)at $θ = 6 0^{\circ}$ , the block will start sliding down the plane and continue to do so at higher angles
(D)at $θ = 6 0^{\circ}$ , the block will start sliding down the plane and on further increasing $θ$ , it will topple at certain $θ$

Answer: (B)

For sliding to occur,

tan θ \geq μ = 3 \approx 1.732

, which gives

θ \geq 6 0^{\circ}

. For toppling to occur, the vertical line through the center of mass must pass outside the base. Taking moments about the edge, toppling occurs when

tan θ \geq \frac{base /2}{height /2} = \frac{5}{7.5} = \frac{2}{3} \approx 0.67

. Since

\frac{2}{3} < 3

, toppling will occur before sliding. The block will remain at rest up to a certain angle and then topple.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option