JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Laws of Motion

A block of mass

m_{1} = 1 k g

another mass

m_{2} = 2 k g

, are placed together (see figure) on an inclined plane with angle of inclination

θ

. Various values of

θ

are given in List I. The coefficient of friction between the block

m_{1}

and the plane is always zero. The coefficient of static and dynamic friction between the block

m_{2}

and the plane are equal to

μ = 0.3

. In List II expressions for the friction on block

m_{2}

are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by

g

.
[Useful information:

tan (5. 5^{\circ}) \approx 0.1; tan (11. 5^{\circ}) \approx 0.2; tan (16. 5^{\circ}) \approx 0.3

]

List I	List II
P. $θ = 5^{\circ}$	1. $m_{2} g sin θ$
Q. $θ = 1 0^{\circ}$	2. $(m_{1} + m_{2}) g sin θ$
R. $θ = 1 5^{\circ}$	3. $μ m_{2} g cos θ$
S. $θ = 2 0^{\circ}$	4. $μ (m_{1} + m_{2}) g cos θ$

(A)P-1, Q-1, R-1, S-3
(B)P-2, Q-2, R-2, S-3
(C)P-2, Q-2, R-2, S-4
(D)P-2, Q-2, R-3, S-3✓ Correct

Answer: (D)

Solution

The critical angle for slipping is when

μ m_{2} g cos θ = (m_{1} + m_{2}) g sin θ

, giving

tan θ = \frac{μ m _{2}}{m _{1} + m _{2}} = \frac{0.3 \times 2}{3} = 0.2

, corresponding to

θ \approx 11. 5^{\circ}

.

**For

θ < 11. 5^{\circ}

** (P, Q): blocks move together without slipping. Friction on

m_{2}

must support both blocks:

f = (m_{1} + m_{2}) g sin θ

. P→2, Q→2.

**For

θ > 11. 5^{\circ}

** (R, S): blocks slip relative to each other. Friction on

m_{2}

is kinetic:

f = μ m_{2} g cos θ

. R→3, S→3.

Answer: P-2, Q-2, R-3, S-3.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Laws of Motion

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