JEE Advanced (Matrix Match & Integer Answer): Laws of Motion | Chapter-wise Tests for JEE Main & Advanced PDF Download

Match the Following

DIRECTIONS (Q. No. 1) : Following question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q.1. A block of mass m₁ = 1 kg another mass m₂ = 2 kg, 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 m₁ and plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to μ = 0.3. In List-II expressions for the friction on block m₂ 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°) ≈ 0.1; tan (11.5°) ≈ 0.2; tan (16.5°) ≈ 0.3] 

Code: 

(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

Ans. (d)

Solution. If (m₁ + m₂) sin θ < μN₂ the bodies will be at rest i.e., (m₁ + m₂)g sin θ < μm₂ g cos θ

⇒ tan θ < 0.2 i.e., 

If the angle θ < 11.5° the frictional force is less than 

µN₂  = µm₂g = 0.3 × 2 × g = 0.6 g 

and is equal to (m₁ + m₂)g sin θ 

At θ = 11.5° the bodies are on the verge of moving, f = 0.6 g 

At θ > 11.5° the bodies start moving and f = 0.6 g 

The above relationship is true for (d).

Integer Value Correct Type

Q.1. A block is moving on an inclined plane making an angle 45° with the horizontal and the coefficient of friction is μ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 μ, then N is                    (2011)

Ans. 5

Solution.

The pushing force F₁ = mg sinθ + f 

∴ F₁ = mg sin θ + μ mg cos θ  = mg (sin θ + μ cos θ) 

The force required to just prevent it from sliding down 

F₂ = mg sin θ – µN = mg (sin θ – μ cos θ) 

Given ,   F₁ = 3F₂ 

∴ sin θ + μcos θ = 3(sin θ – μ cos θ)
∴ 1 + μ = 3(1 – μ)  [∴ sin θ = cos θ]
∴ 4μ = 2
∴ μ = 0.5
∴ N = 10 μ = 5