Laws of Motion 3 Question 8

8. A block of mass m1=1kg another mass m2=2kg 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 the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to μ=0.3.

(2014 Adv.)

In List II expressions for the friction on the 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]

List I List II
P. θ=5 1. m2gsinθ
Q. θ=10 2. (m1+m2)gsinθ
R. θ=15 3. μm2gcosθ
S. θ=20 4. μ(m1+m2)gcosθ

Codes

(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

Show Answer

Answer:

Correct Answer: 8. (d)

Solution:

  1. Block will not slip if

(m1+m2)gsinθμm2gcosθ

3sinθ310(2)cosθtanθ1/5θ11.5

(P) θ=5 friction is static

f=(m1+m2)gsinθ

(Q) θ=10 friction is static

f=(m1+m2)gsinθ

(R) θ=15 friction is kinetic

f=μm2gcosθ

(S) θ=20 friction is kinetic

f=μm2gcosθ



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