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JEE Advanced · Physics · 5. Laws of Motion

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 . 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: tan5.5o0.1;tan11.5o0.2;tan16.5o0.3 ]

List IList II
A.θ=5oP.m2gsinθ
B.θ=10oQ.m1+m2gsinθ
C.θ=15oR.μm2gcosθ
D.θ=20oS.μm1+m2gcosθ

  1. A a-q;b-q;c-r;d-r;
  2. B a-q;b-s;c-p;d-r;
  3. C a-q;b-p;c-s;d-r;
  4. D a-r;b-p;c-s;d-q;
Verified Solution

Answer & Solution

Correct Answer

(A) a-q;b-q;c-r;d-r;

Step-by-step Solution

Detailed explanation

Condition for not sliding,
fmax>m1+m2gsinθ
μN>m1+m2gsinθ
0.3 m2 gcosθ30sinθ
630tanθ
15tanθ
0.2 tanθ
for P, Q
f=m1+m2gsinθ
For R and S
F=fmax=μm2gsinθ

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