Q60 / 60 · JEE Advanced 2014 (Paper 2) · Physics

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 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: tan⁡5.5o≈0.1;tan⁡11.5o≈0.2;tan⁡16.5o≈0.3 ]
Figure
 

 List I List II
A.θ=5oP.m2gsin⁡θ
B.θ=10oQ.m1+m2gsin⁡θ
C.θ=15oR.μm2 gcos⁡θ
D.θ=20oS.μm1+m2gcos⁡θ

  • Correct answer
    A
    a-q;b-q;c-r;d-r;
  • B
    a-q;b-s;c-p;d-r;
  • C
    a-q;b-p;c-s;d-r;
  • D
    a-r;b-p;c-s;d-q;

Solution

Condition for not sliding,
fmax>m1+m2gsin⁡θ
μN>m1+m2gsin⁡θ
0.3 m2 gcos⁡θ≥30sin⁡θ
6≥30tan⁡θ
15≥tan⁡θ
0.2 ≥tan⁡θ
∴ for P, Q
f=m1+m2gsin⁡θ
For R and S
F=fmax⁡=μm2gsin⁡θ

Figure

Hence, the correct answer is option: 'A'.

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