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Question

# In below figure two blocks are placed on fixed inclined plane and friction coefficient between the blocks and inclined plane are μ1 and μ2 respectively as shown in figure. Value of μ1 and μ2 are given in column 1 and value of a1 and a2 are given in column 2 and column 3 respectively. Column 1 Column 2 Column 3(i)μ1=0.8(i)a1=2 m/s2(P)a2=2 m/s2 μ2=0.8 (ii)μ1=0.5(ii)a1=0.8 m/s2(Q)a2=0.8 m/s2 μ2=0.5 (iii)μ1=0.5(ii)a1=4.4 m/s2(R)a2=4.4 m/s2 μ2=0.2 (iv)μ1=0.5(iv)a1=0(S)a2=0 μ2=0.8 Which of the following combination is correct?

A

(III) (i) (Q)

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B

(III) (iv) (Q)

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C

(III) (i) (R)

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D

(III) (ii) (R)

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Solution

## The correct option is C (III) (i) (R) (I) As μ1 and μ2>tanθ ∴ Both block will remain at rest Hence, a1=a2=0 and N=0 (II) a1=a2=g(sin37∘−μ cos37∘) ⇒a1=a2=2m/s2 and N=0 (III) a1=g(sin37∘–μ1cos37∘) a1=2m/s2 a2=g(sin37∘−μ2cos37∘) a2=4.4m/s2 and N=0 (IV) a1=a2=4g sin37∘−μ12g cos37∘−μ22g cos37∘4 ⇒a1=a2=0.8m/s2 and N≠0

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