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Two block of masses m_1=5/\sqrt3 kg and m_2=5kg are released from rest on a frictionless inclined plane as shown in fig then

Option: 1

The block A moves down the plane


Option: 2

The block B moves down the plane


Option: 3

Both blocks remain at rest.


Option: 4

Both blocks move down the plane.


Answers (1)

best_answer

\\ \text{Free body diagram of given system :}

\\ \text{Component of weight responsible for downward movement of the blocks is } mg\sin\theta

\\ \text{For block A :} \\ mg\sin\theta= mg\sin60^{\circ}= \frac{5}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=2.5 N \\ \\ \text{For block B :} \\ mg\sin\theta= mg\sin30^{\circ}= 5\times \frac{{1}}{2}=2.5 N

\\ \text{Since,} mg\sin60^{\circ}=mg\sin30^{\circ} \text{is same. Hence Both blocks remain at rest. }

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