In the arrangement shown in the figure at a particular instant, the roller is coming down with a speed of 12 m s-1 and C is moving up with 4 ms-1. At the same instant, it is also known that w.r.t. pulley P, blockr4 is moving down with speed 3 ms-1. Determine the motion of block B (velocity) w.r.t. ground.

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4 ms-1 in downward direction

3 ms-1 in upward direction

7 ms-1 in downward direction

7 ms-1 in upward direction

answer is D.

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Detailed Solution

Using constraint theory,l1+l2+l3= constant ⇒v1+2v2+v3=0Take downward as positive and upward as -ve. So 12+2(−4)+v3=0v3= velocity of pulley P=−4ms−1=4ms−1 in upward direction v→AP=−v→BP⇒ v→AP=−v→B−v→Pv→B=v→P−v→AP=−4−(3)=−7ms−1i.e., block B will moves up with speed 7 ms-1.

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