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In fig shown block B moves down with a velocity 10m/s.The velocity of A in the position shown is
Its answer is 12.5 m/sec till body A will remain at surface The ideal weight of a tempting the sets of question is breaking the component of velocity in the direction of string and then equating it equal to velocity of body BNet velocity of body A be vTake its component in the direction of string as v cos 37So 10 = 4/5 vV = 12.5
If velocity of A is V then velocity of B is 2VVelocity of A= 2× velocity of BV cos37° = 2×10V×4÷5 = 20V = 20×5÷4V=25m/s
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