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Question: A heavy uniform rope hangs vertically from the ceiling, with its lower end free. A disturbance on th...

A heavy uniform rope hangs vertically from the ceiling, with its lower end free. A disturbance on the rope travelling upward from the lower end has a velocity v at a distance x from the lower end. Then –

A

vµ 1/x

B

v µ x

C

v µ x\sqrt{x}

D

v µ 1/x\sqrt{x}

Answer

v µ x\sqrt{x}

Explanation

Solution

v = Tm\sqrt{\frac{T}{m}} and T = mxlg\frac{mx}{\mathcal{l}}g

v = gx\sqrt{gx} or v µ x\sqrt{x}