Q152 / 170 · Manipal (MET) 2021 · Physics

A heavy uniform rope hangs vertically from the ceiling, with its lower end free. A disturbance on the rope travelling upwards from the lower end has a velocity $v$ at a distance $x$ from the lower end such that :

  • A
    $v \propto x$
  • Correct answer
    B
    $v \propto \sqrt{x}$
  • C
    $v \propto \frac{1}{x}$
  • D
    $v \propto \frac{1}{\sqrt{x}}$

Solution

Let $m=$ mass per unit length of rope $T=$ tension in the rope at a distance $x$ from the lower end $\therefore T=(\mathrm{mg}) \mathrm{x}=$ weight of $x$ metre of rope As $v=\sqrt{\frac{T}{m}}$ $\therefore \quad v=\sqrt{\frac{m g x}{m}}=\sqrt{g x}$ i.e., $\quad v \propto \sqrt{x}$

Hence, the correct answer is option: 'B'.

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