Q109 / 150 · MHT CET 2021 (21 Sep Shift 2) · Physics

A body at rest falls through a height ' $h$ ' with velocity ' $\mathrm{V}$ '. If it has to fall down further for its velocity to become three times, the distance travelled in that interval is

  • Correct answer
    A
    8 h
  • B
    6 h
  • C
    4 h
  • D
    12 h

Solution

The body acquired velocity $\mathrm{V}$ when it falls through a height $\mathrm{h}$, starting from rest. $$ \begin{aligned} & \therefore \mathrm{V}^2=2 \mathrm{gh} \\ & \therefore \mathrm{h}=\frac{\mathrm{V}^2}{2 \mathrm{~g}} \end{aligned} $$ If it falls further and attains velocity $3 \mathrm{~V}$ and if the total height through which it falls is h', then $$ \begin{aligned} & \left(3 \mathrm{~V}^2\right)=2 \mathrm{gh}^{\prime} \\ & \therefore 9 \mathrm{~V}^2=2 \mathrm{gh}^{\prime} \\ & \therefore \mathrm{h}^{\prime}=\frac{9 \mathrm{~V}^2}{2 \mathrm{~g}}=9 \mathrm{~h} \\ & \therefore \mathrm{h}^{\prime}-\mathrm{h}=9 \mathrm{~h}-\mathrm{h}=8 \mathrm{~h} \end{aligned} $$

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

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