Q105 / 200 · NEET 2010 (Screening) · Physics

A man of $50 \mathrm{~kg}$ mass is standing in a gravity free space at a height of $10 \mathrm{~m}$ above the floor. He throws a stone of $0.5 \mathrm{~kg}$ mass downwards with a speed $2 \mathrm{~ms}^{-1}$. When the stone reaches the floor, the distance of the man above the floor will be

  • A
    $9.9 \mathrm{~m}$
  • Correct answer
    B
    $10.1 \mathrm{~m}$
  • C
    $10 \mathrm{~m}$
  • D
    $20 \mathrm{~m}$

Solution

$m r=$ constant $$ \begin{aligned} m_1 r_1 & =m_2 r_2 \\ r_2 & =\frac{m_1 r_1}{m_2} \\ & =\frac{0.5 \times 10}{50}=0.1 \end{aligned} $$ The distance of the man above the floor (total height $)=10+0.1=10.1 \mathrm{~m}$.

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

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