Q125 / 160 · TS EAMCET 2006 · Physics

A bullet of mass $10 \mathrm{~g}$ is fired horizontally with a velocity $1000 \mathrm{~ms}^{-1}$ from a rifle situated at a height $50 \mathrm{~m}$ above the ground. If the bullet reaches the ground with a velocity $500 \mathrm{~ms}^{-1}$, the work done against air resistance in the trajectory of the bullet is : $\left(g=10 \mathrm{~ms}^{-2}\right)$

  • A
    $5005 \mathrm{~J}$
  • B
    $3755 \mathrm{~J}$
  • Correct answer
    C
    $3750 \mathrm{~J}$
  • D
    $17.5 \mathrm{~J}$

Solution

From equation of motion, $v^2=u^2-2 a s$ $(500)^2=(1000)^2-2 \times a \times s$ $s=\frac{(1000)^2-(500)^2}{2 a}=\frac{375000}{a}$ $\therefore$ Work done against air resistance $=F S$ $=m a \times s$ $=\frac{10}{1000} a \times \frac{375000}{a}$ $=3750 \mathrm{~J}$

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

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