Q124 / 180 · JIPMER 2015 · Physics

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of $12 \mathrm{~m} / \mathrm{s}^2$. Find the displacement of the block during the first 0.2 s after the start. (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

  • A
    30 cm
  • B
    zero
  • Correct answer
    C
    20 cm
  • D
    25 cm

Solution

The acceleration (downward) of the elevator is $12 \mathrm{~m} / \mathrm{s}^2$ which is greater than gravitational acceleration $g$. So, the block will act as a freely falling body. From the motion of the block $\begin{aligned} s & =u t+\frac{1}{2} g t^2 \\ & =0 \times t+\frac{1}{2} \times 10 \times(0.2)^2 \\ & =\frac{1}{2} \times 10 \times 0.2 \times 0.2 \\ & =20 \mathrm{~cm}\end{aligned}$

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

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