Q124 / 160 · AP EAMCET 2023 (16 May Shift 2) · Physics

Two bodies of masses ' $\mathrm{m}_1{ }^{\prime}$ and ' $\mathrm{m}_2$ ' are dropped from two different heights $h_1$ and $h_2$ respectively. The ratio of the times taken by the two masses to touch the ground is (neglect air resistance)

  • A
    $\frac{\mathrm{h}_1}{\mathrm{~h}_2}$
  • B
    $\frac{m_1 h_1}{m_2 h_2}$
  • C
    $\frac{\mathrm{m}_1 \mathrm{~h}_2}{\mathrm{~m}_2 \mathrm{~h}_1}$
  • Correct answer
    D
    $\sqrt{\frac{\mathrm{h}_1}{\mathrm{~h}_2}}$

Solution

from the equation of motion $\begin{aligned} & \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\ & \text { take } \mathrm{u}=0 \mathrm{~S}_1=\mathrm{h}_1, \mathrm{~S}_2=\mathrm{h}_2 \\ & \mathrm{~h}_1=\frac{1}{2} \mathrm{~g} \mathrm{t}_1{ }^2 ; \mathrm{h}_2=\frac{1}{2} \mathrm{gt}_2{ }^2 \\ & \Rightarrow \frac{\mathrm{t}_1}{\mathrm{t}_2}=\sqrt{\frac{\mathrm{h}_1}{\mathrm{~h}_2}} \end{aligned}$

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

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