Q132 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Physics

Let the escape speed of an object on the earth's surface be $V_0$. The object is projected out with speed $5 V_0$. The speed of the object far away from the earth will be

  • Correct answer
    A
    $2 \sqrt{6} V_0$
  • B
    $4 V_0$
  • C
    $2 \sqrt{3} V_0$
  • D
    $3 \sqrt{2} V_0$

Solution

By law of conservation of mechanical energy $\begin{aligned} & \frac{1}{2} \mathrm{~m}\left(5 \mathrm{~V}_0\right)^2-\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mV}^2 \\ & \Rightarrow \quad \frac{1}{2} \mathrm{~m} \times 25 \times \mathrm{V}_0^2-\frac{2 \mathrm{GM}}{\mathrm{R}} \cdot \frac{\mathrm{m}}{2}=\frac{1}{2} \mathrm{mV}^2 \\ & \Rightarrow \quad \frac{1}{2} \mathrm{mV}_0^2 \times 25-\mathrm{V}_0^2 \cdot \frac{\mathrm{m}}{2}=\frac{1}{2} \mathrm{mV}^2 \\ & \Rightarrow \quad 24 \times \frac{1}{2} \mathrm{mV}_0^2=\frac{1}{2} \mathrm{mV}^2 \Rightarrow \mathrm{V}^2=24 \mathrm{~V}_0^2 \Rightarrow \mathrm{V}=2 \sqrt{6} \mathrm{~V}_0\end{aligned}$

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

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