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Gravitation question

2023 · Q129
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Gravitation question

2023 · Q129

NEETPhysicsGravitationMCQ+4 / −1

The escape velocity of a body on the earth surface is 11.2 km/s11.2 \mathrm{~km} / \mathrm{s}11.2 km/s. If the same body is projected upward with velocity 22.4 km/s22.4 \mathrm{~km} / \mathrm{s}22.4 km/s, the velocity of this body at infinite distance from the centre of the earth will be:

  1. A
    11.22 km/s11.2 \sqrt{2} \mathrm{~km} / \mathrm{s}11.22​ km/s
  2. B
    Zero
  3. C
    11.2 km/s11.2 \mathrm{~km} / \mathrm{s}11.2 km/s
  4. D
    11.23 km/s11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}11.23​ km/s
View written solutionFree

Correct answer: D

$$V_{\infty}=\sqrt{V^2-V_e^2}$$

Given than

$$\mathrm{V}=2 \mathrm{~V}_e$$

So,

$$\begin{aligned} & \mathrm{V}_{\infty}=\sqrt{\left(2 \mathrm{~V}_e\right)^2-\mathrm{V}_e^2} \\ & \mathrm{~V}_{\infty}=\sqrt{3} \mathrm{~V}_e=11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}\end{aligned}$$

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