BITSAT 2024 | Gravitation Question 1 | Physics

A body which is initially at rest at a height $ R $ above the surface of the Earth of radius $ R $, falls freely towards the Earth, then its velocity on reaching the surface of the Earth is

$ \sqrt{(2 g R)} $

$ \sqrt{(g R)} $

$ \sqrt{\frac{3}{2} g R} $

$ \sqrt{(4 g R)} $

BITSAT 2023

MCQ (Single Correct Answer)

-1

A man of mass $m$ starts falling towards a planet of mass $$M$$ and radius $$R$$. As he reaches near to the surface, he realizes that will pass through a small role in the planet. As he enters the role, we sees that the planet is really made of two pieces a spherical shell of negligible thickness of mass $$3 M / 4$$ and a point mass $$M / 4$$ at the centre. Change in the force of gravity experienced by the man is

$$\frac{3}{4} \frac{G M m}{R^2}$$

$$\frac{1}{3} \frac{G M m}{R^2}$$

$$\frac{4}{3} \frac{G M m}{R^2}$$

BITSAT 2022

MCQ (Single Correct Answer)

-1

What will be the acceleration due to gravity at a depth d, where g is acceleration due to gravity on the surface of earth?

$${g \over {{{\left[ {1 + {d \over R}} \right]}^2}}}$$

$$g\left[ {1 - {{2d} \over R}} \right]$$

$${g \over {{{\left[ {1 - {d \over R}} \right]}^2}}}$$

$$g\left[ {1 - {d \over R}} \right]$$

BITSAT 2021

MCQ (Single Correct Answer)

-1

From a solid sphere of mass M and radius R, a spherical portion of radius $${R \over 2}$$ is removed as shown in the figure.

BITSAT 2021 Physics - Gravitation Question 6 English

Taking gravitational potential V = 0 at r = $$\infty$$, the potential at the centre of the cavity thus formed is

$$ - {{GM} \over {2R}}$$

$$ - {{GM} \over {R}}$$

$$ - {{2GM} \over {3R}}$$

$$ - {{2GM} \over {R}}$$

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