MHT CET 2024 11th May Evening Shift | Gravitation Question 10 | Physics

A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into space with velocity ' $n V_{\mathrm{e}}$ ' $(\mathrm{n}<1)$. The maximum height from the surface of earth to which a body can reach is

$\frac{\mathrm{n}^2 \mathrm{R}}{\left(1-\mathrm{n}^2\right)}$

$\frac{\mathrm{n}^2 \mathrm{R}^2}{(1-\mathrm{n})}$

$\frac{n R^2}{\left(1+n^2\right)}$

$\frac{\mathrm{n}^2 \mathrm{R}^2}{(1+\mathrm{n})}$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

$\frac{\mathrm{R}}{9}$

$\frac{\mathrm{R}}{3}$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

The depth 'd' at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth)

$R\left(\frac{n}{n-1}\right)$

$R\left(\frac{n-2}{n-1}\right)$

$R\left(\frac{2 n-1}{n}\right)$

$\mathrm{R}\left(\frac{\mathrm{n}-1}{2 \mathrm{n}-1}\right)$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

Assuming that the earth is revolving around the sun in circular orbit of radius R , the angular momentum is directly proportional to $\mathrm{R}^{\mathrm{n}}$. The value of ' $n$ ' is

1.5

0.5

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