MHT CET 2024 2nd May Morning Shift | Gravitation Question 45 | Physics

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

$\frac{\mathrm{R}}{\sqrt{\mathrm{n}}}$

$\mathrm{R} \cdot \sqrt{\mathrm{n}}$

$\quad(\sqrt{n}+1) R$

$(\sqrt{\mathrm{n}}-1) \mathrm{R}$

MHT CET 2024 2nd May Morning Shift

MCQ (Single Correct Answer)

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The magnitude of gravitational field at distance ' $r_1$ ' and ' $r_2$ ' from the centre of a uniform sphere of radius ' $R$ ' and mass ' $M$ ' are ' $F_1$ ' and ' $F_2$ ' respectively. The ratio ' $\left(F_1 / F_2\right)$ ' will be (if $r_1>R$ and $r_2

$\mathrm{\frac{R^2}{r_1 r_2}}$

$\frac{\mathrm{R}^3}{\mathrm{r}_1 \mathrm{r}_2^2}$

$\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}$

$\frac{\mathrm{R}^4}{\mathrm{r}_1^2 \mathrm{r}_2^2}$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

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Earth is assumed to be a sphere of radius R. If '$$\mathrm{g}_\phi$$' is value of effective acceleration due to gravity at latitude $$30^{\circ}$$ and '$$g$$' is the value at equator, then the value of $$\left|g-g_\phi\right|$$ is ($$\omega$$ is angular velocity of rotation of earth, $$\cos 30^{\circ}=\frac{\sqrt{3}}{2}$$ )

$$\frac{1}{4} \omega^2 \mathrm{R}$$

$$\frac{3}{4} \omega^2 \mathrm{R}$$

$$\omega^2 \mathrm{R}$$

$$\frac{1}{2} \omega^2 \mathrm{R}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

A body (mass $$\mathrm{m}$$ ) starts its motion from rest from a point distant $$R_0\left(R_0>R\right)$$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $$\mathrm{G}=$$ universal constant of gravitation, $$\mathrm{M}=$$ mass of earth, $$\mathrm{R}$$ = radius of earth)

$$2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$$

$$\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$$

$$\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$$

$$2 \mathrm{GM}\left[\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$$

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