MHT CET 2023 14th May Morning Shift | Gravitation Question 35 | Physics

Considering earth to be a sphere of radius '$$R$$' having uniform density '$$\rho$$', then value of acceleration due to gravity '$$g$$' in terms of $$R, \rho$$ and $$\mathrm{G}$$ is

$$g=\sqrt{\frac{3 \pi R}{\rho G}}$$

$$\mathrm{g}=\sqrt{\frac{4}{3} \pi \rho \mathrm{GR}}$$

$$\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}$$

$$g=\frac{G M}{\rho R^2}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The value of acceleration due to gravity at a depth '$$d$$' from the surface of earth and at an altitude '$$h$$' from the surface of earth are in the ratio

$$1: 1$$

$$\frac{R-2 h}{R-d}$$

$$\frac{R-d}{R-2 h}$$

$$\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}-\mathrm{h}}$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

-0

If two planets have their radii in the ratio $$x: y$$ and densities in the ratio $$m: n$$, then the acceleration due to gravity on them are in the ratio

$$\frac{n y}{m x}$$

$$\frac{m y}{n x}$$

$$\frac{n x}{m y}$$

$$\frac{m x}{n y}$$

MHT CET 2023 13th May Evening Shift

MCQ (Single Correct Answer)

-0

A mine is located at depth $$R / 3$$ below earth's surface. The acceleration due to gravity at that depth in mine is ($$R=$$ radius of earth, $$g=$$ acceleration due to gravity)

$$g$$

$$3 g$$

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

$$\frac{g}{3}$$

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