1
MHT CET 2023 14th May Morning Shift
+1
-0

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

A
$$g=\sqrt{\frac{3 \pi R}{\rho G}}$$
B
$$\mathrm{g}=\sqrt{\frac{4}{3} \pi \rho \mathrm{GR}}$$
C
$$\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{GR}$$
D
$$g=\frac{G M}{\rho R^2}$$
2
MHT CET 2023 14th May Morning Shift
+1
-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

A
$$1: 1$$
B
$$\frac{R-2 h}{R-d}$$
C
$$\frac{R-d}{R-2 h}$$
D
$$\frac{\mathrm{R}-\mathrm{d}}{\mathrm{R}-\mathrm{h}}$$
3
MHT CET 2023 13th May Evening Shift
+1
-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

A
$$\frac{n y}{m x}$$
B
$$\frac{m y}{n x}$$
C
$$\frac{n x}{m y}$$
D
$$\frac{m x}{n y}$$
4
MHT CET 2023 13th May Evening Shift
+1
-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)

A
$$g$$
B
$$3 g$$
C
$$\frac{2 g}{3}$$
D
$$\frac{g}{3}$$
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