MHT CET 2021 20th September Evening Shift | Gravitation Question 39 | Physics

The depth at which acceleration due to gravity becomes $$\frac{\mathrm{g}}{\mathrm{n}}$$ is [ $$\mathrm{R}$$ = radius of earth, $$\mathrm{g}=$$ acceleration due to gravity, $$\mathrm{n}=$$ integer $$]$$

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

$$\frac{(\mathrm{n}-1)}{\mathrm{nR}}$$

$$\frac{\mathrm{Rn}}{(\mathrm{n}-1)}$$

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

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

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

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

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

$$\frac{R}{n}$$

$$\frac{\mathrm{R}}{\mathrm{n}^2}$$

MHT CET 2021 20th September Morning Shift

MCQ (Single Correct Answer)

-0

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is 'V'. For the satellite orbiting at an altitude of half the earth's radius, the orbital velocity is

$$\frac{3}{2}$$V

$$\sqrt{\frac{3}{2}}$$V

$$\sqrt{\frac{2}{3}}$$V

$$\frac{2}{3}$$V

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

The mass of earth is 81 times the mass of the moon and the distance between their centres is $$R$$. The distance from the centre of the earth, where gravitational force will be zero is

$$\frac{R}{4}$$

$$\frac{R}{2}$$

$$\frac{9 R}{10}$$

$$\frac{R}{81}$$

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