MHT CET 2025 19th April Morning Shift | Gravitation Question 14 | Physics

Two particles of equal mass ' $m$ ' move in a circle of radius ' $r$ ' under the action of their mutual gravitational attraction. The speed of each particle will be ( $\mathrm{G}=$ Universal gravitational constant)

$\sqrt{\frac{\mathrm{Gm}}{4 \mathrm{r}}}$

$\sqrt{\frac{\mathrm{Gm}}{\mathrm{r}}}$

$\sqrt{\frac{\mathrm{Gm}}{2\mathrm{r}}}$

$\sqrt{\frac{4 \mathrm{Gm}}{\mathrm{r}}}$

MHT CET 2024 16th May Evening Shift

MCQ (Single Correct Answer)

-0

A boy weighs 72 N on the surface of earth. The gravitational force on a body due to earth at a height equal to half the radius of earth will be

32 N

48 N

96 N

162 N

MHT CET 2024 16th May Morning Shift

MCQ (Single Correct Answer)

-0

A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$ '. The quantity $\mathrm{T}^2 \rho$ is equal to ( $\mathrm{G}=$ universal gravitational constant)

$\frac{4 \pi^2}{G}$

$\frac{3 \pi^2}{G}$

$\frac{3 \pi}{G}$

$\frac{\pi}{G}$

MHT CET 2024 16th May Morning Shift

MCQ (Single Correct Answer)

-0

The speed with which the earth would have to rotate about its axis so that a person on the equator would weigh $\frac{3}{5}$ th as much as at present weight is ( $\mathrm{g}=$ gravitational acceleration, $\mathrm{R}=$ equatorial radius of the earth)

$\sqrt{\frac{2 g}{5 R}}$

$\sqrt{\frac{3 \mathrm{~g}}{5 \mathrm{R}}}$

$\sqrt{\frac{5 R}{3 g}}$

$\sqrt{\frac{3}{5}} \mathrm{gR}$

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