MHT CET 2025 22nd April Morning Shift | Gravitation Question 7 | Physics

Two satellites P and Q go round a planet in circular orbits having radii ' $3 R$ ' and ' $R$ ' respectively. If the speed of satellite $P$ is ' $2 V$ ', the speed of the satellite $Q$ will be

$2 \sqrt{3} \mathrm{~V}$

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

$\frac{\mathrm{V}}{2}$

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

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

The radii of circular orbits of two satellites $A$ and $B$ of the earth are ' $4 R$ ' and ' $R$ ' respectively, where $R$ is the radius of earth. If the speed of satellite $B$ is 6 V , then the speed of satellite $A$ will be

3 V

4 V

12 V

$\frac{3}{4} \mathrm{~V}$

MHT CET 2025 21st April Morning Shift

MCQ (Single Correct Answer)

-0

Water rises up to height ' $x$ ' in a capillary tube immersed vertically in water. When the whole arrangement is taken to a depth ' $d$ ' in a mine, the water level rises up to height Y . If R is the radius of the earth then the ratio $\mathrm{Y}: \mathrm{x}$ is

$\quad \mathrm{R}:(\mathrm{R}+\mathrm{d})$

$\mathrm{R}:(\mathrm{R}-\mathrm{d})$

$\mathrm{R}:(\mathrm{R}-\mathrm{d})^2$

$\mathrm{R}:(\mathrm{R}+\mathrm{d})^2$

MHT CET 2025 21st April Morning Shift

MCQ (Single Correct Answer)

-0

The total energy of a circularly orbiting satellite is

half the kinetic energy of the satellite.

half the potential energy of the satellite.

twice the kinetic energy of the satellite.

twice the potential energy of the satellite.

PREVIOUS NEXT

MHT CET Subjects