MHT CET 2025 22nd April Morning Shift | Gravitation Question 6 | Physics

Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' $2 R$ ' above earth's surface. The value of $x$ will be $(R=$ radius of earth)

MHT CET 2025 22nd April Morning Shift

MCQ (Single Correct Answer)

-0

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

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

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

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

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

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