MHT CET (PCB) 2025 9th April Evening Shift | Gravitation Question 1 | Physics

An object is released from a distance ' $r$ ' from the centre of the earth. The velocity of the object at the time of striking the earth will be ( $R=$ radius of earth, $r>R, g=$ acceleration due to gravity)

$[2 \mathrm{gR}(\mathrm{r}-\mathrm{R}) / \mathrm{r}]^{\frac{1}{2}}$

$\left[g R(r-R) / r^2\right]^{\frac{1}{2}}$

$\quad\left[2 \mathrm{gR}(\mathrm{r}-\mathrm{R}) / \mathrm{r}^2\right]^{\frac{1}{2}}$

$\left[g R^2(r-R) / r^2\right]^{\frac{1}{2}}$

MHT CET (PCB) 2025 9th April Morning Shift

MCQ (Single Correct Answer)

-0

The value of acceleration due to gravity (g) becomes $\left(\frac{g}{3}\right)$ at height ' $h$ ' above the earth's surface. If ' R ' is the radius of earth, the height h will be equal to

$\sqrt{3} \mathrm{R}$

3 R

$(\sqrt{3}-1) \mathrm{R}$

$(\sqrt{3}+1) R$

MHT CET (PCB) 2024 22th April Evening Shift

MCQ (Single Correct Answer)

-0

The period of revolution of planet A around the sun is 8 times that of $B$. The distance of $A$ from the sun is how many times greater than that of $B$ from the sun?

MHT CET (PCB) 2024 22th April Evening Shift

MCQ (Single Correct Answer)

-0

A satellite of mass ' $m$ ' is orbiting the earth of radius ' $R$ ' at height ' $h$ ' from the surface of earth. The total energy of the satellite is ( $\mathrm{g}=$ acceleration due to gravity at the earth's surface)

$-\frac{\mathrm{mgR}^2}{2(\mathrm{R}+\mathrm{h})}$

$+\frac{m g R^2}{2(R+h)}$

$\frac{2 \mathrm{mgR}^2}{(\mathrm{R}+\mathrm{h})}$

$\quad-\frac{2 \mathrm{mgR}^2}{(\mathrm{R}+\mathrm{h})}$

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