MHT CET 2024 11th May Morning Shift | Gravitation Question 12 | Physics

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

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

$R\left(\frac{n-2}{n-1}\right)$

$R\left(\frac{2 n-1}{n}\right)$

$\mathrm{R}\left(\frac{\mathrm{n}-1}{2 \mathrm{n}-1}\right)$

MHT CET 2024 11th May Morning Shift

MCQ (Single Correct Answer)

-0

Assuming that the earth is revolving around the sun in circular orbit of radius R , the angular momentum is directly proportional to $\mathrm{R}^{\mathrm{n}}$. The value of ' $n$ ' is

1.5

0.5

MHT CET 2024 10th May Evening Shift

MCQ (Single Correct Answer)

-0

The height ' h ' from the surface of the earth at which the value of ' $g$ ' will be reduced by $64 \%$ than the value at surface of the earth is ( $\mathrm{R}=$ radius of the earth)

$\frac{1}{3} R$

$\frac{2}{3} R$

$\frac{3}{2} R$

$\mathrm{2 R}$

MHT CET 2024 10th May Evening Shift

MCQ (Single Correct Answer)

-0

A body starts from rest from a distance $\mathrm{R}_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( $R=$ radius of earth, $M=$ mass of earth)

$2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

$\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$

$\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

$2 \mathrm{GM} \sqrt{\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$

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