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NEET Physics : Gravitation

Short Answer Type

131.

Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes 1/4th that at earth's surface ?

Multiple Choice Questions

132.

The height vertically above the earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth)

8 R
9 R
10 R
20 R

133.

Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m respectively from the origin. The resulting gravitational potential due to this system at the origin will be

− G
$- \frac{8}{3} G$
$- \frac{4}{3} G$
− 4 G

134.

If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth's surface to a height equal to the radius R of the earth is

$\frac{mgR}{4}$
$\frac{mgR}{2}$
mgR
2mgR

135.

A body of mass m taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

mg2R
$\frac{2}{3} mgR$
3 mgR
$\frac{1}{3} mgR$

136.

Average distance of the Earth from the Sun is L₁. If one year of the Earth = D days, one year of another planet whose average distance from the Sun is L₂ will be

$D {(\frac{L_{2}}{L_{1}})}^{2} days$
$D {(\frac{L_{2}}{L_{1}})}^{3 / 2} days$
$D {(\frac{L_{2}}{L_{1}})}^{2 / 3} days$
$D (\frac{L_{2}}{L_{1}}) days$

137.

An artificial satellite moves in a circular orbit around the earth. Total energy of the satellite is given by E. The potential energy of the satellite is

- 2E
2E
$\frac{2 E}{3}$
$- \frac{2 E}{3}$

138.

Two particles of mass m₁ and m₂ approach each other due to their mutual gravitational attraction only. Then

accelerations of both the particles are equal
acceleration of the particle of mass m₁ is proportional to m₁
acceleration of the particle of mass m₁ is proportional to m₂
acceleration of the particle of mass m₁ is inversely proportional to m₁

139.

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the Earth is (Here R is the radius of the earth)

$(\frac{n}{n + 1}) mgR$
$(\frac{n}{n - 1}) mgR$
nmgR
$\frac{mgR}{n}$

140.

A stone falls freely under gravity. It covers distances h₁, h₂, and h₃, in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂, and h₃ is

h₁ = 2h₂ = 3h₃
$h_{1} = \frac{h_{2}}{3} = \frac{h_{2}}{5}$
h₂ = 3h₁ and h₃ = 3h₂
h₁ = h₂ = h₃

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Subject

NEET Physics : Gravitation

Short Answer Type

Multiple Choice Questions