NEET Physics Questions: Gravitation

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The gravitational force Fg between two objects does not depend on




Two sphere of mass $m_1$ and $m_2$ are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be




A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational force suddennly disappears, the satellite will




Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is =..................




The distance of the moon and earth is D the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero




One can easily “ Weight the earth ” by calculating the mass of earth using the formula (in usual notation)




Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of $\sqrt 2 m$ from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............




Which of the following statement about the gravitational constant is true




Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is 1/3rd of the force between band C, Then the distance C form A is = m




The gravitational force between two point masses $m_1$ ans $m_2$ at separation r is given by$ F = G { m_1 m_2 \over r^2 } $. The constant k




As we go from the equator to the poles, the value of g




If R is the radius of the earth and g the acceleration due to gravity on the earth’s surface, the mean density of the earth is =




The radius of the earth is 6400 km and $g=10ms^{-2}$. In order that a body of 5 kg weights zero at the equator, the angular speed of the earth is = rad/sec




The time period of a simple pendulum on a freely moving artificial satellite is




The value of g on he earth surface is $980 cm/sec^2$. Its value at a height of 64 km from the earth surface is …………..$cm5^{–2} $