A pellet of mass 1 g is moving with an angular velocity of 1 rad/s along a circle of radius 1 m the centrifugal force is
0.1 dyne
1 dyne
10 dyne
100 dyne
Two point objects of masses 1.5 g and 2.5 g respectively area at a distance of 16 cm apart, the centre of gravity is at a distance x from the object of mass 1.5 g where x is
10 cm
6 cm
13 cm
3 cm
The weight of a body on surface of earth is 12.6 N. When it is raised to a height half the radius of earth its weight will be
2.8 N
5.6 N
12.5 N
25.2 N
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 ?
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
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)
nmgR
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
mgR
2mgR
B.
The potential energy on the surface of earth would be equal to mgR. So, the change in potential energy would be equal to
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
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
h2 = 3h1 and h3 = 3h2
h1 = h2 = h3
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
3 mgR