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
D.
100 dyne
Given that m = 1 × 10-3 kg ; ω = 1 rad/s and r = 1 m
Hence, centrifugal force = mω2r = 10-3 N
1 N = 105 dyne
∴ Centrifugal force = 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
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