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)

• $\left(\frac{\mathrm{n}}{\mathrm{n} + 1}\right) \mathrm{mgR}$

• $\left(\frac{\mathrm{n}}{\mathrm{n} - 1}\right) \mathrm{mgR}$

• nmgR

• $\frac{\mathrm{mgR}}{\mathrm{n}}$

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

• $\frac{\mathrm{mgR}}{4}$

• $\frac{\mathrm{mgR}}{2}$

• 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

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^2 \mathrm{days}$

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^{3/2} \mathrm{days}$

• $\mathrm{D} {\left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right)}^{2/3} \mathrm{days}$

• $\mathrm{D} \left(\frac{{\mathrm{L}}_2}{{\mathrm{L}}_1}\right) \mathrm{days}$

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

• ${\mathrm{h}}_1 = \frac{{\mathrm{h}}_2}{3} = \frac{{\mathrm{h}}_2}{5}$

• 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

• $\frac{2}{3} \mathrm{mgR}$

• 3 mgR

• $\frac{1}{3} \mathrm{mgR}$