The gravitational force on a body of mass 1.5 kg situated at a point is 45 N. The gravitational field intensity at that point is:

1. 30 N/kg

2. 67.5 N/kg

3. 46.5 N/kg

4. 43.5 N/kg

Two-point masses having mass m and 2m are placed at distance d. The point on the line joining point masses, where the gravitational field intensity is zero will be at a distance:

1. $\frac{2\mathrm{d}}{\sqrt{3}+1}$ from point mass "2m".

2. $\frac{2\mathrm{d}}{\sqrt{3}-1}$ from point mass "2m".

3. $\frac{\mathrm{d}}{1+\sqrt{2}}$ from point mass "m".

4. $\frac{\mathrm{d}}{1-\sqrt{2}}$ from point mass "m".

At what height above the surface of the earth, the value of "g" decreases by 2%?

[radius of the earth is 6400 km]

1. 32 km

2. 64 km

3. 128 km

4. 1600 km

During the motion of a man from the equator to the pole of the earth, its weight will (neglect the effect of change in the radius of the earth)

1. increase by 0.34%.

2. decrease by 0.34%.

3. increase by 0.52%.

4. decrease by 0.52%.

If R is the radius of earth and g is the acceleration due to gravity on the earth's surface, then mean density of the earth is:

1. $\frac{4\mathrm{\pi G}}{3\mathrm{gR}}$

2. $\frac{3\mathrm{\pi R}}{4\mathrm{gG}}$

3. $\frac{3\mathrm{g}}{4\mathrm{\pi RG}}$

4. $\frac{\mathrm{\pi Rg}}{12\mathrm{G}}$

The value of g at the surface of the earth is 9.8 m/s². Then the value of 'g' at a place 480 km above the surface of the earth will be nearly (radius of the earth is 6400 km):

1. 9.8 m/s²

2. 7.2 m/s²

3. 8.5 m/s²

4. 4.2 m/s²

If the change in the value of 'g' at a height 'h' above the surface of the earth is the same as at a depth x below it, then: (x and h being much smaller than the radius of the earth)

1. $\mathrm{x}=\mathrm{h}$

2. $\mathrm{x}=2\mathrm{h}$

3. $\mathrm{x}=\frac{\mathrm{h}}{2}$

4. $\mathrm{x}={\mathrm{h}}^2$

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

1. Remains the same

2. Decreases

3. Increases

4. First increase and then decrease

What should be the angular speed with which the earth has to rotate on its axis so that a person on the equator would weigh $\frac{3}{5}$th as much as he weighs at present?

1. $\sqrt{\frac{2\mathrm{g}}{5\mathrm{R}}}$

2. $\sqrt{\frac{2\mathrm{R}}{5\mathrm{g}}}$

3. $\frac{2\sqrt{\mathrm{R}}}{\sqrt{5\mathrm{g}}}$

4. $\frac{2\mathrm{g}}{5\mathrm{R}}$

The acceleration due to gravity on a planet is 1.96 m/s². If it is safe to jump from a height of 3 m on the earth, the corresponding height on the planet will be:

1. 3 m

2. 6 m

3. 9 m

4. 15 m