A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at^a⁄₂ distance from the centre, will be

\(-\frac{3GM}{a}\)
\(-\frac{2GM}{a}\)
\(-\frac{GM}{a}\)
\(-\frac{4GM}{a}\)

Solution

Potential at the given point = Potential at the point due to the shell + Potential due to the particle

\(=-\frac{GM}{a}-\frac{2GM}{a}=-\frac{3GM}{a}\)

The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If the speed of satellite A is 3 V, then the speed of satellite B will be

3 V/4
6 V
12 V
3 V/2

Solution

The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB.If t₁ is the time for the planet to move from C to D and t₂ is the time to move from A to B then :

t₁ = 4t₂
t₁ = 2t₂
t₁ = t₂
t₁ > t₂

Solution

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

¹⁄₂
¹⁄_√2
2
√2

Solution

Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g’, then

g’= g/9
g’= 27g
g’=9g
g’=3g

Solution

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

½ R
2 R
4 R
1/4 R

Solution

If g_E and g_M are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces, one will find the ratio \(\frac{electronic\, charge\, on\, the\, moon}{electronic\, charge\, on\, the\, earth}\) to be

g_M/g_E
1
zero
g_E/g_M

Solution

Electronic charge does not depend on acceleration due to gravity as it is a universal constant.So, electronic charge on earth

= electronic charge on moon

∴ Required ratio = 1.

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R.The period of second satellite is larger than the first one by approximately

0.5%
1.0%
1.5%
3.0%

Solution

The change in potential energy,when a body of mass m is raised to a height nR from the earth’s surface is (R = radius of earth)

\(mgR\left ( \frac{n}{n-1} \right )\)
nmgR
\(mgR\left ( \frac{n^{2}}{n^{2}+1} \right )\)
\(mgR\left ( \frac{n}{n+1} \right )\)

Solution

A planet of mass m moves around the sun of mass Min an elliptical orbit. The maximum and minimum distance of the planet from the sun are r₁ and r₂ respectively. The time period of planet is proportional to

\(r_{1}^{2/5}\)
(r₁+r₂)^3/2
(r₁-r₂)^3/2
r^3/2

Solution

T² ∝ r³, where r = mean radius =\(\frac{r_{1}+r_{2}}{2}\)

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