Central Forces MSQ - IIT JAM MCQ

The correct answers are: a^3/2, m^–1/2

Use arguments of symmetry as the yz plane divides the objects symmetrically.

The correct answers are: The gravitational force due to this object at the origin is zero, The gravitational potential is the same at all points of the circle y² + z² = 36, The gravitational potential is the same at all points of the circle y² + z² = 4

It will fall because mg is acting on it towards the centre of planet and initial velocity is zero. It’ll move in straight line.

 By energy conservation

using this we get  V = f(r)
Now use 

R' = radius of the planet.
In the final expression (on in the beginning itself)



The correct answers are: It will fall on the planet, The time of fall of the satellite on the planet is nearly 

A) At ∞ both V and E are zero.
B) Let, V_∞​=GM/R​ for an unit mass. 

So, V_R​=0 i.e. at the radius R of solid sphere(mass M) and E_R​=GM/R²​
C) Inside a spherical shell V =GM/R​ and E =0
Thus, all the above are correct.

For, h<<R force is constant and is equal to mg

Therefore, work done is mgh in moving by distance h

For, h=R

gravitational potential at x is − GM/x

Work done in moving from x=R to x=2R is − GMm/2R + GMm/R

​= GMm/2R

which is equal to  1/2 mgR

Let ω₀ = the angular velocity of the earth about its axis.

Let ω = the angular velocity of the satellite

For a satellite rotating from west to east (the same as the earth), the relative angular velocity, ω₁ = ω - ω₀
The period of rotation relative to the earth 
For a satellite rotating from east to west (opposite to the earth), the relative angular velocity,  ω₂ = ω + ω₀.
The correct answers are: 1.6hours if it is rotating from west to east, 24/17hours if it is rotating from east to west

F ∝ T,  if r < R, and

  if r > R.

The correct answers are: 

The correct answers are: r^3/2, m^–1/2

The correct answers are: l^3/2, m^–1/2