## A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is l , the minimum velocity of projection of the particle so as to escape is equal to

Q: A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is l, the minimum velocity of projection of the particle so as to escape is equal to
(A)√(Gm/l)
(B)√(Gm/2l)
(C )√(2Gm/l)
(D) 2√(2Gm/l)

Solution :

The gravitational potential at the mid-point P is
V = V1 + V2
= -Gm/(l/2)-Gm/(l/2) = -4Gm/l
 The gravitational potential energy
= U = -4Gmm0/l
 , m0 = mass of particle
When it is projected with a speed v, it just escapes to infinity, and the potential & kinetic energy will become zero.
 ΔKE + ΔPE = 0
 = 0 – ½mv² +{0-(-4Gmm0/l)}= 0
v = 2√2Gm/l

## Two massive particles of masses M & m (M > m) are separated by a distance l . They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass m is

Q: Two massive particles of masses M & m (M > m) are separated by a distance l . They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass m is
(A)√GMm/(M+m)l
(B) √GM²/(M+m)l
(C)√Gm/l
(D)√Gm²/(M+m)l
Solution : The system rotates about the centre of mass. The gravitational force acting on the particle m accelerates it towards the centre of the circular path, which has the radius
r = Ml /(M+m)

 F = mv²/r
GMm/l² = mv²/(Ml /(M+m))
By Solving ,

## A body is fired with velocity of magnitude √gR< V< √2gR at an angle of 30° with the radius vector of earth. If at the highest point the speed of the body is v/4 , the maximum height attained by the body is equal to

Q: A body is fired with velocity of magnitude √gR< V<√2gR  at an angle of 30° with the radius vector of earth. If at the highest point the speed of the body is v/4, the maximum height attained by the body is equal to
(A) v²/8g
(B) R
(C) √2 R
(D) none of these
where R = radius of earth. (Use the appropriate assumptions)
Solution :

Conservation of angular momentum of the body about O yields
mv (sin30°) R = mv'(R + h)
 ½ vR = ¼ v(R + h) 
Hence , h = R.

## A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time

Q: A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time
(A) 1.5 hrs.
(B) 1.6 hrs. if it is rotating from west to east.
(C) 24/17 hrs. if it is rotating from west to east.
(D) 24/17 hrs. if it is rotating from east to west.

Solution : Let ωo = the angular velocity of earth above its axis = 2π/24  rad/hr
Let ω = angular velocity of satellite.
  = 2π/1.5
For a satellite rotating from west to east. (same as earth), the relative angular velocity ω1 = ω – ωo
 Time period of rotation relative to earth =2π/ω1 = 1.6 h
Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ω2 = ω + ωo
Time period of rotation relative to earth =2π/ω2 = 24/17 hrs.

## Two containers of equal volume contain the same gas at pressure p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to

Q: Two containers of equal volume contain the same gas at pressure p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to
(a)p1/T1 + p2/T2
(b)½ [p1/T1 + p2/T2]
(c)[p1T2 + p2T1]/(T1+T2)
(d)[p1T2 – p2T1]/(T1-T2)
Solution : (b)
For a closed system, the total mass of gas or the number of moles remains constant.
P1V = n1RT1 ,     p2V = n2RT2
p(2V) = (n1 + n2)RT
p(2V) = (p1V/RT1 + p2V/RT2)RT
p/T =1/2[p1/T1 + p2/T2]