## A light cylindrical vessel is kept on a horizontal surface. Its base area is A. A hole of cross sectional area a is made just at its bottom side. The minimum coefficient of friction necessary for sliding of the vessel due to the impact force of the emerging liquid is (a<< A)

Q:A light cylindrical vessel is kept on a horizontal surface. Its base area is A. A hole of cross sectional area ‘ a ‘ is made just at its bottom side. The minimum coefficient of friction necessary for sliding of the vessel due to the impact force of the emerging liquid is (a<< A)
(A) varying
(B) a/A
(C) 2a/A
(D) None of these

Solution :

The velocity of efflux of the liquid is given as
v =√2gy
The impact force of the emerging liquid on the vessel + liquid content is equal to
F = v dm/dt = vaρv = aΡv²
F = aρ(2gy) = 2aρgy
The force of friction = f = F = 2aρgy
μN = 2aρgy
μ(Aρgy) = 2a ρgy
μ = 2a/A

## 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.
ω= 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.