1. A point mass is placed inside a thin spherical shell of radius R and mass m at a distance R/2 from the centre of the shell. The gravitational force exerted by the shell on the point mass is ________
(c). Gravitational force and gravitational intensity at point inside a hollow spherical shell is zero.
(d). Escape velocity ve=√(2GM/R)
=√((2×6.67×10-11×1.99×1030)/6.96×108 ) = 618 km/s
3. A satellite of mass m is orbiting close to the surface of Earth (Radius R = 6400km) has a kinetic energy K. The corresponding kinetic energy of the satellite to escape from the earth’s gravitational field is __________
(b) Kinetic energy of a satellite close to earth’s surface
K = ( GMm)/2R => ( GMm)/R=2K
Escape K.E = 1/2mv2 = (1/2)m x (√(2GM/R))2= GMm/R = 2K
4. The height from the Earth surface at which the value of acceleration due to gravity reduces to 1/4th of its value at earth’s surface (assume Earth to be a sphere of radius 6400km) _________
(a) g= g0 (R2/(R+h)2)
Here g = ¼ g0
∴1/4=R2/(R+h)2 => 1/2= R/(R+h)
=> h= R= 6400km
5. A particle of mass m is rotating in a plane in circular path of radius r. Its angular momentum is L. the centripetal force acting on the particle is _________
(c) L= mvr, v = L/mr
Centripetal Force (F) = (mv2)/r = m/r (L/mr)2= L2/(mr3 )
6. A cyclist riding a bicycle at a speed of 14√3 m/s takes a turn around a circular road of radius 20√3 m without skidding. Given g= 9.8m/s2, what is his inclination to the vertical?
(d) tanθ = v2/rg = (14√3)2/(20√3 X 9.8)=> tanθ= √3 => θ = 600
7. Two particles having a mass M and m are moving in circular path having radius R and r. If their time periods are same, then the ratio of angular velocity will be _________.
(c) As ω = 2π/T and T is same, therefore ω must be same. i.e., ω1/ ω2 = 1
8. A cyclist is riding with a speed of 36km/hr. As he approaches a circular turn on the road of radius 100m, he applies brakes and reduces his speed at the constant rate of 0.50m/s2 every second. What is the magnitude of the net acceleration of the cyclist on the circular turn?
(d) Here , v = 36km/hr = 36 x 5/18 = 10m/s
r = 100m
Centripetal acceleration ,
ac = v2/r = (10)2/100 = 1m/s2
Tangential acceleration, at = 0.5m/s2
Magnitude of net acceleration (ac2+at2)1/2 = √5/2 m/s2
9. Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the Earth, if it weighed 250N on the surface?
(b) g' = g(1- d/R) or mg' = mg (1-d/R)
Here mg = 250N , d = R/2
Therefore mg' = 250(1- R/2) = 125N
R
10.If a body is projected up from the Earth's surface with a velocity 20 % of the escape velocity, how high will it rise?
(d) The kinetic energy supplied to the body is ½ m[0.2×√(2GM/R)]2 = 0.04 GMm/R.
The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is h, the increase in the gravitational potential energy is – GMm/(R+h) – (– GMm/R) = GMm/R – GMm/(R+h)
[Note that the gravitational potential energy is negative]
Therefore, we have
0.04 GMm/R = GMm/R – GMm/(R+h)
Or, 0.04/R = 1/R – 1/(R+h) = h/[R(R+h)]
This gives h = 0.04(R+h) from which h = 0.04 R/0.96 = R/24.
11.Two satellites have orbital radii R and 1.01 R. Their orbital periods differ by _____
(b) We have from Kepler’s third law, T2 α R3 where T and R are the orbital period and orbital radius respectively.
Therefore, T = kR3/2 where k is the constant of proportionality.
Taking logarithms, ln T = ln k + (3/2) ln R
Differentiating, dT/T = 0 + (3/2) dR/R. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius.
In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5%
12.If the gravitational force between two point masses were inversely proportional to the nth power of the distance between them, what would be the relation between the orbital period (T) of a planet around the sun and the mean distance (r) of the planet from the sun?
(c)The centripetal force required for the circular motion is supplied by the gravitational pull so that we have
mrω2 = GMm/rn where m is the mass of the planet, ω is its orbital angular velocity, M is the mass of the sun and G is the gravitational constant.
The above equation gives ω2 = GM/rn+1
Since ω = T/2π, we have
(T/2π)2= GM/rn+1
Therefore, T2 α rn+1
13.In uniform circular motion of a particle which one among the following does not remain constant?
(d) Linear momentum
14. A body weighs 63kgwt on the surface of Earth.Its weight on the surface of Mars will be ______(Mass of Mars = 1/9 mass of Earth, Radius of Mars = 1/2 Radius of Earth)
(b) g = GM/R2 , therefore weight = mg = GMm/R2
so,
weight on Mars = (Mass of Mars/Mass of Earth) x (radius of Earth/radius of Mars)2 x weight on Earth
= (1/9) x (2)2 x 63 = 28kgwt
15.A particle is originally at rest at the highest point of a smooth vertical circle of radius R. If the particle is slightly displaced, the distance below the highest point where the particle will leave the circle is _______
The point where it leaves the circle reaction force(N)=0
therefore mgcosθ = mv2/R
or cosθ =v2/Rg
Now applying conservation of energy
mgh = (1/2)mv2
or v2 = 2gh
therefore cosθ = 2h/R
by geometry
cosθ = R-h/R,
therefore R-h/R = 2h/R or h = R/3
16.The speed of an artificial satellite moving in an orbit of radius r around the earth is increased by 41.4%. The satellite will _______
(d)The escape velocity is √2 times the orbital speed.
The speed v of the satellite becomes 1.414 times the initial value on increasing the speed by 41.4%. Now, 1.414 v = √2 v. Therefore the new speed is the escape speed so that the satellite will escape into the outer space.
17.A body revolves n times in a circle of radius π cm in one minute. Its linear velocity is _____
(c) v = rω = r2πf
Given, f = n/60 r.p.s,
Therefore v = 2π x π x n/60= 2π2n/60 m/s
18.A particle is moving in a circle of radius r centered at O with a constant speed v. What is the change velocity in moving from A to B if ∠ AOB = π/3?
(a) The change in velocity = (v2 + v2 - 2v2cosθ)1/2
= 2vsin(θ/2)
Therefore change in velocity = 2vsin(π/6)
19.If the Earth of Radius R, while rotating with angular velocity ω becomes stand still, what will be the effect on the weight of a body of mass m at latitude of 450?
(d) Effect of latitude on weight is given by,
mg' = m(g - Rω2cos2λ) where λ is the latitude angle
So, if Earth stopped rotating ω=0 ,weight will increase by Rω2cos2450=Rω2/2
20. For a particle in circular motion, the centripetal acceleration is _______.
(d) maybe more or less than its tangential acceleration
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