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This question contains Statement -1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement – I:
For a mass M kept at the centre of a cube of side ‘a’, the flux of gravitational field passing through its sides is 4π GM.
and
Statement – II
If the direction of a field due to a point source is radial and its dependence on the distance ‘r’ for the source is given as 1/r2, its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface
Statement – 1is false, Statement – 2 is true.
Statement – 1is true, Statement – 2 is true; Statement -2 is correct explanation for Statement-1.
Statement – 1 is true, Statement – 2 is true; Statement -2 is not a correct explanation for Statement-1.
Statement – 1 is true, Statement – 2 is False.
B.
Statement – 1is true, Statement – 2 is true; Statement -2 is correct explanation for Statement-1.
g = GM/r2
Average density of the earth
does not depend on g
is a complex function of g
is directly proportional to g
is inversely proportional to g
C.
is directly proportional to g
A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected?
moment of inertia
angular momentum
angular velocity
rotational kinetic energy
B.
angular momentum
In free space, neither acceleration due to gravity for external torque act on the rotating solid sphere. Therefore, taking the same mass of sphere if the radius is increased then a moment of inertia, rotational kinetic energy and angular velocity will change but according to the law of conservation of momentum, angular momentum will not change.
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
gx
D.
The gravitational force exerted on satellite at a height x is
where Me = mass of earth Since, gravitational force provides the necessary centripetal force, so,
The time period of an earth satellite in circular orbit is independent of
the mass of the satellite
radius of its orbit
both the mass and radius of the orbit
neither the mass of the satellite nor the radius of its orbit.
A.
the mass of the satellite
Time period of satellite
where R + h = orbital radius of satellite,
Me = mass of earth Thus, time period does not depend on the mass of the satellite.
If g is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is
2 mgR
mgR/2
mgR/4
mgR
B.
mgR/2
Gravitational potential energy of body on earth's surface
At a height h from earth's surface, its value is
where Me = mass of earth
m = mass of body
R = radius of earth
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period planet in circular orbit of radius R around the sun will be proportional to
Rn
A.
The necessary centripetal force required for a planet to move round the sun = gravitational force exerted on it
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then
d =1/2 km
d = 1 km
d =3/2 km
d = 2 km
D.
d = 2 km
Above the earth surface
Which of the following statements is correct?
(a) Centre of the mass of a body always coincides with the centre of gravity of the body.
(b) Centre of the mass of a body is the point at which the total gravitational torque on the body is zero
(c) A couple on a body produce both translational and rotational motion in a body.
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.
(b) and (d)
(a) and (b)
(b) and (c)
(c) and (d)
A.
(b) and (d)
Centre of mass may or may not coincide with centre of gravity.
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:
Keep floating at the same distance between them
Move towards each other
Move away from each other
Will become stationary
B.
Move towards each other
Both the astronauts are in the condition of weightness. Gravitational force between them pulls towards each other
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?
Raindrops will fall faster
Walking on the ground would become more difficult
'g' on the Earth will not change
Time period of a simple pendulum on the Earth would decrease
C.
'g' on the Earth will not change
If universal gravitational constant becomes tem times then G' = 10 G
Acceleration due to gravity,
So, Acceleration due to gravity increases.
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Two 20 g flatworms climb over a very thin wall, 10 cm high. One of the worms is 20 cm long, the other is wider and only 10 cm long. which of the following statement is correct regarding them?
20 cm worm has done more work against gravity
10 cm worm has done more work against gravity
Both worms have done equal work against gravity
Ratio of work done by both the worms is 4:5
B.
10 cm worm has done more work against gravity
The work done against gravity can be calculated from the increased in height of the centre of mass. The centre of mass of a worm folded in two is located at the middle of either half, i.e., at a point one-quarter of the worm's total length from one end.
Thus, the centre of mass of the narrow flat worm travels 5 cm up the wall; while that of the broad worm moves 7.5 cm. Hence, work done by 10 cm worm is more against gravity. Also, the ratio of the work done is 2:3.
A rocket is intended to leave the Earth's gravitational field. The fuel in its main engine is a little less than the amount that is necessary and an auxiliary engine, (only capable of operating for a short time) has to be used as well. When is it best to switch on the auxiliary engine?
at take-off
When the rocket has nearly stopped with respect to the Earth
It doesn't matter
Can't say
A.
at take-off
The rocket has to reach the highest possible total energy. If the zero level of gravitational potential energy is infinitely far away, then the energy of the rocket standing on the surface on the earth is negative. The energy released during the operation of the engines increases the total energy of the rocket, and the rocket can leave the earth's gravitational field if the sum of its potential and kinetic energies becomes positive.
The energy released in the course of the operation of the principal and auxiliary engines increase the total energy of the rocket and its ejected combustion products by a fixed value; this increase is independent of the moment when the engines are scratched on. However, the speed at which the combustion products fall back to the earth does depend on the timing of the rocket's operations. Indeed, if the auxiliary engine starts working when the rocket is at a greater height, the combustion fall further and their speed and total energy are higher when they hit the ground. This means that the sooner the auxiliary engine is switched on. the higher, the energy ultimately acquired by the rocket. The same concept is valid for the principal engine, and if the only energy consideration applies, it is best to operate the engines for the shortest time and at the highest thrust.
The planets with radii R1 and R2 have densities ρ1, ρ2 respectively. Their atmospheric pressure is p1 and p2 respectively. Therefore, the ratio of masses of their atmospheres, neglecting variation of g within the limits of the atmosphere is
D.
Acceleration due to gravity
Atmospheric pressure can be given by
P = W/S
Where w = weight of the atmosphere,
S = surface area of the planet
A thin symmertical double convex lens of refractive index μ2 = 1.5 is placed between a medium of refractive index μ1 = 1.4 to the left and another medium of refractive index μ3 = 1.6 to the right. Then, the system behaves as
a convex lens
a concave lens
a glass plate
a convexo-concave lens
C.
a glass plate
For the first refracting surface,
For a parallel beam of light;
For second refracting surface,
Hence, the combination behaves as a glass plate.
Write the SI unit of Gravitational constant
Therefore, we can obtain G as,
Since SI unit of
Force = Newton
Mass = Kilogram
Distance, r = metre
G = Nm²kg⁻²
OR
G = kg⁻¹ m³s⁻²
Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to
R
A.
Kinetic energy of the satellite
So the kinetic energy of the satellite is inversely proportional
to R. Therefore R increases then KE decreases.
Earth is revolving around the sun. If the distance of the earth from the sun is reduced to 1/4th of the present distance then the length of present day will be reduced by :
C.
Law of periods:- The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.
From Kepeler's law
T2 ∝ R3
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' = 3g
g' = 9g
g' = 27g
A.
g' = 3g
The acceleration due to gravity on the planet can be found using the relation
⇒ g ∝ R
Two satellites of earth, S1 and S2 are moving in the same orbit. The mass of S1 is four times the mass of S2. Which one of the following statements is true ?
The time period of S1 is four times that of S2
The potential energies of earth and satellite in the two cases are equal
S1 and S2 are moving with the same speed
The kinetic energies of the two satellites are equal
C.
S1 and S2 are moving with the same speed
When two satellites of earth are moving in same orbit, then time period of both are equal.
From Kepler's third law
T2 ∝ r3
Time period is independent of mass, hence their time periods will be equal.
The potential energy and kinetic energy are mass dependent, hence the PE and KE of satellites are not equal.
But, if they are orbiting in a same orbit, then they have equal orbital speed.
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is
2
B.
Given:- Kinetic energy of satellite is half of its potential energy.
Potential energy of satelite
Kinetic energy of satellite
Thus
A satellite orbiting the earth in a circular orbit of radius R completes one revolution in 3h. If orbital radius of geostationary satellite is 36000 km, orbital radius of earth is
6000 km
9000 km
12000 km
15000 km
B.
9000 km
Time period of satellite
T2 =
T2 = [ since GM = gR2 ]
⇒ T =
⇒ T ∝ r3/2
∴
Given T1 = 3h , T2 = 24 h
(geostationary satellite) r1 = R, r2 = 36000 km
⇒
⇒ R =
⇒ R =
⇒ R = 9000 km
A launching vehicle carrying an artificial satellite of mass m is set for launch on the surface of the earth of mass M and radius R. If the satellite is intended to move in a circular orbit of radius 7R, the minimum energy required to be spent by the launching vehicle on the satellite is (Gravitational constant= G)
B.
The energy of artificial satellite at the surface of the earth
E1 =
When the satellite is intended to move in a circular orbit of radius 7R, then energy of artificial satellite
E2 =
The minimum energy required
=
=
E =
The escape velocity for the earth is 11.2 km/s. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is
280 km/s
56.0 km/s
112 km/s
56 km/s
B.
56.0 km/s
Escape velocity is the minimum speed needed to a free object to escape speed from the gravitational influence of a massive body.
The escape velocity of the planet
Vescape =
Vescape =
∴
where Vp - velocity of planet, Ve - velocity of earth
⇒
⇒
⇒ VP = 5 × 11.2
⇒ VP = 56 km/s
Change in acceleration due to gravity is same upto a height h from each other the earth surface and below depth x then relation between x and his ( h and x<<<Re )
x = h
x = 2h
x =
x = h2
B.
x = 2h
Variation in g at the height h from the earth surface
g' = g
Variation in g at a depth x below the earth surface
g'' = g
Here d = x i.e depth
But variations are same
∴ g
∴ x = 2h
A uniform ring of mass m and radius a is placed directly above a uniform sphere of mass m and of equal to radius. The centre of the ring is at a distance a from the centre of the sphere. The gravitational force (Fl exerted by the sphere on the ring is
A.
Now, the gravitational field due to the ring at a distance d = a on its axis is
E =
E =
E =
The force in a particle of mass M placed here is
Force on the sphere = Mass of sphere × gravitational field
F = M E
F =
This is also the force due to the sphere in the ring.
A body of mass 2m is placed on earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of h, where h = 4R.
C.
Potential energy,
U =
= [ m = 2m]
At earth's surface, r = R
∴ Ue =
Now, if a body is taken to height, h = 4R, then the potential energy is given by
Uh = [ r = h + R ]
=
Thus, change in gravitational potential energy,
ΔU = Uh Ue
=
=
ΔU = [ Gm = gR2 ]
ΔU =
A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 x 107 m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is
∼ 0.13 × 1014 kg m2 s-1
∼ 1.3 × 1014 kg m2 s-1
∼ 0.58 × 1014 kg m2 s-1
∼ 2.58 × 1014 kg m2 s-1
C.
∼ 0.58 × 1014 kg m2 s-1
As the satellite is moving in equatorial plane with orbital radius 4 x 107 m.
∴ Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution
T = 24 h
= 24 × 3600 s
T = 86400 s
Velocity of satellite
v =
Angular momentum, L = mvr
L = m r
= r2
∴ L = × ( 4 × 107 )2
L = 0.58 × 1014 kg m2 s-1
The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is ( here R is the radius of the earth )
mgR
mgR
nmgR
A.
mgR
Gravitational potential energy of mass m at any point at a distance r from the centre of earth is
U =
At the surface of earth r = R,
∴ Us =
= - mgR
At the height h = nR from the surface of earth
r = R + h
= R + nR
r = R ( 1 + n )
∴ Uh =
=
Change in gravitational potential energy is
ΔU = Uh Us
=
=
ΔU = m gR
Assertion: Two bodies of masses M and m (M > m) are allowed to fall from the same height if the air resistance for each be the same then both the bodies will reach the earth simultaneously.
Reason: For same air resistance, acceleration of both the bodies will be same.
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false.
D.
If both assertion and reason are false.
The force acting on the body of mass M are its weight Mg acting vertically downwards and air resistance F acting vertically upward.
∴ Acceleration of the body
Now, M > m, therefore, the body with larger mass will have greater acceleration and it will reach the ground first.
Assertion: Kepler's second law can be understood by conservation of angular momentum principle.
Reason: Kepler's second law is related with areal velocity which can further be proved to be based on conservation of angular momentum as (dA/dt) = ( r2ω / 2 )
If both assertion and reason are true and reason is the correct explanation of assertion.
If both assertion and reason are true but reason is not the correct explanation of assertion.
If assertion is true but reason is false.
If both assertion and reason are false
A.
If both assertion and reason are true and reason is the correct explanation of assertion.
According to Kepler's law,
= constant
As,
∴ = constant
∴ L = constant
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Time period of pendulum, on a satellite orbiting the earth, is
zero
infinity
D.
infinity
On an artificial satellite orbiting the earth the acceleration is given by towards the centre of the earth.
Now for a body of mass m on the satellite the graviational force due to earth is towards the centre of the earth.
Let the reaction force on the surface of the satellite be N, then
⇒ N = 0
That is on the satellite there is a state of weightlessness or g = 0
T = 2 = ∞
Hence T = ∞
The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M to transfer it from a circular orbit of radius R, to another of radius R2 (R2 > R1 ) is
GmM
GmM
2 GmM
GmM
D.
GmM
Additional kinetic energy provided to satelite
KE =
∴ KE =
A person is standing in an elevator. In which situation he finds himself weightless?
Elevator is going up with constant speed.
Elevator is moving down with constant speed.
Elevator is accelerating up with g.
Elevator is accelerating down with g.
D.
Elevator is accelerating down with g.
Elevator is accelerating down with g.
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