Q1: The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is: [2024]
(a) 19.6 m s−2
(b) 9.8 m s−2
(c) 4.9 m s−2
(d) 3.92 m s−2
Ans: (d)
The acceleration due to gravity (g) on a planet is given by the formula:
where:
In this question, we know that:
The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius Rp is also half the radius of the earth Re. Therefore, Rp
Using the formula for acceleration due to gravity and substituting the above information:
Simplify the expression:
Knowing that is the acceleration due to gravity on Earth we substitute this value into the equation:
Thus, the acceleration due to gravity on the planet is 3.92m/s2, which corresponds to:
Option D:
(a)
(b)
(c)
(d)
Ans: (a)
Q2: If R is the radius of the earth and g is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be : [2023]
(a)
(b)
(c)
(d)
Ans: (c)
Q3:Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant) [2023]
(a)
(b)
(c)
(d)
Ans: (c)
Let electric field at point Q be zero
So,
Q4: A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity [2023]
(a) T
(b) T2
(c) T3
(d) √T
Ans: (b)
Time period of satellite above earth surface
Q1: A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is [2022]
(a) 50 N/kg
(b) 20 N/kg
(c) 180 N/kg
(d) 0.05 N/kg
Ans: (a)
F = mEg
3 = 60/1000 Eg
Eg = 50 N/kg
Q2: Match List - I with List - II
Choose the correct answer from the options given below:
(a) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(b) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
(c) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
(d) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) [2022]
Ans: (b)
Q3: A gravitational field is present in a region and a mass is shifted from A to B through different paths as shown. If W1, W2 and W3 represent the work done by the gravitational force along the respective paths, then : [2022]
(a) W1 < W2 < W3
(b) W1 = W2 = W3
(c) W1 > W2 > W3
(d) W1 > W3 > W2
Ans: (b)
Since the gravitational field is conservative in nature hence the work done would depend only on the initial and final positions and not on the path followed by the mass.
Hence, W1 = W2 = W3
Q4: In a gravitational field, the gravitational potential is given by, V = -K/x (J/Kg). The gravitational field intensity at point (2, 0, 3) m is
(a) +K/4
(b) +K/2
(c) -K/2
(d) -K/4 [2022]
Ans: (d)
Q1: The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is: [2021]
(a) 3v
(b) 4v
(c) v
(d) 2v
Ans: (b)
Q2: A particle of mass 'm' is projected with a velocity v = kVe (k < 1) from the surface of the earth. (Ve = escape velocity) The maximum height above the surface reached by the particle is : [2021]
(a)
(b)
(c)
(d)
Ans: (b)
=
Q1: A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth? [2020]
(a) 30 N
(b) 24 N
(c) 48 N
(d) 32 N
Ans: (d)
Q1: A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth? [2019]
(a) 150 N
(b) 200 N
(c) 250 N
(d) 100 N
Ans: (d)
Solution:
Acceleration due to gravity at a depth d from surface of earth
Where g = acceleration due to gravity at earth's surface
Multiplying by mass 'm' on both sides of (1)
Q2:The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :
(a) mgR
(b) 2mgR
(c) 1/2mgR
(d) 3/2mgR [2019]
Ans: (c)
Initial potential energy at Earths surface is
Final potential energy at height h = R :
work done = change in PE
w = Uf - Ui
Q1: If the mass of the Sun were ten times smaller and the universal gravitational constant were ten time larger in magnitude, which of the following is not correct ? [2018]
(a) Raindrops will fall faster
(b) Walking on the ground would become more difficult
(c) Time period of a simple pendulum on the Earth would decrease
(d) 'g' on the Earth will not change
Ans: (d)
Solution:
Raindrops will fall faster, Walking on the ground would become more difficult. Time period of a simple pendulum on the earth would decrease.
Q2: The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then
(a) KA < KB < KC
(b) KA > KB > KC
(c) KB < KA < KC
(d) KB > KA > KC [2018]
Ans: (b)
Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.
Point A is perihelion and C is aphelion.
Clearly, VA > VB > VC
So, KA > KB > KC
Q1: 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:- [2017]
(a) d = 1 km
(b)
(c) d = 2 km
(d)
Ans: (c)
Solution:
Q2: Two astronauts are floating in gravitational free space after having lost contact with their spaceship.The two will :- [2017]
(a) Move towards each other.
(b) Move away from each other.
(c) Will become stationary
(d) Keep floating at the same distance between them.
Ans: (a)
Solution: Astronauts move towards each other under mutual gravitational force.
Q1: At what height from the surface of earth the gravitational potential and the value of g are -5.4 x 10-7J kg-2 and 6.0 ms-2 respectively ? Take the radius of earth as 6400 km. [2016]
(a) 2000 km
(b) 2600 km
(c) 1600 km
(d) 1400 km
Ans: (b)
Solution:
-GM/r = 5.4 x 10-7
-GM/r2 = 6.0
dividing both the equations, r = 9000 km.
so height from the surface = 9000 - 6400 = 2600 km
Q2: The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is : [2016]
(a) 1:√2
(b) 1:2
(c) 1:2√2
(d) 1:4
Ans: (c)
Solution:
So the answer is option 3
Q3: A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth's surface, is
(a)
(b)
(c)
(d)
Ans: (b)
Total mechanical energy of satellite is given by
E = – GMm/2r
Here, r = R + h
And, GM = goR2
So, E = – mgoR2/2(R+h)
Q4: Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by
(a)
(b)
(c)
(d) [2016]
Ans: (b)
Since, acceleration due to gravity is given as
Q1: Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T2 = Kr3. here K is constant.
If the masses of sun and planet are M and m respectively then as per Newton?s law of gravitation force of attraction between them is F= GMm/r2 , here G is gravitational constant. The relation between G and K is described as : [2015]
(a) K=1/G
(b) GK4π2
(c) GMK=4π2
(d) K=G
Ans: (c)
Solution:
Q2: Two spherical bodies of mass M and 5 M and radii R and 2 R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is : [2015]
(a) 1.5 R
(b) 2.5 R
(c) 4.5 R
(d) 7.5 R
Ans: (d)
Solution:
Let centre of mass C at a distance x1 from m1 and x2 from m2
Since the masses are moving under mutual attraction the position of centre of mass remains constant.
When the masses are in contact, let x'1 and x'2 be the distance of their centres from the centre of mass.
Q3: A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × 106 m above the surface of earth. If earth's radius is 6.38 × 106 m and g = 9.8 ms−2, then the orbital speed of the satellite is [2015]
(a) 9.13 km s−1
(b) 6.67 km s−1
(c) 7.76 km s−1
(d) 8.56 km s−1
Ans: (c)
The orbital speed of the satellite is
where R is the earth’s radius, g is the acceleration due to gravity on earth’s surface and h is the height above the surface of earth.
Here, R = 6.38 × 106m, g = 9.8 m s–2 and h = 0.25 × 106 m
= 7.76 × 103 m s–1 = 7.76 km s–1
Q4: A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,
(a) the linear momentum of S remains constant in magnitude.
(b) the acceleration of S is always directed towards the centre of the earth.
(c) the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.
(d) the total mechanical energy of S varies periodically with time. [2015]
Ans: (b)
The gravitational force on the satellite will be aiming towards the centre of the earth so acceleration of the satellite will also be aiming towards the centre of the earth.
Q1: Dependence of intensity of gravitational field (E) of earth with distance (r) from centre of earth is correctly represented by: [2014]
(a)
(b)
(c)
(d)
Ans: (b)
For a point inside the earth i.e. r < R
where M and R be mass and radius of the earth respectively. At the centre, r = 0
For a point outside the earth i.e. r > R,
On the surface of the earth i.e, r = R
The variation of E with distance r from the centre is as shown in the figure.
Q2: A block hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × 1024 kg) have to be compressed to be a black hole ? [2014]
(a) 10−2 m
(b) 100 m
(c) 10-9 m
(d) 10-6m
Ans: (a)
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1. What is the formula for gravitational force? |
2. How does the gravitational force between two objects change with distance? |
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