2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 16 InterviewSolution

751.	Statement I: The magnitude of the gravitational potential at the surface of solid sphere is less than that of the centre of sphere. Statement II: Due to the solid sphere, the gravitational potential is the same within the sphere.A. Statement I is True, Statement II is True: Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True: Statement II is Not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - C `V_("in")=(GM)/(2R^(3))[3R^(2)-r^(2)]` `V_("in")=3/2V_(s)` `V_("in")gtV_(s)` `V` is not same everywhere as indicated by `V_("in")`.

Discussion

752.	A ring of radius `R = 4m` is made of a highly dense material. Mass of the ring is `m_(1) = 5.4 xx 10^(9) kg` distributed uniformly over its circumference. A highly dense particle of mass `m_(2) = 6 xx 10^(8) kg` is placed on the axis of the ring at a distance `x_(0) = 3 m` from the centre. Neglecting all other forces, except mutual gravitational interacting of the two. Caculate (i) displacemental of the ring when particle is at the centre of ring, and (ii) speed of the particle at that instant.
Answer» Correct Answer - A::C Let `x` be the displacemental of ring. Then displacemental of the particle is `x_(o) - x`, or `(3.0 - x) m`. Centre of mass will not move. Hence, `(5.4 xx 10^(9)) x = (6 xx 10^(8)) (3 - x)` Solving, we get `x = 0.3 m` Apply conservation of linear momentum and conservation of mechanical energy.

Discussion

753.	Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. if the gravitational force of attraction between the planet and the star is proportational to `R^(-5//2)`, then (a) `T^(2)` is proportional to `R^(2)` (b) `T^(2)` is proportional to `R^(7//2)` (c) `T^(2)` is proportional to `R^(3//3)` (d) `T^(2)` is proportional to `R^(3.75)`.A. `r^(3)`B. `r^(2)`C. `r^(2.5)`D. `r^(3.5)`
Answer» Correct Answer - D `(mv^(2))/r=(GMm)/R^(5//2) [:.Fprop1/r^(5//2)]` or `romega^(2)=(GM)/r^(5//2)` or `r(4pi^(2))/(T^(2))=(GM)/r^(5//2)` or `T^(2)propr^(5//2)` or `T^(2)propr^(3.5)`

Discussion

754.	Write true or false for the following statements:The inertia of an object depends upon its mass.
Answer» True Explanation: Yes, the inertia of the system depends on its mass. Inertia is the property to resist the change in motion. The more the mass of the object, the more is the resistance to the change in velocity. F = m × a For same force, if mass is more, then acceleration is less and thus mass is the measure of inertia.

754.

Write true or false for the following statements:The inertia of an object depends upon its mass.

Answer»

True

Explanation: Yes, the inertia of the system depends on its mass.
Inertia is the property to resist the change in motion. The more the mass of the object, the more is the resistance to the change in velocity.

F = m × a

For same force, if mass is more, then acceleration is less and thus mass is the measure of inertia.

Discussion

755.	Write true or false for the following statements:The value of g is zero at the centre of the earth.
Answer» True Explanation: The value of g is zero at the centre of earth. This is because at the centre of earth the object is surrounded from masses from all sides. The total force acting on the object is thus zero and thus acceleration due to gravity is also zero.

755.

Write true or false for the following statements:The value of g is zero at the centre of the earth.

Answer»

True

Explanation: The value of g is zero at the centre of earth. This is because at the centre of earth the object is surrounded from masses from all sides. The total force acting on the object is thus zero and thus acceleration due to gravity is also zero.

Discussion

756.	Name the scientist who gave the magnitude of buoyant force acting on a solid object immersed in a liquid.
Answer» Correct Answer - Archimedes

Discussion

757.	Why the value of 'g' is more at the poles than at the equator?
Answer» The shape of the earth is oblate. Therefore, the radius of the earth is more at the equator than at the poles. Moreover, due to rotation of the earth, everybody at the equator experiences centrifugal force away from the centre of the earth. In contrast, there is no centrifugal force at the poles. Due to these two reasons the value of 'g' is more at the poles than at the equator.

Discussion

758.	Show that the weight of all bodies is zero at the centre of the earth.
Answer» The value of acceleration due to gravity at a depth 'd' below the surface of radius R is given bt 'g' = g(1 - d/R). At the centre of the earth, d = R. Therefore, g' = 0. The weight of a body of mass 'm' at the centre of the earth = mg' = m x 0 = 0

Discussion

759.	The density of gold is `19 g //cm^(2)`. Find the volume of 95 g of gold.
Answer» Correct Answer - `5 cm^(3)`

Discussion

760.	Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?
Answer» Correct Answer - A::B `nu_(e) = sqrt((2 GM)/(R))` or `nu_(e) prop sqrt((M)/(R))`

Discussion

761.	Two small bodies of masses 2.00 kg and 4.00 kg are kept at rest at a separation of 2.0 m. Where should a particle of mass 0.10 kg be placed to experience no net gravitational force from these bodies? The particle is placed t this point.What is the gravitational potential energy of the system of three particle with usual reference level?
Answer» Correct Answer - A::B::C::D Let k0.1 kg mass is xm form 2 kg mass and `(2-x)m` from 4 kg mass. `-.gt (2xx0.1)/x^2=-(4xx0.1)/((2-x^2)^2)` or `0.2/x^2=0.4/((2-x^2)^2)` or 1/x^2=2/((2-x^2)^2)` or `(2-x)^2=2x^2` `=(2-x)^2=2x^2` `or 2-x=sqrt2 x` `x(sqrt2+1)=2` or` x=2/2.414` `=0.83m from 2kg mass`.

Discussion

762.	A body in the uniform circular motion then the parts 1 and 2 represents A) Time, Velocity B) Speed, Velocity C) Velocity, Acceleration D) Acceleration, Velocity
Answer» C) Velocity, Acceleration

Discussion

763.	Two electrons, each having mass 9.1 × 10-31 kg, are separated by a distance 10 Å. Calculate the gravitational force between them.
Answer» Given, mass of electron (me) = 9.1 × 10^-31 kg Distance between them(r) = 10 Å =10 × 10^-10 m G = 6.673 × 10^-11 N m² kg^-2 The gravitational force between them (Fg) is given by, F_g= G x \(\frac{(m_e)(m_e)}{r^2}\) F_g= 6.67 x 10^-11x \(\frac{(9.1\times10^{-31})^2}{(10\,\times\,10^{-10})^2}\) = 5.52 x 10^-53 N

763.

Two electrons, each having mass 9.1 × 10-31 kg, are separated by a distance 10 Å. Calculate the gravitational force between them.

Answer»

Given,

mass of electron (me) = 9.1 × 10^-31 kg

Distance between them(r) = 10 Å =10 × 10^-10 m

G = 6.673 × 10^-11 N m² kg^-2

The gravitational force between them (Fg) is given by,

F_g= G x \(\frac{(m_e)(m_e)}{r^2}\)

F_g= 6.67 x 10^-11x \(\frac{(9.1\times10^{-31})^2}{(10\,\times\,10^{-10})^2}\) = 5.52 x 10^-53 N

Discussion

764.	Why moon does not have an atmosphere?
Answer» The moon’s gravity is 1/6th of the gravity on earth. It is not strong enough to hold the atmosphere on its surface from going back to space.

Discussion

765.	State true/false. If false, correct the statement: “Earth exerts a greater force of attraction on apple than apple exerts on the earth.”
Answer» False Earth exerts the force of attraction on apple equal to the force apple exerts on the earth.

765.

State true/false. If false, correct the statement: “Earth exerts a greater force of attraction on apple than apple exerts on the earth.”

Answer»

False

Earth exerts the force of attraction on apple equal to the force apple exerts on the earth.

Discussion

766.	Two blocks of masses m each are hung from a balances as shown in the figure. The scale pan `A` is at height `H_(1)` whereas scale pan `B` is at height `H_(2)` Net torque of weights acting on the system about point `C` will be (length of the rod is `l` and `H_(1)` & `H_(2)` are lt ltR ) `(H_(1) gt H_(2))` .A. `mg((1-2H_(1))/(R))l`B. `(mg)/(R)(H_(1)-H_(2))l`C. `(2mg)/(R)(H_(1)+H_(2))l`D. `2"mg"(H_(2)H_(1))/(H_(1)+H_(2))l`
Answer» Correct Answer - B

Discussion

767.	Consider two solid uniform spherical objects of the same density `rho`. One has radius `R` and the other has radius `2R`. They are in outer space where the gravitational fields from other objects are negligible. If they are arranged with their surface touching, what is the contact force between the objects due to their traditional attraction?A. `Gpi^(2)rho^(4)`B. `128/81Gpi^(2)R^(2)rho^(2)`C. `128/81 Gpi^(2)`D. `128/87piR^(2)G`
Answer» Correct Answer - B `m_(1)=4/3piR^(3)rho, m_(2)=4/3pi(2R)^(3)rho`, Distance between centres of two sphericals objects, `r=3R`. `F=(Gm_(1)m_(2))/(r^(2))=G(4/3piR^(3)rho)(8xx4/3piR^(2)rho)xx1/((3R)^(2))` `128/81Gpi^(2)R^(4)rho^(2)`

Discussion

768.	When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant. Q. A particle of mass `m` is projected from the surface of earth with velocity `v_(0)` at angle `theta` with horizontal suppose `h` be the maximum height of particle from surface of earth and v its speed at that point then v is `v_(0)costheta`A. `v_(0) cos theta`B. `gt v_(0) cos theta`C. `lt v_(0) cos theta`D. zero
Answer» Correct Answer - C

Discussion

769.	When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant. Q. Maximum height `h` of the particle isA. `=(v_(0)^(2) sin^(2) theta)/(2g)`B. `gt(v_(0)^(2) sin^(2) theta)/(2g)`C. `lt(v_(0)^(2) sin^(2) theta)/(2g)`D. can be greater than or less than `(v_(0)^(2) sin^(2) theta)/(2g)`
Answer» Correct Answer - B

Discussion

770.	When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant. Q. Maximum height `h` of the particle isA. `=(v_(0)^(2)sin^(2)theta)/(2g)`B. `gt(v_(0)^(2)sin^(2)theta)/(2g)`C. `lt(v_(0)^(2)sin^(2)theta)/(2g)`D. can be greater than or less than `(v_(0)^(2)sin^(2)theta)/(2g)`
Answer» Correct Answer - B By applying conservation energy `(1)/(2)mv_(0)^(2)-(GM_(e)m)/(r)=(1)/(2)mv^(2)-(GM_(e)m)/((R+h))` Solving above equation `hgt(v_(0)^(2)sin^(2)theta)/(2g)` Alternate: As height increases gravitational force decreases and hence the acceleration therefore height will be more than `H=(v_(0)^(2)sin^(2)theta)/(2g)`

Discussion

771.	A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. `gx`B. `(gR)/(R-x)`C. `(gR^(2))/(R-x)`D. `((gR^(2))/(R+x))^(1//2)`
Answer» Correct Answer - D For the satellite, the gravitational force provides the necessary centripetal force, i.e., `(GM_(e)m)/((R+x)^(2))=(mv_(0)^(2))/((R+x))`

Discussion

772.	When a particle is projected from the surface of earth its mechanical energy and angular momentum about center of earth at all time are constant. Q. A particle of mass `m` is projected from the surface of earth with velocity `v_(0)` at angle 0 with horizontal suppose `h` be the maximum height of particle from surface of earth and v its speed at that point then v is`v_(0)cos0`A. `gtv_(0)costheta`B. `ltv_(0)costheta`C. zeroD. `lt(v_(0)^(2)sin^(2)theta)/(2g)`
Answer» Correct Answer - B By applying conservation of angular momentum `mv_(0)Rcostheta=mv(R+h)` `v=(v_(0)Rcostheta)/(R+h)((R)/(R+h)lt1)impliesv_(0)costhetagtv`

Discussion

773.	If `d` is the distance between the centre of the earth of mass `M_(1)` and the moon of mass `M_(2)`, then the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escape isA. `sqrt(G(M_(1)+M_(2)))/(d)`B. `sqrt((G(M_(1)+M_(2)))/(2d))`C. `sqrt((2G(M_(1)+M_(2)))/(d))`D. `sqrt((4G(M_(1)+M_(2)))/d)`
Answer» Correct Answer - D Using law of conservation of energy, `1/2mv_(e)^(2)=(2Gm)/d(M_(1)+M_(2))`

Discussion

774.	The gravitational field in a region due to a certain mass distribution is given by `vec(E)=(4hati-3hatj)N//kg`. The work done by the field in moving a particle of mass `2kg` from `(2m,1m)` to `(2/3m,2m)` along the line `3x+4y=10` isA. `-25/3 N`B. `-50/3 N`C. `25/3N`D. zero
Answer» Correct Answer - B `W=m(vec(E).vec(dr))=m(vec(E).(vec(r)_(2)-vec(r)_(1)))`

Discussion

775.	Two points mass `m` and `2m` are kept at a distance `a`. Find the speed of particles and their relative velocity of approach when separation becomes `a//2`.
Answer» Reduced mass `mu=(m.2m)/(m+2m)=(2m)/(3)` Changining in potential energy is `U_(1)-U_(2)=-(Gm.2m)/(a)-(-(Gm.2m)/(a//2))` `=(2Gm^(2))/(a)` `(1)/(2)mu v_(1//2)^(2)=(2Gm^(2))/(a)implies(1)/(2)=sqrt((6Gm)/(a))` `v_(1//2)=v_(1)+v_(2)=sqrt((6Gm)/(a))` ......(i) `mv_(1)=2mv_(2)` `v_(1)=2v_(2)` `v_(1)+v_(2)=3v_(2)=sqrt((6Gm)/(a))implies v_(2)=(1)/(3)sqrt((6Gm)/(a))=sqrt((2Gm)/(3a))` `v_(1)=2v_(2)=2aqrt((2Gm)/(3a))`

Discussion

776.	Choose the odd one out :`983m//s^(2), 977m//s^(2),980m//s^(2),9.83m//s^(2)`
Answer» `9.83m//s^(2)`. Others are in CGS system.

Discussion

777.	Choose the odd one out :`9.83m//s^(2),98.3m//s^(2),983m//s, 98.03m^(2)s`
Answer» `9.83m//s^(2)`. Others are not values of g.

Discussion

778.	A satellite orbits the earth at a height of `3.6xx10^(6)m` from its surface. Compute it’s a kinetic energy, b. potential energy, c. total energy. Mass of the satellite `=500kg` mass of the earth `=6xx10^(24)`kg, radius of the earth `=6.4xx10^(6), G=6.67xx10^(-11)Nm^(2)kg^(-2)`.
Answer» Here `r=R+h=6.4xx10^(6)+3.6xx10^(6)=10^(7)m` Orbital velocity of the satellite around the earth is given by `v_(0)=sqrt((GM)/((R+h)))=sqrt((6.67xx10^(-11)xx6xx10^(24))/(10^(7)))` `=sqrt(6.67xx6xx10^(6))ms^(-1)` a. `KE` of the satellite `=1/2mv_(0)^(2)` `=1/2xx500xx6.67x6xx10^()6=10^(10)J` b. `PE` of the satellite `=-(GMm)/(R+H)` `=-(6.67xx10^(-11)xx(6xx10^(24))xx500)/(10^(7))` `=-2xx10^(10)J` c. Total energy `=KE+PE=10^(10)-2xx10^(10)=-10^(10)J`

Discussion

779.	An artificial satellite revolves round the earth at a height of `1000 km`. The radius of the earth is `6.38xx10^(3)km`. Mass of the earth `6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2)`. Find the orbital speed and period of revolution of the satellite.
Answer» Here `h=1000 km=1000xx10^(3)m=10^(6)m` `r=R+h=6.38xx10^(6)+10^(6)=7.38xx10^(6)m` Orbital speed, `v_(0)=sqrt((GM)/(R+h))=sqrt((6.67xx10^(-11)xx6xx10^(24))/(7.38xx10^(6)))=7364ms^(-1)` Time period `T=(2pir)/(v_(0))=(2xx(22/7)xx(7.38xx10^(6)))/(7.364xx10^(3))=6297s`

Discussion

780.	`6xx10^(24)kg`: Mass of the Earth :: `6.4xx10^(6)m`: ______________.
Answer» Radius of the Earth Mass is measured in kg and distance is measured in metres.

Discussion

781.	The time period of an earth satellite in circular orbit is independent ofA. the mass of the satelliteB. radius of its orbitC. both the mass and radius of the orbitD. neither the mass of the satellite nor the radius of its orbit
Answer» Correct Answer - A

Discussion

782.	What is the acceleration of free fall?
Answer» When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms⁻², which is constant for all objects (irrespective of their masses).

Discussion

783.	What do we call the gravitational force between the Earth and an object?
Answer» Gravitational force between the earth and an object is known as the weight of the object.

Discussion

784.	In the usual notation, the acceleration due to gravity at a height h from the surface of the earth is ……..(a) \(g=\frac{GM}{(R\,+\,h)}\)(b) \(g=\frac{GM}{\sqrt{R\,+\,h}}\)(c) \(g=\frac{GM}{(R\,+\,h)^2}\)(d) g = GM (R + h)2
Answer» Answer is (c) g = \(\frac{GM}{(R+h)^2}\)

Discussion

785.	a) Is the acceleration due to gravity of earth ‘g’ a constant? Discuss. b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 1022 kg and a radius of 1.74 × 106m. Which satellite do you think it could be?
Answer» a) No, the value of acceleration due to gravity is no constant at all the places on the surface of the earth. This is because the radius of the earth is minimum at the poles while maximum at the equator. The value of g is maximum at the poles and minimum at the equator. The value of g increases as we go from the centre of the earth towards the poles and the value of g decreases as we go inside the earth. b) Acceleration due to gravity, g = G M/R² Mass, M = 7.4 × 10²² kg Radius, R = 1.74 × 10⁶m Gravitational constant, G = 6.7× 10^-11Nm²/kg² Therefore, substituting the values we get g = 1.637 m/s^2.

785.

a) Is the acceleration due to gravity of earth ‘g’ a constant? Discuss. b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 1022 kg and a radius of 1.74 × 106m. Which satellite do you think it could be?

Answer»

a) No, the value of acceleration due to gravity is no constant at all the places on the surface of the earth. This is because the radius of the earth is minimum at the poles while maximum at the equator. The value of g is maximum at the poles and minimum at the equator. The value of g increases as we go from the centre of the earth towards the poles and the value of g decreases as we go inside the earth.

b) Acceleration due to gravity,

g = G M/R²

Mass, M = 7.4 × 10²² kg

Radius, R = 1.74 × 10⁶m

Gravitational constant, G = 6.7× 10^-11Nm²/kg²

Therefore, substituting the values we get g = 1.637 m/s^2.

Discussion

786.

Differentiate between the mass of an object and its weight.

Answer»

Mass	Weight
Mass is the quantity of matter contained in the body.	Weight is the force of gravity acting on the body.
It is the measure of inertia of the body.	It is the measure of gravity.
Mass is a constant quantity.	Weight is not a constant quantity. It is different at different places.
It only has magnitude.	It has magnitude as well as direction.
Its SI unit is kilogram (kg).	Its SI unit is the same as the SI unit of force, i.e., Newton(N)

Discussion

787.	a) What do you understand by the term ‘acceleration due to gravity of earth’? b) What is the usual value of acceleration due to gravity of earth? c) State the SI unit of acceleration due to gravity.
Answer» a) Acceleration due to gravity is the uniform acceleration produced by the body that is falling freely due to gravitational force of the earth. b) Usual value of acceleration due to gravity of earth is 9.8 m/s^2. c) SI unit of acceleration due to gravity is m/s².

787.

a) What do you understand by the term ‘acceleration due to gravity of earth’? b) What is the usual value of acceleration due to gravity of earth? c) State the SI unit of acceleration due to gravity.

Answer»

a) Acceleration due to gravity is the uniform acceleration produced by the body that is falling freely due to gravitational force of the earth.

b) Usual value of acceleration due to gravity of earth is 9.8 m/s^2.

c) SI unit of acceleration due to gravity is m/s².

Discussion

788.	An object has mass of 20 kg on the earth. What will be its a) mass b) weight on the moon.
Answer» a) The mass of an object on the moon will be 20 kg. b) The weight of the object on the moon will be (20)(1.6) = 32N.

788.

An object has mass of 20 kg on the earth. What will be its a) mass b) weight on the moon.

Answer»

a) The mass of an object on the moon will be 20 kg.

b) The weight of the object on the moon will be (20)(1.6) = 32N.

Discussion

789.	a) State the universal law of gravitation. Name the scientist who gave this law. b) Define gravitational constant. What are the units of gravitational constant?
Answer» a) The universal law of gravitation states that every body in the universe attracts every other body with a force F which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. This law was given by Sir Isaac Newton. F = G × (m × M)/d² b) The gravitational constant G is equal to the force of gravitation which exists between the two bodies of unit masses that are kept at a unit distance from each other. The unit of gravitational constant is Nm²/kg². G = F × d²/(m × M)

789.

a) State the universal law of gravitation. Name the scientist who gave this law. b) Define gravitational constant. What are the units of gravitational constant?

Answer»

a) The universal law of gravitation states that every body in the universe attracts every other body with a force F which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

This law was given by Sir Isaac Newton.

F = G × (m × M)/d²

b) The gravitational constant G is equal to the force of gravitation which exists between the two bodies of unit masses that are kept at a unit distance from each other.

The unit of gravitational constant is Nm²/kg².

G = F × d²/(m × M)

Discussion

790.	Which is more fundamental, the mass of a body or its weight? Why?
Answer» The mass of the body is more fundamental than the weight because mass of the body is constant and does not vary from place to place.

Discussion

791.	What do you mean by acceleration due to gravity?
Answer» When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s².

Discussion

792.	a) Define mass of a body. What is the SI unit of mass? b) Define weight of a body. What is the SI unit of weight? c) What is the relation between mass and weight of a body?
Answer» a) The mass of the body is defined as the quantity of matter that is contained in it. The SI unit of mass is kg. b) The weight of the body is defined as the force with which the body gets attracted towards the centre of the earth. The SI unit is N. c) The relation between mass and weight of a body is W = mg.

792.

a) Define mass of a body. What is the SI unit of mass? b) Define weight of a body. What is the SI unit of weight? c) What is the relation between mass and weight of a body?

Answer»

a) The mass of the body is defined as the quantity of matter that is contained in it. The SI unit of mass is kg.

b) The weight of the body is defined as the force with which the body gets attracted towards the centre of the earth. The SI unit is N.

c) The relation between mass and weight of a body is W = mg.

Discussion

793.	What do you mean by free fall?
Answer» Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall.

Discussion

794.	How much is the weight of an object on the moon as compared to its weight on the earth? Give reason for your answer.
Answer» The weight of an object on the moon as compared to its weight on the earth is one-sixth of its weight on the earth. This is because the acceleration due to gravity on the moon is one-sixth of that on the earth.

Discussion

795.	What is the intensity of gravitational field at the center of spherical shell?(a)\(\frac{Gm}{r^2}\) (b) g (c) zero (d) None of these
Answer» Correct answer is (c) zero

Discussion

796.	The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?
Answer» The moon too revolves around the sun along with the earth.

Discussion

797.	A thin spherical shell of mass `M` and radius `R` has a small hole. A particle of mass `m` released at its mouth. ThenA. The particle wil execute simple harmonic motion inside the shellB. The particle will oscillate inside the shell, but the oscillations are not simple harmonicC. The particle will not oscillate, but the speed of the particle will go on increasingD. None of these
Answer» Correct Answer - D

Discussion

798.	At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface ( R = radius of earth)A. `2 R`B. `R`C. `1.414 R`D. `0.414 R`
Answer» Correct Answer - D

Discussion

799.	A spy satellite `S_(1),` travelling above the equator is taking pictures at quick intervals. The satellite is travelling from west to east and is ready with picture around the whole equator in 8 hours. Another similar satellite `S_(2),` travelling in the same plane is travelling from east to west and is able to take pictures around the whole equator in 6 hours.Find the ratio of radii of the circular paths of the satellite `S_(1)` and `S_(2).`
Answer» Correct Answer - `(r_(1))/(r_(2))=(3/4)^(2//3)`

Discussion

800.	A comet is going around the sun in an elliptical orbit with a period of 64 year. The closest approach of the comet to the sun is 0.8 AU [AU = astronomical unit]. Calculate the greatest distance of the comet form the sun.
Answer» Correct Answer - 31.2 AU

Discussion

Explore topic-wise InterviewSolutions in .

Write true or false for the following statements:The inertia of an object depends upon its mass.

Write true or false for the following statements:The value of g is zero at the centre of the earth.

Name the scientist who gave the magnitude of buoyant force acting on a solid object immersed in a liquid.

Why the value of 'g' is more at the poles than at the equator?

Show that the weight of all bodies is zero at the centre of the earth.

The density of gold is `19 g //cm^(2)`. Find the volume of 95 g of gold.

Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?

A body in the uniform circular motion then the parts 1 and 2 represents A) Time, Velocity B) Speed, Velocity C) Velocity, Acceleration D) Acceleration, Velocity

Two electrons, each having mass 9.1 × 10-31 kg, are separated by a distance 10 Å. Calculate the gravitational force between them.

Why moon does not have an atmosphere?

State true/false. If false, correct the statement: “Earth exerts a greater force of attraction on apple than apple exerts on the earth.”

A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. `gx`B. `(gR)/(R-x)`C. `(gR^(2))/(R-x)`D. `((gR^(2))/(R+x))^(1//2)`

The gravitational field in a region due to a certain mass distribution is given by `vec(E)=(4hati-3hatj)N//kg`. The work done by the field in moving a particle of mass `2kg` from `(2m,1m)` to `(2/3m,2m)` along the line `3x+4y=10` isA. `-25/3 N`B. `-50/3 N`C. `25/3N`D. zero

Two points mass `m` and `2m` are kept at a distance `a`. Find the speed of particles and their relative velocity of approach when separation becomes `a//2`.

Choose the odd one out :`983m//s^(2), 977m//s^(2),980m//s^(2),9.83m//s^(2)`

Choose the odd one out :`9.83m//s^(2),98.3m//s^(2),983m//s, 98.03m^(2)s`

A satellite orbits the earth at a height of `3.6xx10^(6)m` from its surface. Compute it’s a kinetic energy, b. potential energy, c. total energy. Mass of the satellite `=500kg` mass of the earth `=6xx10^(24)`kg, radius of the earth `=6.4xx10^(6), G=6.67xx10^(-11)Nm^(2)kg^(-2)`.

An artificial satellite revolves round the earth at a height of `1000 km`. The radius of the earth is `6.38xx10^(3)km`. Mass of the earth `6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2)`. Find the orbital speed and period of revolution of the satellite.

`6xx10^(24)kg`: Mass of the Earth :: `6.4xx10^(6)m`: ______________.

The time period of an earth satellite in circular orbit is independent ofA. the mass of the satelliteB. radius of its orbitC. both the mass and radius of the orbitD. neither the mass of the satellite nor the radius of its orbit

What is the acceleration of free fall?

What do we call the gravitational force between the Earth and an object?

In the usual notation, the acceleration due to gravity at a height h from the surface of the earth is ……..(a) \(g=\frac{GM}{(R\,+\,h)}\)(b) \(g=\frac{GM}{\sqrt{R\,+\,h}}\)(c) \(g=\frac{GM}{(R\,+\,h)^2}\)(d) g = GM (R + h)2

a) Is the acceleration due to gravity of earth ‘g’ a constant? Discuss. b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 1022 kg and a radius of 1.74 × 106m. Which satellite do you think it could be?

Differentiate between the mass of an object and its weight.

a) What do you understand by the term ‘acceleration due to gravity of earth’? b) What is the usual value of acceleration due to gravity of earth? c) State the SI unit of acceleration due to gravity.

An object has mass of 20 kg on the earth. What will be its a) mass b) weight on the moon.

a) State the universal law of gravitation. Name the scientist who gave this law. b) Define gravitational constant. What are the units of gravitational constant?

Which is more fundamental, the mass of a body or its weight? Why?

What do you mean by acceleration due to gravity?

a) Define mass of a body. What is the SI unit of mass? b) Define weight of a body. What is the SI unit of weight? c) What is the relation between mass and weight of a body?

What do you mean by free fall?

How much is the weight of an object on the moon as compared to its weight on the earth? Give reason for your answer.

What is the intensity of gravitational field at the center of spherical shell?(a)\(\frac{Gm}{r^2}\) (b) g (c) zero (d) None of these

The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?

At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface ( R = radius of earth)A. `2 R`B. `R`C. `1.414 R`D. `0.414 R`

A comet is going around the sun in an elliptical orbit with a period of 64 year. The closest approach of the comet to the sun is 0.8 AU [AU = astronomical unit]. Calculate the greatest distance of the comet form the sun.