2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 17 InterviewSolution

801.	The radius of a planet is `1/4` of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surfaceA. `1//sqrt2`B. `sqrt2`C. `2sqrt2`D. 2
Answer» Correct Answer - a `(v_(e_(2)))/(v_(e_(1)))=sqrt((g_(2)R_(2))/(g_(1)R_(1)))=sqrt((1)/(4)xx2)=(1)/(sqrt2)`.

Discussion

802.	Potential energy of a 3kg body at the surface of a planet is `-54 J`, then escape velocity will be :A. 18 m/sB. 162 m/sC. 36 m/sD. 6 m/s
Answer» Correct Answer - D

Discussion

803.	Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet isA. `- 5000 J`B. `- 1000 J`C. `-2400 J`D. `-10000 J`
Answer» Correct Answer - A

Discussion

804.	Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet isA. 5000JB. `-5000J`C. `-2400J`D. `-1000J`
Answer» Correct Answer - B Escape velocity `(v_(e ))=sqrt((2GM)/(R )) " " therefore v_(e )^(2)=(2GM)/(R )` `therefore (GM)/(R )=(v_(e )^(2))/(2)" "` ……(1) and gravitational P.E. (U) `=-((GM)/(R ))m` `=-((v_(e)^(2))/(2))m " "` …. From (1) `therefore U=-((100xx100)/(2)xx1)=-5000 J`

Discussion

805.	The binding energy of an artifical satellite moving in a circular orbit is `4xx10^(9)J` . What is the potential energy of the satellite?A. `-4xx10^(9)`JB. `-8xx10^(9)J`C. `2xx10^(9)J`D. `6xx10^(9)`J
Answer» Correct Answer - B `B.E.=-(1)/(2)(P.E.)` `therefore P.E. = -2xx4xx10^(9)=-8xx10^(9)J`

Discussion

806.	The gravitational potential energy of a body of mass ‘ m ’ at the earth’s surface `-mgR_(e)`. Its gravitational potential energy at a height `R_(e)` from the earth’s surface will be (Here `R_(e)` is the radius of the earth)A. `2 mgR_(e)`B. `2 mgR_(e)`C. `1/2 mgR_(e)`D. `-1/2 mgR_(e)`
Answer» Correct Answer - D

Discussion

807.	Assertion : A particle is projected upwards with speed `upsilon` and it goes to a heigth `h`. If we double the speed then it will move to height `4h`. Reason : In case of earth, acceleration due to gravity g varies as `g prop(1)/(r^(2))` (for `r ge R`)A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explantion of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - D `h = (u^(2))/(2g)` or `h prop u^(2)` cannot be applied in this case. Beause, for heigher value of `upsilon`, acceleration due to gravity g does not remain constant.

Discussion

808.	The percentage change in the acceleration of the earth towards the Sun from a total eclipse of the Sun to the point where the Moon is on a side of earth directly opposite to the Sun isA. `(M_(s))/(M_(m)) (r_(2))/(r_(1))xx100`B. `(M_(s))/(M_(m))((r_(2))/(r_(1)))xx100`C. `2((r_(1))/(r_(2)))^(2) (M_(m))/(M_(s))xx100`D. `((r_(1))/(r_(2)))^(2)(M_(m))/(M_(s))xx100`
Answer» Correct Answer - C During total eclipse total attraction due to the sun and the moon. `F_(1)=(GM_(s)M_(e))/(r_(1)^(2))+(GM_(m)M_(e))/(R_(2)^(2)` When the Moon goes on the opposite side to the earth, the effective force of attraction is `F_(2)=(GM_(s)M_(e))/(r_(1)^(2))-(GM_(m)M_(e))/(r_(2)^(2))` Change in force `/_F=F_(1)-F_(2)=(2GM_(m)M_(e))/(r_(2)^(2)` change in acceleration of the earth `/_a=(/_F)/(M_(e))=(2Gmm)/(r_(2)^(2))` Average force on the earth `F_(av)=(F_(1)+F_(2))/2=(GM_(s)M_(e))/(r_(1)^(2))` Average acceleration of the earth `a_(av)+F_(av)/(M_(e))=(GM_(s))/(r_(1)^(2))` Percentage change in acceleration is `(/_a)/a_(av)xx100=(2GM_(m))/(r_(2)^(2))xx(r_(1)^(2))/(GM_(s))xx100=2((r_(1))/(r_(2)))^(2)(M_(m))/(M_(s))xx100`

Discussion

809.	Assertion : There are two identical spherical bodies fixed in two positions as shows. While moving from `A` to `B` gravitational potential first increases then decreases. Reason : At centre point of `A` and `B` field strength will be zero.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explantion of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - B If a mass `m` is displaced from centre along the line `AB`, force it is away from the centre . So, it is in ustable equilibrium position at centre. So, potential energy and hence the potential at centre is maximum.

Discussion

810.	A star `2.5` times the mass of the sun is reduced to a size of `12km` and rotates with a speed of `1.5rps.` Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun `=2xx10^(30) kg`)
Answer» Acceleration due to gravity, `g=(GM)/(R^(2))` `=(6.67xx10^(-11)xx2.5xx2xx10^(30))/((12000)^(2))=2.3xx10^(12) ms^(-2)` Centrifugal acceleration `=r omega^(2)=r(2pif)^(2)=12000(2pixx1.5)^(2)=1.1xx10^(6) ms^(-2)` Since, `g gt r omega^(2)`, the body will remain stuck with the surface of the star.

Discussion

811.	Why is it that we can learn more about the shape of the earth by studying the motion of an artificial satellite than by studying the motion of the Moon?
Answer» We can learn more about the shape of the earth by studying the motion of an artificial satellite than by studying the motion of the moon because the motion of the satellite is governed by the value of `g` at each of its positions. Since the moon is very fary away, the variation of `g` near the surface of the earth is averaged out at the position of the moon.

Discussion

812.	An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is ……..(a) 98 N (b) zero (c) 49 N (d) 9.8 N
Answer» (b) zero The lift is in freefall. It and its contents will experience apparent weightlessness just like astronauts. The spring balance reading will change from 100 N to zero.

812.

An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is ……..(a) 98 N (b) zero (c) 49 N (d) 9.8 N

Answer»

(b) zero

The lift is in freefall. It and its contents will experience apparent weightlessness just like astronauts.

The spring balance reading will change from 100 N to zero.

Discussion

813.	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.A. The acceleration of `S` is always directed towards the centre of the earth.B. The angular momentum of `S` about the centre of the earth changes in direction, but its magnitude remains constant.C. The total mechanical energy of `S` varies periodically with time.D. The linear momentum of `S` remains constant in magnitude.
Answer» Correct Answer - A::C Force on the satellite is always towards the earth, therefore, acceleration of satellite `S` is always directed towards the centre of the earth. Net torque of this gravitational force `F` about the centre of the earth is zero. Therefore, angular momentum (both in magnitude and direction) fo `S` about the centre of the earth is constant throughout. Since force `F` is conservative in nature, therefore mechanical energy of the satellite remain constant. Speed of `S` is maximum when it is nearest to the earth and minimum when it is farthest.

Discussion

814.	The magnitude of the gravitational field at distance `r_(1)` and `r_(2)` from the centre of a uniform sphere of radius `R` and mass `M` are `F_(1)` and `F_(2)` respectively. Then:A. `(F_(1))/(F_(2))=(r_(1))/(r_(2))` if `r_(1)ltR` and `r_(2)ltR`B. `(r_(2)^(2))/(r_(2))` if `r_(1)gtR` and `r_(2)gtR`C. `(F_(1))/(F_(2))=(r_(1))/(r_(2))` if `r_(1)gtR` and `r_(2)gtR`D. `(F_(1))/(F_(2))=(r_(1)^(2))/(r_(2)^(2)` if `r_(1)ltR` and `r_(2)ltR`
Answer» Correct Answer - A::B For `rgtR,` the gravitational field is `F=GM//r^(2)` `:. F_(1)=(GM)/(r_(1)^(3))` and `F_(2)=(GM)/(r_(2)^(2))implies(F_(1))/(F_(2))=(r_(2)^(2))/(R_(1)^(2))` For `rgt gtR` The gravitational field is `F=(GM)/(R^(3))xxr` `:. F_(1)=(GM)/R^(3)xxr_(1)` and `F_(2)=(GM)/(R^(3))xxr_(2)` `implies (F_(1))/(F_(2))=(r_(1))/(r_(2))`

Discussion

815.	A satellite is revolving round the earth in circular orbitA. if mass of earth is made four times, keeping other factors constant, orbital speed of satellite will become two timesB. corresponding to change in part(a), times period of satellite will become halfC. when value of `G` is made two times orbital speed increases and time period decreasesD. `G` has no effect on orbital speed and time period
Answer» Correct Answer - A::B::C `nu = sqrt((Gm)/(r ))` or `nu prop sqrt(GM)` `T = (2 pi)/(sqrt(GM))r^(3//2)` or `T prop (1)/(sqrt(GM))`

Discussion

816.	Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.
Answer» Correct Answer - A::B::C `W = DeltaU = U_(f) - U_(i)` `= 3[(-Gmm)/(2a)] -3 [(-Gmm)/(a)]` `= (3 Gmm)/(2a) = (3 Gm^(2))/(2a)`

Discussion

817.	A spherically symmetric gravitational system of particles has a mass density` rho={(rho_0,for, r,lt,R),(0,for,r,gt,R):}` where`rho_0` is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed v as a function of distahce `r(0ltrltOO)` form the centre of the system is represented byA. B. C. D.
Answer» Correct Answer - C

Discussion

818.	If the earth shrinks of half of its radius its mass remaining same the weight of the object on the earth will change toA. 2 timesB. 6 timesC. 4 timesD. no change at all
Answer» Correct Answer - c `R_(2)=(1)/(2)R_(1)` `(g_(2))/(g_(1))=((R_(2))/(R_(1)))^(2)=(1)/(4)` `mg_(2)=(g_(1))/(4)m` `F_(2)=(F_(1))/(4)` `therefore F_(1)=4F_(2).`

Discussion

819.	A body of mass 5 kg is taken to the centre of the earth. What will be its (i) mass, (ii) weight there.
Answer» Mass does not change, weight at centre of earth will be 0 because g = 0.

Discussion

820.	Why is gravitational potential energy negative ?
Answer» Because it arises due to attractive force of gravitation.

Discussion

821.	The escape velocity of a body from the earth is `V_(e)` if the radius of earth contracts to `1//4` th of its value keeping the mass of the earth constant the escape velocity will beA. doubledB. halvedC. tripledD. unaltered
Answer» Correct Answer - A Escape velocity `V_(e)=sqrt(2GM)/(R ),R=(R )/(4)` opr `V_(e )=2 sqrt(2GM)/(R )` since G and M are constant hence `V_(e )=2v_(e )`

Discussion

822.	A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r(rltR), then the correct option(s) is (are)A. `P(r = 0)=0`B. `(P(r = 3R//4))/(P(r=2R//3))=(63)/(80)`C. `(P(r = 3R//5))/(P(r=2R//5))=(16)/(21)`D. `(P(r=R//2))/(P(r=R//3))=(20)/(27)`
Answer» Correct Answer - B::C

Discussion

823.	The force of attraction between two unit point masses separated by a unit distance is called: a) gravitational potential b) acceleration due to gravity c) gravitational field strength d) universal gravitational constant
Answer» The correct answer is d) universal gravitational constant

Discussion

824.	The atmosphere consisting of a large number of gases is held to the earth by: a) winds b) clouds c) earth’s magnetic fieldd) gravity
Answer» The correct answer is d) gravity

Discussion

825.	The value of g on the surface of the moon: a) is the same as on the earth b) is less than that on the earth c) is more than that on the earth d) keeps changing day by day
Answer» The correct answer is b) is less than that on the earth

Discussion

826.	The mass of moon is about 0.012 times that of earth and its diameter is about 0.25 times that of earth .The value of G on the moon will beA. less than that of earthB. more than that on the earthC. same as that on the earthD. about one sixth of that on the earth
Answer» Correct Answer - C

Discussion

827.	The mass of moon is about 0.012 times that of the earth and its diameter is about 0.25 times that of earth. The value of G on the moon will be:a) less than that on the earth b) more than that on the earth c) same as that on the earth d) about one-sixth of that on the earth
Answer» The correct answer is c) same as that on the earth

Discussion

828.	A uniform ring of mass m is lying at a distance `sqrt(3)` a from the centre of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Find the magnitude of gravitational force between them . A. `(GMm)/(8R^(2))`B. `(GMm)/(3R^(2))`C. `sqrt3(GMm)/(R^(2))`D. `sqrt3(GMm)/(8R^(2))`
Answer» Correct Answer - D

Discussion

829.	The gravitational potential of two homogeneous spherical shells `A` and `B` of same surface density at their respective centres are in the ratio `3:4`. If the two shells collapse into a single one such that surface charge density remains the same, then the ratio of potential at an internal point of the new shell to shell `A` is equal toA. `3:2`B. `4:3`C. `5:3`D. `5:6`
Answer» Correct Answer - C `4pi^(2)rho=4pir_(1)^(2)rhorArr r^(2)=r_(1)^(2)+r_(2)^(2)` `V=(-GM)/r=-(G4pir^(3)rho)/r` `V=-4pirGrhorArr Vpropr` `(V_(1))/(V_(2))=(r_(1))/(r_(2))=3/4rArr (r_(1)^(2))/(r_(2)^(2))=9/16` `r_(1)^(2):r_(2)^(2):r^(2)=r_(1)^(2):r_(2)^(2):(r_(1)^(2)+r_(2)^(2))=9:16:(9+16)` `rArr r_(1):r_(2):r=3:4:5=V_(1):V_(2):V`

Discussion

830.	The gravitational potential of two homogeneous spherical shells `A` and `B` of same surface density at their respective centres are in the ratio `3:4`. If the two shells collapse into a single one such that surface charge density remains the same, then the ratio of potential at an internal point of the new shell to shell `A` is equal toA. `3 : 5`B. `4 : 5`C. `5 : 3`D. `5 : 4`
Answer» Correct Answer - C

Discussion

831.	The gravitational potential of two homogeneous spherical shells `A` and `B` of same surface density at their respective centres are in the ratio `3:4`. If the two shells collapse into a single one such that surface charge density remains the same, then the ratio of potential at an internal point of the new shell to shell `A` is equal toA. `3:2`B. `4:3`C. `5:3`D. `3:5`
Answer» Correct Answer - C

Discussion

832.	The atmosphere is held to the earth byA. WindsB. GravityC. CloudsD. None of the above
Answer» Correct Answer - B

Discussion

833.	The weight of a body at the centre of the earth isA. ZeroB. InfiniteC. Same as on the surface of earthD. None of the above 6
Answer» Correct Answer - A

Discussion

834.	What is the intensity of gravitational field of the centre of a spherical shellA. `Gm//r^(2)`B. `g`C. ZeroD. None of these
Answer» Correct Answer - C

Discussion

835.	There is an infinite thin flat sheet with mass density `eta` per unit area. Find the gravitational force, due to sheet, on a point mass m located at a distance x from the sheet.Consider a large flat horizontal sheet of material density r and thickness t, placed on the surface of the earth. The density `phi`of the earth is `phi_(0)` If it is found that ravitational field intensity just between the sheet is larger than field just above it, prove that rho_(@)lt(3)/(2)rho.`
Answer» Correct Answer - (a) `G2pisigma m`

Discussion

836.	(a) What is meant by the terms mass and weight?(b) Are they vector of scalar quantities? Why?(c) The mass of a body is 30 kg. What is its weight on earth? (g = 9.8 m/s2)(d) What is its weight on the moon? (g = 1.62 m/s2)
Answer» (a) 1. The quantity of matter present in a body is called mass. 2. The force of attraction exerted by earth on a body is referred as weight. (b) Mass is a scalar quantity (No direction) Weight is a vector quantity (Possess both magnitude and direction) (c) Weight = mg = 30 × 9 = 294 N (d) Weight on the moon = 30 × 1.62 = 48N

836.

(a) What is meant by the terms mass and weight?(b) Are they vector of scalar quantities? Why?(c) The mass of a body is 30 kg. What is its weight on earth? (g = 9.8 m/s2)(d) What is its weight on the moon? (g = 1.62 m/s2)

Answer»

(a) 1. The quantity of matter present in a body is called mass.

2. The force of attraction exerted by earth on a body is referred as weight.

(b) Mass is a scalar quantity (No direction) Weight is a vector quantity (Possess both magnitude and direction)

(c) Weight = mg = 30 × 9 = 294 N

(d) Weight on the moon = 30 × 1.62 = 48N

Discussion

837.	If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is the mutual force of attraction between them?
Answer» F = \(\frac{Gm_1m_2}{d^2}\) m₁ = 40 kg, m₂ = 60 kg, d = 0.5 m = \(\frac{G\times40\times60}{(0.5)^2}\) = \(\frac{G\times 2400}{0.25}\) = \(\frac{6.67\times10^{-11}\times2400}{0.25}\) = 64032 x 10^-11 = 6.4032 x 10^-7N

837.

If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is the mutual force of attraction between them?

Answer»

F = \(\frac{Gm_1m_2}{d^2}\)

m₁ = 40 kg,

m₂ = 60 kg,

d = 0.5 m

= \(\frac{G\times40\times60}{(0.5)^2}\)

= \(\frac{G\times 2400}{0.25}\)

= \(\frac{6.67\times10^{-11}\times2400}{0.25}\)

= 64032 x 10^-11

= 6.4032 x 10^-7N

Discussion

838.	If the earth is regarded as a hollow sphere, then what is the weight of an object below the surface of earth?
Answer» Zero. If the earth is regarded as a hollow sphere, then the weight of an object below the surface of earth is Zero.

838.

If the earth is regarded as a hollow sphere, then what is the weight of an object below the surface of earth?

Answer»

Zero.

If the earth is regarded as a hollow sphere, then the weight of an object below the surface of earth is Zero.

Discussion

839.

Observe the figure and complete the tableAttracting bodiesForce of gravitationA, BB, CC, A

Answer»

The table is given below:

Attracting bodies	Force of gravitation
A, B	\(\frac{G\times20\times30}{10^2}=6\,G\)
B, C	\(\frac{G\times30\times60}{15^2}=8\,G\)
C, A	\(\frac{G\times20\times60}{20^2}=3\,G\)

Discussion

840.	The orbital velocity of the stationary satellite of earth is nearly…
Answer» 30 kms^-1 . The orbital velocity of the stationary satellite of earth is nearly 30 kms^-1.

840.

The orbital velocity of the stationary satellite of earth is nearly…

Answer»

30 kms^-1 .

The orbital velocity of the stationary satellite of earth is nearly 30 kms^-1.

Discussion

841.	Fill in the blanks with appropriate words and write the completed sentences :i. According to Kepler’s second law, the line joining the planet and the Sun …….. in equal intervals of time.ii. According to Kepler’s third law T2 ∝ rn, where n = ……..iii. For a freely falling object, we can write Newton’s second equation of motion as ……..
Answer» i. According to Kepler’s second law, the line joining the planet and the Sun Sweeps equal areas in equal intervals of time. ii. According to Kepler’s third law T² ∝ rⁿ, where n = 3 iii. For a freely falling object, we can write Newton’s second equation of motion as S = \(\frac{1}{2}\)gt²

841.

Fill in the blanks with appropriate words and write the completed sentences :i. According to Kepler’s second law, the line joining the planet and the Sun …….. in equal intervals of time.ii. According to Kepler’s third law T2 ∝ rn, where n = ……..iii. For a freely falling object, we can write Newton’s second equation of motion as ……..

Answer»

i. According to Kepler’s second law, the line joining the planet and the Sun Sweeps equal areas in equal intervals of time.

ii. According to Kepler’s third law T² ∝ rⁿ, where n = 3

iii. For a freely falling object, we can write Newton’s second equation of motion as S = \(\frac{1}{2}\)gt²

Discussion

842.	Why do different planets have different escape speeds?
Answer» A escape speed is given by `sqrt(GM//R)` therefore its values are different for different planets because of their different masses and different size.

Discussion

843.	In which case was the reading higher?
Answer» Reading of the spring balance is higher when the stone having greater mass is weighed.

Discussion

844.	On which fundamental law of physics is keplers second law is based?
Answer» Low of conservation of angular momentum.

Discussion

845.	Distance between two bodies is increased to three times its original value. What is the effect on the gravitational force between them?
Answer» Since F ∝ 1/r² r¹ → 3r → Force will be decreased to 1/9 times

845.

Distance between two bodies is increased to three times its original value. What is the effect on the gravitational force between them?

Answer»

Since F ∝ 1/r²

r¹ → 3r

→ Force will be decreased to 1/9 times

Discussion

846.	Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?
Answer» In accordance with the universal law of gravitation both the forces are equal and opposite.

Discussion

847.	Which of the stones experienced greater force of attraction of the earth?
Answer» Greater force of attraction of the earth is experienced in stone having greater mas

Discussion

848.	If a planet of given density were made larger, its force of attraction for an object on its surface would increase because of the greater distance from the object to the centre of the planet. Which effect predominates?
Answer» Due to the rotation of the earth fro m west to east the stone will deviate slightly to the east.

Discussion

849.	A satellite of mass 1000 kg revolves around the earth in a circular path. If the distance between the satellite and the centre of the earth is 40000 km, find the gravitational force exerted on the satellite by the earth.[G = 6.67 × 10-11 N.m2/kg2, mass or the earth = 6 × 1024 kg, radius or the earthe 6.4 × 106 m]
Answer» Answer is 250.1 N

Discussion

850.	The masses of two spheres are 10 kg and 20 kg respectively. If the distance between their centers is 100 m, find the magnitude of the gravitational force between them.
Answer» Answer is 1.334 × 10^-12 N)

Discussion

Explore topic-wise InterviewSolutions in .

The radius of a planet is `1/4` of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surfaceA. `1//sqrt2`B. `sqrt2`C. `2sqrt2`D. 2

Potential energy of a 3kg body at the surface of a planet is `-54 J`, then escape velocity will be :A. 18 m/sB. 162 m/sC. 36 m/sD. 6 m/s

Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet isA. `- 5000 J`B. `- 1000 J`C. `-2400 J`D. `-10000 J`

Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet isA. 5000JB. `-5000J`C. `-2400J`D. `-1000J`

The binding energy of an artifical satellite moving in a circular orbit is `4xx10^(9)J` . What is the potential energy of the satellite?A. `-4xx10^(9)`JB. `-8xx10^(9)J`C. `2xx10^(9)J`D. `6xx10^(9)`J

The gravitational potential energy of a body of mass ‘ m ’ at the earth’s surface `-mgR_(e)`. Its gravitational potential energy at a height `R_(e)` from the earth’s surface will be (Here `R_(e)` is the radius of the earth)A. `2 mgR_(e)`B. `2 mgR_(e)`C. `1/2 mgR_(e)`D. `-1/2 mgR_(e)`

A star `2.5` times the mass of the sun is reduced to a size of `12km` and rotates with a speed of `1.5rps.` Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun `=2xx10^(30) kg`)

Why is it that we can learn more about the shape of the earth by studying the motion of an artificial satellite than by studying the motion of the Moon?

An object of mass 10 kg is hanging on a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is ……..(a) 98 N (b) zero (c) 49 N (d) 9.8 N

Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.

If the earth shrinks of half of its radius its mass remaining same the weight of the object on the earth will change toA. 2 timesB. 6 timesC. 4 timesD. no change at all

A body of mass 5 kg is taken to the centre of the earth. What will be its (i) mass, (ii) weight there.

Why is gravitational potential energy negative ?

The escape velocity of a body from the earth is `V_(e)` if the radius of earth contracts to `1//4` th of its value keeping the mass of the earth constant the escape velocity will beA. doubledB. halvedC. tripledD. unaltered

The force of attraction between two unit point masses separated by a unit distance is called: a) gravitational potential b) acceleration due to gravity c) gravitational field strength d) universal gravitational constant

The atmosphere consisting of a large number of gases is held to the earth by: a) winds b) clouds c) earth’s magnetic fieldd) gravity

The value of g on the surface of the moon: a) is the same as on the earth b) is less than that on the earth c) is more than that on the earth d) keeps changing day by day

The mass of moon is about 0.012 times that of earth and its diameter is about 0.25 times that of earth .The value of G on the moon will beA. less than that of earthB. more than that on the earthC. same as that on the earthD. about one sixth of that on the earth

The mass of moon is about 0.012 times that of the earth and its diameter is about 0.25 times that of earth. The value of G on the moon will be:a) less than that on the earth b) more than that on the earth c) same as that on the earth d) about one-sixth of that on the earth

The atmosphere is held to the earth byA. WindsB. GravityC. CloudsD. None of the above

The weight of a body at the centre of the earth isA. ZeroB. InfiniteC. Same as on the surface of earthD. None of the above 6

What is the intensity of gravitational field of the centre of a spherical shellA. `Gm//r^(2)`B. `g`C. ZeroD. None of these

(a) What is meant by the terms mass and weight?(b) Are they vector of scalar quantities? Why?(c) The mass of a body is 30 kg. What is its weight on earth? (g = 9.8 m/s2)(d) What is its weight on the moon? (g = 1.62 m/s2)

If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is the mutual force of attraction between them?

If the earth is regarded as a hollow sphere, then what is the weight of an object below the surface of earth?

Observe the figure and complete the tableAttracting bodiesForce of gravitationA, BB, CC, A

The orbital velocity of the stationary satellite of earth is nearly…

Why do different planets have different escape speeds?

In which case was the reading higher?

On which fundamental law of physics is keplers second law is based?

Distance between two bodies is increased to three times its original value. What is the effect on the gravitational force between them?

Which is greater the attraction of the earth for 1 kg of aluminum or aluminum or attraction of 1kg of aluminum for the earth?

Which of the stones experienced greater force of attraction of the earth?

If a planet of given density were made larger, its force of attraction for an object on its surface would increase because of the greater distance from the object to the centre of the planet. Which effect predominates?

The masses of two spheres are 10 kg and 20 kg respectively. If the distance between their centers is 100 m, find the magnitude of the gravitational force between them.