2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 3 InterviewSolution

101.	a) What do you mean by the term free fall? b) During a free fall, will heavier objects accelerate more than the lighter ones?
Answer» a) The term free fall refers of the falling of an object towards the earth from a certain height towards the earth under the gravitation force of the earth. b) Acceleration of an object is independent of the mass of the body during free fall.

101.

a) What do you mean by the term free fall? b) During a free fall, will heavier objects accelerate more than the lighter ones?

Answer»

a) The term free fall refers of the falling of an object towards the earth from a certain height towards the earth under the gravitation force of the earth.

b) Acceleration of an object is independent of the mass of the body during free fall.

Discussion

102.	Explain what is meant by the equation: g = G (M/R2) where the symbols have their usual meanings.
Answer» The equation is the acceleration produced by the earth which is also known as acceleration due to gravity. g = G (M/R²⁾ Where, G is the gravitational constant M is the mass of the earth R is the radius of the earth

102.

Explain what is meant by the equation: g = G (M/R2) where the symbols have their usual meanings.

Answer»

The equation is the acceleration produced by the earth which is also known as acceleration due to gravity.

g = G (M/R²⁾

Where,

G is the gravitational constant

M is the mass of the earth

R is the radius of the earth

Discussion

103.	Two artificial satellites of the same mass are moving around the earth in circular orbits of different radii. In comparision to the satellite with lesser orbital radius, the other satellite with higher orbital radius will have:A. greater kinetic energyB. less potential energyC. greater total energyD. less magnitude of angular momentum, about the centre of the circular orbit.
Answer» Correct Answer - C `K.E.=(GMm)/(2R), PE=-(GMm)/R` `T.E.=-(GMm)/(2R)` Angular momentum `mvR=msqrt(GMR)`

Discussion

104.	A and B are two satellite revolving around the earth in circular orbits with radii `R_(A) and R_(B)`. Their periods `T_(A) and T_(B)` are 8 h and 1 h respectively. Then the ratio of `(R_(A)//R_(B))` is equal toA. 4B. 8C. `(8)^(1//3)`D. `sqrt8`
Answer» Correct Answer - a `(R_(A))/(R_(B))=((T_(A))/(T_(B)))^(2//3)=((8)/(1))^(2//3)` `(R_(A))/(R_(B))=4.`

Discussion

105.	What is the height of a geostationary satellite in terms of M, R, G and T? (T=Length of the day and the other symbols have their usual meanings in the following options.)A. `((GMT^(2))/(4pi^(2)))^(1//3)+R`B. `((GMT^(2))/(4pi^(2)))^(1//3)-R`C. `((4pi^(2)GM)/(T^(2)))^(1//3)`D. `((4piGM)/(R^(2)))^(1//3)-R`
Answer» Correct Answer - B `T = 2pi sqrt(((R+h)^(3))/(GM))` `therefore T^(2)=(4pi^(2)(R+h)^(3))/(GM)` `therefore R+h=[(GMT^(2))/(4pi^(2))]^(1//3)` `therefore h=[(GMT^(2))/(4pi^(2))]^(1//3)-R`

Discussion

106.	An artificial satellite is made to move in circular orbits of different radii around the earth. The variations of its K.E.,P.E. and total energy (E) in different orbits is shown in the figure by different curves. Then for satellites A. A represents the K.E., B the P.E. and C the total energyB. A represents the P.E.,Bthe K.E. and C the total energyC. A represents the P.E.,Bthe total energy and C the K.E.D. A represents the total energy, B the K.E. and C the P.E.
Answer» Correct Answer - C The K.E. is always positive. `therefore` The curve C represents the K.E. Both P.E. and total energy are positive but `\|E\|lt\|P.E.\|` As P.E. = 2 (Total energy) `therefore` B represents the total energy and A represents the P.E.

Discussion

107.	(a) What do you mean by the term free fall ? (b) uring a free fall, will heavier objects accelerate more than lighter ones?
Answer» Correct Answer - No

Discussion

108.	The rotation of the earth radius R about its axis speeds upto a value such that a man at latitude angle `60^(@)` feels weightless. The duration of the day in such case will beA. `pi sqrt((R)/(g))`B. `(pi)/(2)sqrt((R)/(g))`C. `(pi)/(3)sqrt((R)/(g))`D. `pi sqrt((g)/(R))`
Answer» Correct Answer - A `0=g-Romega^(2)cos^(2) 60^(@)` or `omega^(2)=(4g)/(R)or omega=2sqrt((g)/(R))` `(2pi)/(T)=2sqrt((g)/(R))rArrT=pisqrt((R)/(g))`

Discussion

109.	The rotation of the earth radius R about its axis speeds upto a value such that a man at latitude angle `60^(@)` feels weightless. The duration of the day in such case will beA. `2pisqrt((R)/(g))`B. `4pisqrt((R)/(g))`C. `2pisqrt((g)/(R))`D. `4pisqrt((g)/(R))`
Answer» Correct Answer - B

Discussion

110.	Find the angular speed of earth so that a body lying at `30^(@)` latitude may become weightless.
Answer» Correct Answer - `sqrt((4g)/(3R))`

Discussion

111.	A particle of mass M is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point P at a distance `a/2` from the centre.
Answer» The gravitatioN/Al potential at the point P due to the particle at the centre is `V_1=-(GM)/(a/2)=-(2GM)/a` the potential at P due to the shell is `V_2=(GM)/a` The net potential at P is `V_1+V_2=-(3GM)/a`.

Discussion

112.	Two planets of radii in the ratio `2 : 3` are made from the materials of density in the ratio `3 : 2`. Then the ratio of acceleration due to gravity `g_(1)//g_(2)` at the surface of two planets will beA. `1`B. `2.25`C. `4//9`D. `0.12`
Answer» Correct Answer - A

Discussion

113.	In order to find time, the astronaut orbiting in an earth satellite should useA. A pendulum clockB. A watch having main spring to keep it goingC. Either a pendulum clock or a watchD. Neither a pendulum clock nor a watch
Answer» Correct Answer - B

Discussion

114.	Due to the greenhouse effect if the ice at the poles will melt, then what would be the effect on the duration of the day on Earth?
Answer» On melting the ice will convert into water which moves away from axis of rotation. Therefore the moment of inertia increases. As a result, the angular velocity of rotation of Earth decrease and hence the Earth will take more time to complete the rotation. Hence the duration of the day will increase.

Discussion

115.	If the earth stops rotating about its axis, by what value of the acceleration due to gravity at equator will change?
Answer» Increase by Rω² where R is the radius of earth and ω is the angular velocity of rotation of earth.

Discussion

116.	If the earth stops rotating the value of ‘g’ at the equator will:A. increaseB. remains sameC. decreaseD. none of these
Answer» If the earth stops rotating, then the value of g at the equator increases and becomes equal to the value at poles. This is because, if the earth will stop rotating, then the equatorial bulge will vanish and earth will have a uniform radius and since, g ∝\(\frac{1}{r^2}\) so g is same through the surface of earth now. Hence, option A is correct.

116.

If the earth stops rotating the value of ‘g’ at the equator will:A. increaseB. remains sameC. decreaseD. none of these

Answer»

If the earth stops rotating, then the value of g at the equator increases and becomes equal to the value at poles. This is because, if the earth will stop rotating, then the equatorial bulge will vanish and earth will have a uniform radius and since, g ∝\(\frac{1}{r^2}\) so g is same through the surface of earth now.

Hence, option A is correct.

Discussion

117.	If the Earth stops rotation on its axis, then what would be the change in the value of gravitational acceleration at equator and poles?
Answer» The value of gravitational acceleration (in latitude), g’ = g – ω²Rcos² λ (i) For poles, λ = 90°, ∴ cos² λ = 0 Therefore g’ = g Therefore, even the Earth stops its rotation, i.e, ω = 0, there will be no change in gravitational acceleration. (ii) For equator, λ = 0. ∴ cos λ = 1 If ω = 0 i.e., Earth stops rotation, then g’ = g Thus the value of gravitational acceleration increase on the equator when Earth stops it’s rotation.

117.

If the Earth stops rotation on its axis, then what would be the change in the value of gravitational acceleration at equator and poles?

Answer»

The value of gravitational acceleration (in latitude), g’ = g – ω²Rcos² λ

(i) For poles, λ = 90°,

∴ cos² λ = 0

Therefore g’ = g

Therefore, even the Earth stops its rotation, i.e, ω = 0, there will be no change in gravitational acceleration.

(ii) For equator, λ = 0.

∴ cos λ = 1

If ω = 0 i.e., Earth stops rotation, then g’ = g

Thus the value of gravitational acceleration increase on the equator when Earth stops it’s rotation.

Discussion

118.	If the earth stops rotating, the value of ‘ g ’ at the equator willA. IncreaseB. Remain sameC. DecreaseD. None of the above
Answer» Correct Answer - A

Discussion

119.	The gravitational field in a region is given by `vecE=(5Nkg^-1)veci+(12nk^-1)vecj.`. a.find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin. B. Find the potential at the points (12m,0) and (0,5m) if the potential at the origin is taken to be zero. C. Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12m,5m). d. Find the change in potential energy if the particle is taken from `(12m,0) to (0,5,m)`.A. `26N`B. `30N`C. `20N`D. `35N`
Answer» Correct Answer - A `vecF =vecEm=(10hati+24hatj)N=sqrt((100)+576)=26N`

Discussion

120.	The gravitational field in a region is given by `vecE=(5Nkg^-1)veci+(12nk^-1)vecj.`. a.find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin. B. Find the potential at the points (12m,0) and (0,5m) if the potential at the origin is taken to be zero. C. Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12m,5m). d. Find the change in potential energy if the particle is taken from `(12m,0) to (0,5,m)`.A. `-225J`B. `-240J`C. `-245J`D. `-250J`
Answer» Correct Answer - B `/_v+int_((12m,5m))^((0,0))Emdr=+int_((12m, 5m))^((0,0))(5hati+12hatj)2dr` `=+(10hati+24hatj)\|r\|_((12m,5m))^((0,0))` `=-(10hati+24hatj)(12hati+5hatj)=-240J`

Discussion

121.	A semicircular wire hs a length L and mass M. A paricle of mass m is placed at the centre of the circle. Find the gravitational attrction on the particle due to the wire.
Answer» Correct Answer - B In the semicircle, we can consider a small element `dtheta` then mass of `R/dtheta=(M/L)Rdtheta` dF=(Gmrdthetam)/(LR^2)` dF_1=2dF` `since =(2GMm)/(LR) sintheta d theta `:. F=int_0^(pi/2) (-2GMm)/(LR) sintheta d theta` =(-2GMm)/(LR) (-1)` `(2GMm)/(LR)=(2GMm)/(LL/pi)` =(2piGMm)/L^2`

Discussion

122.	A satellite is orbiting around the earth. If both gravitational force and centripetal force on the satellite is `F`, then, net force acting on the satellite to revolve around the earth isA. 2 FB. zeroC. FD. `(F)/(2)`
Answer» Correct Answer - C The gravitational force produces the centripetal force.

Discussion

123.	An artificial satellite circled around the earth at a distance of 3400 km. Calculate its orbital velocity and period of revolution. Radius of earth =6400 km and `g=9.8 ms^(-2)`.
Answer» Correct Answer - `6400 ms^(-1)`,9621 s

Discussion

124.	Assertion: For the plantes orbiting around the sun, angular speed, linear speed, K.E. changes with time, but angular momentum remains constant. Reason: No torque is acting on the rotating planet. So its angular momentum is constant.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - A

Discussion

125.	The distance between centre of the earth and moon is 384000 km . If the mass of the earth is `6xx10^(24) kg` and `G=6.66xx10^(-11)Nm^(2)//kg^(2)`. The speed of the moon is nearlyA. 1km/secB. 4 km/secC. 8 km/secD. 11.2 km/sec
Answer» Correct Answer - A

Discussion

126.	A satellite launching statian should beA. near the equatorial regionB. near the polar regionC. on the polar axisD. all the locations are equally good
Answer» Correct Answer - a Earth has the maximum speed in the equatorial region. To take advantage of this, the launch station should be in the equatorial region.

Discussion

127.	What is the usual shape of orbits of planets around the sun ?
Answer» Orbits of planets are elliptical.

Discussion

128.	For motiion of planets in ellipticla orbits around the sun the central force isA. the force on the plantet along the vector joining the sun and the planetB. the force on the sun along the vector joining the sun and the other focus of the ellipseC. the force on the planet along the line joining focus of the ellipseD. none of the above
Answer» Correct Answer - A The central force is always directed towards or away from a fixed point i.e along the position vector of the point of application of the force with respect to the fixed point further the magnitude of a central force F depends on r the distance of the point of appication of the force from the fixed point F=F(r) For motion of planets in elliptical orbits around the sun the central force acts alongs the vector joining the sun and the planet

Discussion

129.	Many planets are revolving around the fixed sun in circular orbits of different radius `(R)` and difference time period `(T)` To estimate the mass of the sun the orbital radius `(R)` and time period `(T)` of planets were noted. Then `"log"_(10)` T v/s `"log"_(10)R` curve was plotted. The curve was found to be approximately straight line (as shown in) having y intercept `=6.0` (Neglect the gravitational interaction among the planets) [Take `G = (20)/(3) xx 10^(-11)` in MKS, `pi^(2)= 10`] Two plantes `A` and `B` having orbital radius `R` and `4R` are initally at the closest position and rotating in the same direction if angular velocity of planet `B` is `omega_(0)` then after how much time will both the planets be again in the closest postion? (Neglect the interaction between planets) .A. `(2pi)/(7omega_(0))`B. `(2pi)/(9omega_(0))`C. `(2pi)/(omega_(0))`D. `(2pi)/(5omega_(0))`
Answer» Correct Answer - A

Discussion

130.	A ball of mass `m` is dropped from a height `h` equal to the radius of the earth above the tunnel dug through the earth as shows in the figure. Choose the correct options. (Mass of earth `= M`) A. Particle will oscillate through the earth to a height `h` on both sidesB. Particle will execute simple harmonic motionC. Motion of the particle is periodicD. Particle passes the centre of earth with a sped `upsilon = sqrt((2GM)/(R ))`
Answer» Correct Answer - A::C::D `K_(i) + U_(i) = K_(f) + U_(f)` `:. 0 - (GMm)/(2R ) = (1)/(2) m nu^(2) - (3)/(2) (GMm)/(R)` or `nu = sqrt((2GM)/(R ))`

Discussion

131.	A diametrical tunnel is dug across the earth. A ball dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is [Given: gravitational potential at the centre of earth `= -3//(2GM//R)`]A. `sqrt(R)`B. `sqrt(gR)`C. `sqrt(2.5gR)`D. `sqrt(7.1gR)`
Answer» Correct Answer - B Gravitational potential at a point on the surface of earth is `-(GM)/R` If the earth is assumed to be a solid sphere, then the gravitational potential at the centre of the earth is `(3/(2GM)//R)`. Decrease in gravitational potential is `R/2xx(GM)/(R^(2))=(Rg)/2` Loss is potential energy is `R/2xx(GM)/(R^(2))xxm` Now gain in kinetic energy `=` loss is potnetial energy. Therefore. `1/2mv^(2)=1/2mgR` or `v=sqrt(gR)`

Discussion

132.	Assertion : The gravitational attraction of moon is much less than that of earth. Reason : Moon is the neutral satellite of the earth.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are truebut reason is not the correct explanation of assertionC. If assertion is true but reason is falseD. If both assertion and reason are false.
Answer» Correct Answer - B

Discussion

133.	In Q.No 6.2, if the planet rotates in counter clockwise direction, then areal velocity has a direaction :A. Given by "Right Hand Thumb Rule"B. Given by "Left Hand Thumb Rule"C. Normal to the plane of orbit upwardsD. Normal to the plane of orbit downwards
Answer» Correct Answer - A::C

Discussion

134.	Assertion (A) : weight of an object is less on moon then earth Reason (R): earth is heavier than moon A) Both A and R are correct, R is the correct explanation of A B) Both A and R are correct, R is not the correct explanation of A C) A is correct, R is not correct D) A is not correct, R is correct
Answer» B) Both A and R are correct, R is not the correct explanation of A

Discussion

135.	Two stones are thrown into air with speeds 20 m/s, 40 m/s respectively. What accelerations are possessed by the objects?
Answer» Stone -1 : Initial velocity u = 20 m/s After a time t, it reaches to ground then final velocity v = 0 m/s Acceleration = \(\frac{0\,-\,20}{t}=\frac{-20}{t} \,m/s^2\) Stone – 2 : Initial velocity u = 40 m/s Final velocity v = 0 m/s Acceleration = \(\frac{0\,-\,40}{t}=\frac{-40}{t} \,m/s^2\)

135.

Two stones are thrown into air with speeds 20 m/s, 40 m/s respectively. What accelerations are possessed by the objects?

Answer»

Stone -1 :

Initial velocity u = 20 m/s

After a time t, it reaches to ground then final

velocity v = 0 m/s

Acceleration = \(\frac{0\,-\,20}{t}=\frac{-20}{t} \,m/s^2\)

Stone – 2 :

Initial velocity u = 40 m/s

Final velocity v = 0 m/s

Acceleration = \(\frac{0\,-\,40}{t}=\frac{-40}{t} \,m/s^2\)

Discussion

136.	If there is no gravity your weight becomes A) equal to your mass B) zero C) more than your mass D) less than your mass
Answer» Correct option is B) zero

Discussion

137.	Give an example for the motion of an object of zero speed and with non-zero acceleration?
Answer» Protons and neutrons inside the nucleus.

Discussion

138.	The direction of “g” is ……………. A) towards sky B) towards earth C) horizontal to the earthD) no direction
Answer» B) towards earth

Discussion

139.	An apple falls because of the gravitational attraction of the earth. What is the gravitational attraction of apple on the earth?
Answer» We know that acceleration due to gravity, g = \(\frac{GM}{R^2}\) Here M is the mass of the earth. In this mass of the object which is freely falling has no effect on ‘g’. Hence the gravitational attraction of apple on the earth is negligible.

Discussion

140.	An apple falls because of the gravitational attraction of earth. The gravitational attraction of apple on the earth is ………………. to gravitational force of attraction of earth on the apple. A) greater than B) less than C) equal D) greater than or equal
Answer» Correct option is C) equal

Discussion

141.	An apple of mass 100g falls from a tress because of gravitational attraction between the earth and the apple. If the magnitude of force exerted by the earth on the apple be F1 and the magnitude of force exerted by the apple on the earth be F2, then:a) F1 is very much greater than F2 b) F2 is very much greater than F1 c) F1 is only a little greater than F2 d) F1 and F2 are exactly equal
Answer» The correct answer is d) F1 and F2 are exactly equal

Discussion

142.	An apple of mass 100 g falls from a tree because of gravitational attraction between the earth and the apple. If the magnitude of force exerted by the earth on the apple be `F_(1)` and the magnitude of force exerted by the apple on the earth be `F_(2)` then `F_(1)` is very much greater than `F _(1)` is only a little greater than FA. `F_(1)` is very much greater than `F_(2)` thenB. `F_(2)` is very much greater than `F_(1)`C. `F_(1)` is only a little greater than `F_(2)`D. `F_(1)` and `F_(2)` are exactly equal
Answer» Correct Answer - c

Discussion

143.	The gravitational force exerted by the Sun on the Moon is about twice as great as the gravitational force exerted by the earth on the Moon, but still Moon is not escaping from the gravitational influence of the earth. Mark the option which correctly explains the above system.A. Escape speed is independent of the direction in which it is projected.B. The rotational effect of the earth plays a role in computation of escape speed, however small it may be.C. A body thrown in the eastward direction has less escape speed.D. None of the above
Answer» Correct Answer - C A minimum amount of energy equla to total energy of the moon earth system has to be given to break (unbound) the system. The sun is exerting force on the moon but not providing any energy.

Discussion

144.	If Earth attracts a body with a forcce of 10N, then the body attracts the Earth with ____________N.A. less than `10^(20)N`B. `10^(20)N`C. greater than `10^(20)N`D. `10^(-20)N`
Answer» Correct Answer - B `10^(20)N`

Discussion

145.	An apple falls from a tree because of gravitational between the earth and apple. If `F_(1)` is the magnitude of force exerted by the earth on the apple and `F_(2)` is the magnitude of force exerted by apple on earth, thenA. the accelerations of the apple and the earth are equal is magnitudeB. force of attraction of earth on apple is greater than the force of apple on earthC. force of attraction of apple on earth is greater than the force of earth on appleD. both apple and earth apply equal and opposite forces on each other
Answer» Correct Answer - d

Discussion

146.	A satellite circles a planet of unknown mass in circular orbit of radius `2xx10^(7)m`. The magnitude of the gravitational forcce exerted on the satellite by the planet is 80 N . The kinetic energy of satellite in this orbit in joule isA. `9xx10^(8)`B. `8xx10^(8)`C. `7xx10^(8)`D. `6xx10^(8)`
Answer» Correct Answer - b `"K.E. "=(1)/(2)"F r"`

Discussion

147.	A small satellite revolves around a heavy planet in a circular orbit. At a point in its orbit an impulse acts suddenly and instantaneously increases its kinetic energy `K` times without change in its direction of motion. The ratio of maximum to the minimum distance from the planet is [Assume mass of satellite is negligible small compared to that of planet]A. `K/(K+2)`B. `K/(K-2)`C. `(K+2)/K`D. `(2-K)/K`
Answer» Correct Answer - B `-int_(P_(c))^(0)dp=4/3Grho_(r)^(2)int_(0)^(R)rdr` ltBrgt `P_(c)=4/3piGrho_(r)^(2)(R^(2))/2` `P_(c)=(GM_(0)^(2))/(R_(0)^(4))xx(3/(2pi))`

Discussion

148.	Two satellities `s_(1)` a & `s_(2)` of equal masses revolves in the same sense around a heavy planet in coplaner circular orbit of radii `R` & `4R`A. the ratio of period of revolution `s_(1)` and `s_(2)` 1:4B. their velocities are in the 2:1C. their angular momentum about the planet are in the ration 2:1D. the ratio of angular velocities of `s_(2)` w.r.t. `s_(1)` when all three are in the same line is 4:1
Answer» Correct Answer - B `T prop R^(3//2) rArr (T_(1))/(T_(2))=1/8` `V prop R^(-1//2) rArr (v_(1))/(v_(2))=1/2`

Discussion

149.	The value of acceleration due to gravity will be 1% of its value at the surface of earth at a height of `(R_(e )=6400 km)`A. 6400 kmB. 577600 kmC. 2560 kmD. 6400 km
Answer» Correct Answer - B

Discussion

150.	What will be the acceleration due to gravity at a distance of 3200 km below the surface of the earth ? (Take `R_(e)=6400` km)
Answer» Correct Answer - `4.9 m//s^(2)` Hint : `g=g(1-(d)/(R_(e)))` (Take d = 3200 km and `R_(e)=6400` km)

Discussion

Explore topic-wise InterviewSolutions in .

a) What do you mean by the term free fall? b) During a free fall, will heavier objects accelerate more than the lighter ones?

Explain what is meant by the equation: g = G (M/R2) where the symbols have their usual meanings.

A and B are two satellite revolving around the earth in circular orbits with radii `R_(A) and R_(B)`. Their periods `T_(A) and T_(B)` are 8 h and 1 h respectively. Then the ratio of `(R_(A)//R_(B))` is equal toA. 4B. 8C. `(8)^(1//3)`D. `sqrt8`

(a) What do you mean by the term free fall ? (b) uring a free fall, will heavier objects accelerate more than lighter ones?

The rotation of the earth radius R about its axis speeds upto a value such that a man at latitude angle `60^(@)` feels weightless. The duration of the day in such case will beA. `pi sqrt((R)/(g))`B. `(pi)/(2)sqrt((R)/(g))`C. `(pi)/(3)sqrt((R)/(g))`D. `pi sqrt((g)/(R))`

The rotation of the earth radius R about its axis speeds upto a value such that a man at latitude angle `60^(@)` feels weightless. The duration of the day in such case will beA. `2pisqrt((R)/(g))`B. `4pisqrt((R)/(g))`C. `2pisqrt((g)/(R))`D. `4pisqrt((g)/(R))`

Find the angular speed of earth so that a body lying at `30^(@)` latitude may become weightless.

A particle of mass M is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point P at a distance `a/2` from the centre.

Two planets of radii in the ratio `2 : 3` are made from the materials of density in the ratio `3 : 2`. Then the ratio of acceleration due to gravity `g_(1)//g_(2)` at the surface of two planets will beA. `1`B. `2.25`C. `4//9`D. `0.12`

In order to find time, the astronaut orbiting in an earth satellite should useA. A pendulum clockB. A watch having main spring to keep it goingC. Either a pendulum clock or a watchD. Neither a pendulum clock nor a watch

Due to the greenhouse effect if the ice at the poles will melt, then what would be the effect on the duration of the day on Earth?

If the earth stops rotating about its axis, by what value of the acceleration due to gravity at equator will change?

If the earth stops rotating the value of ‘g’ at the equator will:A. increaseB. remains sameC. decreaseD. none of these

If the Earth stops rotation on its axis, then what would be the change in the value of gravitational acceleration at equator and poles?

If the earth stops rotating, the value of ‘ g ’ at the equator willA. IncreaseB. Remain sameC. DecreaseD. None of the above

A semicircular wire hs a length L and mass M. A paricle of mass m is placed at the centre of the circle. Find the gravitational attrction on the particle due to the wire.

A satellite is orbiting around the earth. If both gravitational force and centripetal force on the satellite is `F`, then, net force acting on the satellite to revolve around the earth isA. 2 FB. zeroC. FD. `(F)/(2)`

An artificial satellite circled around the earth at a distance of 3400 km. Calculate its orbital velocity and period of revolution. Radius of earth =6400 km and `g=9.8 ms^(-2)`.

The distance between centre of the earth and moon is 384000 km . If the mass of the earth is `6xx10^(24) kg` and `G=6.66xx10^(-11)Nm^(2)//kg^(2)`. The speed of the moon is nearlyA. 1km/secB. 4 km/secC. 8 km/secD. 11.2 km/sec

A satellite launching statian should beA. near the equatorial regionB. near the polar regionC. on the polar axisD. all the locations are equally good

What is the usual shape of orbits of planets around the sun ?

A diametrical tunnel is dug across the earth. A ball dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is [Given: gravitational potential at the centre of earth `= -3//(2GM//R)`]A. `sqrt(R)`B. `sqrt(gR)`C. `sqrt(2.5gR)`D. `sqrt(7.1gR)`

In Q.No 6.2, if the planet rotates in counter clockwise direction, then areal velocity has a direaction :A. Given by "Right Hand Thumb Rule"B. Given by "Left Hand Thumb Rule"C. Normal to the plane of orbit upwardsD. Normal to the plane of orbit downwards

Assertion (A) : weight of an object is less on moon then earth Reason (R): earth is heavier than moon A) Both A and R are correct, R is the correct explanation of A B) Both A and R are correct, R is not the correct explanation of A C) A is correct, R is not correct D) A is not correct, R is correct

Two stones are thrown into air with speeds 20 m/s, 40 m/s respectively. What accelerations are possessed by the objects?

If there is no gravity your weight becomes A) equal to your mass B) zero C) more than your mass D) less than your mass

Give an example for the motion of an object of zero speed and with non-zero acceleration?

The direction of “g” is ……………. A) towards sky B) towards earth C) horizontal to the earthD) no direction

An apple falls because of the gravitational attraction of the earth. What is the gravitational attraction of apple on the earth?

An apple falls because of the gravitational attraction of earth. The gravitational attraction of apple on the earth is ………………. to gravitational force of attraction of earth on the apple. A) greater than B) less than C) equal D) greater than or equal

If Earth attracts a body with a forcce of 10N, then the body attracts the Earth with ____________N.A. less than `10^(20)N`B. `10^(20)N`C. greater than `10^(20)N`D. `10^(-20)N`

A satellite circles a planet of unknown mass in circular orbit of radius `2xx10^(7)m`. The magnitude of the gravitational forcce exerted on the satellite by the planet is 80 N . The kinetic energy of satellite in this orbit in joule isA. `9xx10^(8)`B. `8xx10^(8)`C. `7xx10^(8)`D. `6xx10^(8)`

The value of acceleration due to gravity will be 1% of its value at the surface of earth at a height of `(R_(e )=6400 km)`A. 6400 kmB. 577600 kmC. 2560 kmD. 6400 km

What will be the acceleration due to gravity at a distance of 3200 km below the surface of the earth ? (Take `R_(e)=6400` km)