2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 13 InterviewSolution

601.	Match the entries given in Column A with appropriate ones in Column B
Answer» Correct Answer - `A : e" "B : d" "C : I" "D : c" "E : b" "f : j" "G : a" "H : g" "I : f" "J : h`

Discussion

602.	The gravitational field in a region is given by the equation `E=(5i + 12j) N//kg`. If a particle of mass `2kg` is moved from the origin to the point `(12m, 5m)` in this region, the change in the gravitational potential energy is
Answer» `dV=-vec(E) . vec(dr)` `=-(5i+12j).(12i+5j)=-(60+60)=-120` Change in gravitational potential energy `dU=mdV =2(-120)=-240 J`

Discussion

603.	Two satellites are moving at heights of R and 5R above the surface of the earth of radius R. The ratio of their velocitirs `((V_(1))/(V_(2)))` isA. `sqrt(5):1`B. `1 : 1`C. `sqrt(3) : sqrt(2)`D. `sqrt(3) : 1`
Answer» Correct Answer - D `(v_(1))/(v_(2))=(sqrt((GM)/(2R)))/(sqrt((GM)/(6R)))=(sqrt(3))/(1)`

Discussion

604.	Two satellites of identical masses orbit the Earth at different heights. The ratio of their distance from the centre of earth is d : 1 and the ratio of the acceleration due to gravity at those heights is g : 1. Then the ratio of their orbital velocities is ________ .
Answer» Correct Answer - `sqrt(dg)`

Discussion

605.	Two particles each of mass m are revolving in circular orbits of radius `r=5R` in opposite directions with orbital speed `v_(0)` . They collide perfectly inelastically and fall to the ground. The speed of combined mass on triking the ground will beA. `2sqrt(2)v_(0)`B. `sqrt(2)v_(0)`C. `2v_(0)`D. `v_(0)`
Answer» Correct Answer - A

Discussion

606.	The condition for a uniform spherical mass `m` of a radius `r` to be a black hole is [`G` =gravitational constant and `g`=acceleration due to gravity]A. `(2Gm//r)^(1//2)lec`B. `(2Gm//r)^(1//2)=c`C. `(2Gm//r)^(1//2)gec`D. `(Gm//r)^(1//2)gec`
Answer» Correct Answer - C

Discussion

607.	The condition for a uniform spherical mass `m` of a radius `r` to be a black hole is [`G` =gravitational constant and `g`=acceleration due to gravity]A. `(2Gm//r)^(1//2) lt c`B. `(2Gm//r)^(1//2) = c`C. `(2Gm//r)^(1//2) ge c`D. `(gm//r)^(1//2) ge c`
Answer» Correct Answer - C

Discussion

608.	Two satellites `M` and `N` go around the earth in circular orbits at heights of `R_(M)` and `R_(N)` respectively from the surrface of the earth. Assuming the earth to be a uniform sphere of radius `R_(E)`, the ratio of velocities of the satellites `(V_(M))/(V_(N))` isA. `((R_(M))/(R_(N)))^(2)`B. `sqrt((R_(N)+R_(E))/(R_(M)+R_(E)))`C. `(R_(N)+R_(E))/(R_(M)+R_(E))`D. `sqrt(R_(N)/(R_(M))`
Answer» Correct Answer - B `V_(0)=sqrt((GM)/(R+h))rArr V_(0)prop1/(sqrt(R+h))`

Discussion

609.	Consider a hypothetical planet which is very long and cylinderical. The density of the planet is `rho`, its radius is `R`. If an object is projected radially outwards from the surface such that it reaches upto a maximum distance of `3R` from the axis then what should be the speed of projection?A. `Rsqrt(2/3pirhoG)`B. `2Rsqrt(pirhoGln3)`C. `Rsqrt(4/3pirhoG)`D. `Rsqrt(2/3pirhoGln3)`
Answer» Correct Answer - B Work done by gravitational force, `W=-2piGrhoR^(2)mint_(R)^(3R)1/rdr` `rArr E=-2piGrhomR^(2)ln3`. Applying work energy theorem, we get `-1/2mv^(2)=-2piGrhomR^(2)ln3 v=2Rsqrt(pirhoGln3)`

Discussion

610.	Consider a hypothetical planet which is very long and cylinderical. The density of the planet is `rho`, its radius is `R`. Assume that the planet is rotating abouts its axis with time period `T`. How far from the axis of the planet do the synchronous telecommunications satellite orbit?A. `RTsqrt(piGrho)`B. `2RTsqrt(piGrho)`C. `RTsqrt(2piGrho)`D. `RTsqrt((Grho)/(2pi))`
Answer» Correct Answer - D `v=romega=((2pi)/T)r, r=(vT)/(2pi)=sqrt((Grho)/(2pi))RT`.

Discussion

611.	The value of acceleration due to gravity on the surface of earth is `x`. At an altitude of `h` from the surface of the earth, its value is `y`. If `R` is the radius of earth, then the value of `h` isA. `(sqrt(x/y)-1)R`B. `(sqrt(y/x)-1)R`C. `sqrt(y/x)R`D. `sqrt(x/y)R`
Answer» Correct Answer - A `x=(GM)/(R^(2)),y=(GM)/((R+h)^(2))`

Discussion

612.	A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `3/4 mv^(2)`B. `mv^(2)`C. `1/2 mv^(2)`D. `-1/2 mv^(2)`
Answer» Correct Answer - D

Discussion

613.	A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `(3)/(4)mv^(2)`B. `(1)/(2)mv^(2)`C. `mv^(2)`D. `-((1)/(2))mv^(2)`
Answer» Correct Answer - D

Discussion

614.	A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `-1/2 Mv^(2)`B. `1/2 Mv^(2)`C. `3/2 Mv^(2)`D. `Mv^(2)`
Answer» Correct Answer - A

Discussion

615.	A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `1/2 mv^(2)`B. `1/4 mv^(2)`C. `-1/4 mv^(2)`D. `-1/2 mv^(2)`
Answer» Correct Answer - D `therefore` Total energy of the satellite is `E=-1/2 (GM_(e)m)/(R_(e))` where `k=1/2(GMM_(e)m)/(R_(e))` `therefore` Total energy =-kinetic energy `E=-1/2 mv^(2)` putting the value of KE in the form of mass of a satellite m and speed v

Discussion

616.	Find the maximum and minimum distances of the planet `A` from the sun `S`, if at a ceration moment of times it was at a distance `r_(0)` and travelling with the velocity `upsilon_(0)`. With the angle between the radius vector and velocity vector being equal to `phi`.
Answer» At minimum and maximum distance velocity vector `(V)` makes an angle of `90^(@)` with radius vector. Hence, from conservation of angular momentum, `mu upsilon_(0) r_(0) sin phi = mr upsilon` ..(i) Hence, `m` is the mass of the plant. From energy conservation law, it follows that `(m upsilon_(0)^(2))/(2) - (GMm)/(r_(0)) = (m upsilon^(2))/(2) - (GMm)/(r)` ..(ii) Hence, `M` is the mass of the sun. Solving Eqs. (i) and (ii) for `r`, we fet two values of `r`, one is `r_(max)` and another is `r_(min)`. So, `r_(max) = (r_(0))/(2 - K) (1 + sqrt(1 - K(2 - K) sin^(2) phi))` and `r_(min) = (r_(0))/(2 - K) (1 - sqrt(1 - K(2 - K) sin^(2) phi))` Here, `K = (r_(0)^(2) upsilon_(0)^(2))/(GM)`

Discussion

617.	A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is `V`. Due to the rotation of planet about its axis the acceleration due to gravity `g` at equator is `1//2` of `g` at poles. The escape velocity of a particle on the planet in terms of `V`.A. `V_(e)=2V`B. `V_(e)=V`C. `V_(e)=V//2`D. `V_(e)=sqrt(3)V`
Answer» Correct Answer - a `g_(e)=g_(p)-Romega^(2)impliesg/2=g-Romega^(2)` `Romega^(2)=g/2impliesR^(2)omega^(2)=(gR)/2` `V^(2)=(gR)/2.....(1)` `V_(e)=sqrt(2gR).....(2)` From `(1)` and `(2)` `V_(e)=sqrt(2xx2V^(2))impliesV_(e)=2V`

Discussion

618.	Two planets `A` and `B` have the same material density. If the radius of `A` is twice that of `B`, then the ratio of the escape velocity `(v_(A))/(v_(B))` isA. `2`B. `sqrt(2)`C. `1//sqrt(2)`D. `1//2`
Answer» Correct Answer - a `V_(e)=sqrt((2GM)/R)impliesV_(e)=sqrt((2G.4/3piR^(3)rho)/R)` `V_(e)=sqrt(8Gpirho)R` `(V_(A))/(V_(B))=(R_(A))/(R_(B))implies(V_(A))/(V_(B))=2/1`

Discussion

619.	The longest and the shortest distance of a planet from the sun are `R_(1)` and `R_(2)`. Distances from sun when it is normal to major axis of orbit isA. `(R_(1)+R_(2))/2`B. `sqrt((R_(1)^(2)+R_(2)^(2))/2)`C. `(R_(1)R_(2))/(R_(1)+R_(2))`D. `(2R_(1)R_(2))/(R_(1)+R_(2))`
Answer» Correct Answer - D `R_(1)=(1+e)a, R_(2)=(1-e)a` `a=(R_(1)+R_(2))/2,R_(1)R_(2)=(1-e^(2))a^(2)` since semi-latus rectum `=(b^(2))/a` `=(a^(2)(1-e^(2)))/(a)=(R_(1)R_(2))/((R_(1)+R_(2))/2)=(2R_(1)R_(2))/(R_(1)+R_(2))`

Discussion

620.	A satellite is launched into a circular orbit of radius `R` around the earth. While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximately:A. `0.5%`B. `1.0%`C. `1.5%`D. `3.0%`
Answer» Correct Answer - C `(T_(2))/(T_(1))=((1.01 R)/R)^(3//2)=(1+0.01)^(1.5)=1+1.5xx0.01` `=1+0.015` `((T_(2))/(T_(1))-1)xx100=1.5%`

Discussion

621.	A satellite is launched into a circular orbit of radius `R` around the earth. While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximately:A. `0.5%`B. `1.5%`C. `1%`D. `3%`
Answer» Correct Answer - B `TpropR^(3//2)rArr (DeltaT)/Txx100=3/2(Deltar)/rxx100`

Discussion

622.	A particle is projected from the surface of earth with intial speed of 4 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km and `g = 9.8 m//s^(2)`
Answer» Correct Answer - 1 `(mghR_(e))/(R_(e)+h)+1/2mv^(2), h=(v^(2))/(2g-(v^(2))/R)=1`

Discussion

623.	A satellite is orbiting just above the surface of a planet of average density `D` with period `T`. If `G` is the universal gravitational constant, the quantity `(3pi)/G` is equal toA. `T^(2)D`B. `3piT^(2)D`C. `3piD^(2)T`D. `D^(2)T`
Answer» Correct Answer - A Using `T=2pisqrt((R^(3))/(GM))=2pisqrt((R^(3))/(Gxx4/3piR^(3)D))` `T^(2)=(4pi^(2)R^(3))/(G4/3piR^(3)D)=(3pi)/(DG)rArr (3pi)/G=T^(2)D`

Discussion

624.	A homogeneous spherical heavenly body has a uniform and very narrow frictionless duct along its diameter. Let mass of the body be `M` and diameter be `D`. A point mass `m` moves smoothly inside the duct. Force exerted on this mass when it is at a distance `s` from the centre of the body is (numerically)A. `(GMm)/(s^(2))`B. `(piGMm)/((D//2)^(3))s`C. `(8GMms)/(D^(3))`D. `(GMm)/((R-s)^(2))`
Answer» Correct Answer - C `M/(4/3pi(D/2)^(3))=(M_(s))/(4/3piS^(3)), F=-(GmM_(S))/(S^(2))`

Discussion

625.	Consider two configurations of a system of three particles of masses `m`, `2m` and `3m`. The work done by gravity in changing the configuration of the system from figure (i) to figure (ii) is A. zeroB. `(6 Gm^(2))/(a){1 + (1)/(sqrt(2))}`C. `(6 Gm^(2))/(a){1 - (1)/(sqrt(2))}`D. `(6 Gm^(2))/(a){2 - (1)/(sqrt(2))}`
Answer» Correct Answer - C `W = - DeltaU = U_(i) - U_(f)` `= - (Gmm)/(a) [{((1)(2))/(1) +((1) (3))/(1) + ((2)(3))/(sqrt(2))}]` `-{((1 xx2))/(1) +((1xx3))/(1) + ((2xx3))/(1)}]` `= (6 GM^(2))/(a)(1- (1)/(sqrt(2)))`

Discussion

626.	Consider two configurations of a system of three particles of masses `m`, `2m` and `3m`. The work done by gravity in changing the configuration of the system from figure (i) to figure (ii) is A. zeroB. `-(6Gm^(2))/a(1+1/(sqrt(2)))`C. `-(6Gm^(2))/a(1-1/(sqrt(2)))`D. `(6Gm^(2))/(a)(2-1/(sqrt(2)))`
Answer» Correct Answer - C `GPE=(-Gm_(1)m_(2))/(r), W=GPE_(2)-GPE_(1)`

Discussion

627.	A ring having non-uniform distribution of mass `M` and radius `R` is being considered. A point mass `m_(0)` is taken slowly towards the ring. In doing so, work done by the external force against the gravitational force exerted by ring is A. `(GMm_(0))/(sqrt(2)R)`B. `(GMm_(0))/R[1/(sqrt(2)-1/(sqrt(5))]`C. `(GMm_(0))/R[1/(sqrt(5))-1/(sqrt(2))]`D. `(GMm_(0))/(sqrt(5)R)`
Answer» Correct Answer - B `W=m[V_(B)-V_(A)]`

Discussion

628.	The ratio of the escape velocity and the orbital velocity isA. `sqrt(2)`B. `1/(sqrt(2))`C. `2`D. `1//2`
Answer» Correct Answer - A

Discussion

629.	A geostationary satellite is at a height `h` above the surface of earth. If earth radius is `R`- A. The minimum colatitude on earth upto which the satellite can be used for communication is `sin^(-1)(R//R+h)`B. The maximum colatitudes on earth upto which the satellite can be used for communication is `sin^(-1)(R//R+h)`C. The area on earth escaped from this satellite is given as `(2piR^(2)(1+sin theta))`D. The area on earth escaped from this satellite is given as `(2piR^(2)(1+cos theta)`
Answer» Correct Answer - a,c

Discussion

630.	A comet is revolving around the sun in an elliptical orbit. Which of the following will remain constant throughout its orbit?A. Kinetic energyB. Potential enegyC. Linear speedD. Angular momentum
Answer» Correct Answer - d

Discussion

631.	If a satellite is moved from one stable circular orbit to a farther stable circular orbit, then the following quantity increasesA. gravitational forceB. gravitational potential energyC. linear orbital speedD. Centripetal acceleration
Answer» Correct Answer - b

Discussion

632.	Ratio of the radius of a planet `A` to that of planet `B` is `r`. The ratio of acceleration due to gravity for the two planets is `x`. The ratio of the escape velocities from the two planets isA. `sqrt(rx)`B. `sqrt(r//x)`C. `sqrt(r)`D. `sqrt(x//r)`
Answer» Correct Answer - A

Discussion

633.	An artificial satellite is in a circular orbit around the earth. The universal gravitational constant starts decreasing at time `t = 0`, at a constant rate with respect to time `t`. Then the satellite has its:A. Path gradually spiralling out, away from the centre of the earthB. Path gradually spiralling in, towards the centre of the earthC. Angular momentum about the centre of the earth remains constantD. Potential energy increases
Answer» Correct Answer - A::C::D

Discussion

634.	If the universal gravitational constant increases uniformly with time, then a satellite in orbit will still maintain itsA. weightB. tangential speedC. period of revolutionD. angular momentum
Answer» Correct Answer - D

Discussion

635.	STATEMENT -1 : Time period of a satellite is inversely propertional to the square root of the mass of planet. STATEMENT -2 : Self gravitational potential energy of earth is positive. STATEMENT -3 : Orbital velocity of a satellite does depend upon the mass of planet.A. T T TB. T F FC. T F TD. F F F
Answer» Correct Answer - B

Discussion

636.	The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
Answer» Data : R_A = RB/2, gB = \(\frac{1}{2}\) g_A, M_B = ? \(g=\frac{GM}{R^2}\) ∴ g_A= \(\frac{GM_A}{R^{2}_A}\) and g_B= \(\frac{GM_B}{R^2_B}\) ∴ \(\frac{g_B}{g_A}=(\frac{M_B}{M_A})(\frac{R_A}{R_B})^2\) ∴ \(\frac{1}{2}(\frac{M_B}{M_A})(\frac{1}{2})^2\) = \(\frac{1}{4}(\frac{M_B}{M_A})\) ∴ \(\frac{M_B}{M_A}=\frac{4}{2}\) = 2 ∴ M_B = 2M_A.

636.

The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?

Answer»

Data : R_A = RB/2, gB = \(\frac{1}{2}\) g_A, M_B = ?

\(g=\frac{GM}{R^2}\)

∴ g_A= \(\frac{GM_A}{R^{2}_A}\) and g_B= \(\frac{GM_B}{R^2_B}\)

∴ \(\frac{g_B}{g_A}=(\frac{M_B}{M_A})(\frac{R_A}{R_B})^2\)

∴ \(\frac{1}{2}(\frac{M_B}{M_A})(\frac{1}{2})^2\) = \(\frac{1}{4}(\frac{M_B}{M_A})\)

∴ \(\frac{M_B}{M_A}=\frac{4}{2}\) = 2

∴ M_B = 2M_A.

Discussion

637.	The values of the acceleration due to gravity on two planets are `g_(1) - g_(2)`, then the two planets must have the sameA. raduisB. massC. `(("mass")/("radius"))^(2)`D. `("mass")/(("radius")^(2))`
Answer» Correct Answer - D For any planet, `g=(GM)/(R^(2))" " therefore g_(1)=g_(2)` `therefore (GM_(1))/(R_(1)^(2))=(GM_(2))/(R_(2)^(2)) " " therefore (M_(1))/(R_(1)^(2))=(M_(2))/(R_(2)^(2))` `therefore` The two planets must have the same value of `("Mass")/(("Radius")^(2))`.

Discussion

638.	A body projected from the surface of the earth attains a height equal to the radius of the earth. The velocity with which the body was projected isA. `sqrt((2GM)/( R))`B. `sqrt((GM)/( R))`C. `sqrt((3GM)/( R))`D. `sqrt((5GM)/(4R))`
Answer» Correct Answer - B Applying the principle of conservation of energy, `E_("surface of the earth")=E_("at a height h")` equal to R `(1)/(2)mv^(2)-(GMm)/(R )=(1)/(2)m(0)^(2)-(GMm)/(R+R)` [`because` The velocity of the highest point = 0] `therefore (1)/(2)mv^(2)=(GMm)/(R )-(GMm)/(2R)=(GMm)/(2R)` `v^(2)=(GM)/(R )` or `v=sqrt((GM)/(R ))`

Discussion

639.	A body has a weight 72 N. When it is taken to a height `h=R`= radius of earth, it would weightA. 72 NB. 36 NC. 18ND. zero
Answer» Correct Answer - C `g=(g)/(1+(h)/(R ))^(2)=(g)/(4)` (at h=R) `therefore w=mg=(mg)/(4) rarr W =(W)/(4)` `therefore w=(72)/(4)=18 N`

Discussion

640.	Mass `M` is split into two parts `m` and `(M-m)`, which are then separated by a certain distance. What is the ratio of `(m//M)` which maximises the gravitational force between the parts ?A. `1:4`B. `1:2`C. `4:1`D. `2:1`
Answer» Correct Answer - B `F=(Gm(M-m))/(x^(2))` for maximum `(dfF)/(dm)=(G)/(x^(2))(M-2m)=0 rarr (m)/(M)=1/2`

Discussion

641.	Mass `M` is split into two parts `m` and `(M-m)`, which are then separated by a certain distance. What is the ratio of `(m//M)` which maximises the gravitational force between the parts ?
Answer» Correct Answer - `[(1)/(2)]`

Discussion

642.	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 earthB. The angular momentum of S about the centre of the earth changes in direction but its magnitude remains constantC. The total mechanical energy of S varies periodically with timeD. The linear momentum of S remains constant in magnitude
Answer» Correct Answer - A

Discussion

643.	The SI unit of gravitational constant isA. NB. JC. `m//s^(2)`D. `Nm^(2)kg^(-2)`
Answer» Correct Answer - D From `F=(Gm_(1)m_(2))/d^(2)` `G=(Fd^(2))/(m_(1)m_(2))=(Nm^(2))/(kg)^(2)=Nm^(2)`

Discussion

644.	g is called universal constant.
Answer» g is called acceleration due to graity.

Discussion

645.	Say True or False.The mass of the Earth is `6.4xx10^(6)kg`.
Answer» Mass of the earth is `6xx10^(24)kg`

Discussion

646.	The gravitational force of attraction between any two objects does not depend uponA. masses of objectsB. distance between objectsC. size and shape of objectsD. all the three above
Answer» Correct Answer - C Gravitational force does not depend upon size and shape of objects.

Discussion

647.	Gravitational force between two bodies existA. when they are not in contact onlyB. when they are not in contact onlyC. any of the above two casesD. can not be predicted
Answer» Correct Answer - c

Discussion

648.	Gravitational force can be called asA. force without any fieldB. force at a distanceC. contact forceD. fictitious force
Answer» Correct Answer - b

Discussion

649.	Assertion : In moving from centre of a solid sphere to its surface, gravitational potential increases. Reason : Gravitational field strength increase.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 `V_(C) = - 1.5(GM)/(R)` `V_(S) = - (GM)/(R)` `E_(C) = 0` and `E_(S) = (GM)/(R^(2))` `C rarr` centre, `S rarr` surface.

Discussion

650.	Assertion : When two masses come closer, their gravitaional potential energy decreases. Reason : In moving attract each other.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 `U = - (Gm_(1)m_(2))/(R)` If `r` decrease, `U` also decreases.

Discussion

Explore topic-wise InterviewSolutions in .

Match the entries given in Column A with appropriate ones in Column B

The gravitational field in a region is given by the equation `E=(5i + 12j) N//kg`. If a particle of mass `2kg` is moved from the origin to the point `(12m, 5m)` in this region, the change in the gravitational potential energy is

Two satellites are moving at heights of R and 5R above the surface of the earth of radius R. The ratio of their velocitirs `((V_(1))/(V_(2)))` isA. `sqrt(5):1`B. `1 : 1`C. `sqrt(3) : sqrt(2)`D. `sqrt(3) : 1`

Two satellites of identical masses orbit the Earth at different heights. The ratio of their distance from the centre of earth is d : 1 and the ratio of the acceleration due to gravity at those heights is g : 1. Then the ratio of their orbital velocities is ________ .

The condition for a uniform spherical mass `m` of a radius `r` to be a black hole is [`G` =gravitational constant and `g`=acceleration due to gravity]A. `(2Gm//r)^(1//2)lec`B. `(2Gm//r)^(1//2)=c`C. `(2Gm//r)^(1//2)gec`D. `(Gm//r)^(1//2)gec`

The condition for a uniform spherical mass `m` of a radius `r` to be a black hole is [`G` =gravitational constant and `g`=acceleration due to gravity]A. `(2Gm//r)^(1//2) lt c`B. `(2Gm//r)^(1//2) = c`C. `(2Gm//r)^(1//2) ge c`D. `(gm//r)^(1//2) ge c`

The value of acceleration due to gravity on the surface of earth is `x`. At an altitude of `h` from the surface of the earth, its value is `y`. If `R` is the radius of earth, then the value of `h` isA. `(sqrt(x/y)-1)R`B. `(sqrt(y/x)-1)R`C. `sqrt(y/x)R`D. `sqrt(x/y)R`

A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `3/4 mv^(2)`B. `mv^(2)`C. `1/2 mv^(2)`D. `-1/2 mv^(2)`

A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `(3)/(4)mv^(2)`B. `(1)/(2)mv^(2)`C. `mv^(2)`D. `-((1)/(2))mv^(2)`

A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `-1/2 Mv^(2)`B. `1/2 Mv^(2)`C. `3/2 Mv^(2)`D. `Mv^(2)`

A satellite moves around the earth in a circular orbit with speed `v`. If `m` is the mass of the satellite, its total energy isA. `1/2 mv^(2)`B. `1/4 mv^(2)`C. `-1/4 mv^(2)`D. `-1/2 mv^(2)`

Find the maximum and minimum distances of the planet `A` from the sun `S`, if at a ceration moment of times it was at a distance `r_(0)` and travelling with the velocity `upsilon_(0)`. With the angle between the radius vector and velocity vector being equal to `phi`.

Two planets `A` and `B` have the same material density. If the radius of `A` is twice that of `B`, then the ratio of the escape velocity `(v_(A))/(v_(B))` isA. `2`B. `sqrt(2)`C. `1//sqrt(2)`D. `1//2`

The longest and the shortest distance of a planet from the sun are `R_(1)` and `R_(2)`. Distances from sun when it is normal to major axis of orbit isA. `(R_(1)+R_(2))/2`B. `sqrt((R_(1)^(2)+R_(2)^(2))/2)`C. `(R_(1)R_(2))/(R_(1)+R_(2))`D. `(2R_(1)R_(2))/(R_(1)+R_(2))`

A satellite is launched into a circular orbit of radius `R` around the earth. While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximately:A. `0.5%`B. `1.0%`C. `1.5%`D. `3.0%`

A satellite is launched into a circular orbit of radius `R` around the earth. While a second is lunched into an orbit of radius `1.01R` The period of the second satellite is longer than the first one by approximately:A. `0.5%`B. `1.5%`C. `1%`D. `3%`

A particle is projected from the surface of earth with intial speed of 4 km/s. Find the maximum height attained by the particle. Radius of earth = 6400 km and `g = 9.8 m//s^(2)`

A satellite is orbiting just above the surface of a planet of average density `D` with period `T`. If `G` is the universal gravitational constant, the quantity `(3pi)/G` is equal toA. `T^(2)D`B. `3piT^(2)D`C. `3piD^(2)T`D. `D^(2)T`

Consider two configurations of a system of three particles of masses `m`, `2m` and `3m`. The work done by gravity in changing the configuration of the system from figure (i) to figure (ii) is A. zeroB. `-(6Gm^(2))/a(1+1/(sqrt(2)))`C. `-(6Gm^(2))/a(1-1/(sqrt(2)))`D. `(6Gm^(2))/(a)(2-1/(sqrt(2)))`

The ratio of the escape velocity and the orbital velocity isA. `sqrt(2)`B. `1/(sqrt(2))`C. `2`D. `1//2`

A comet is revolving around the sun in an elliptical orbit. Which of the following will remain constant throughout its orbit?A. Kinetic energyB. Potential enegyC. Linear speedD. Angular momentum

If a satellite is moved from one stable circular orbit to a farther stable circular orbit, then the following quantity increasesA. gravitational forceB. gravitational potential energyC. linear orbital speedD. Centripetal acceleration

Ratio of the radius of a planet `A` to that of planet `B` is `r`. The ratio of acceleration due to gravity for the two planets is `x`. The ratio of the escape velocities from the two planets isA. `sqrt(rx)`B. `sqrt(r//x)`C. `sqrt(r)`D. `sqrt(x//r)`

If the universal gravitational constant increases uniformly with time, then a satellite in orbit will still maintain itsA. weightB. tangential speedC. period of revolutionD. angular momentum

STATEMENT -1 : Time period of a satellite is inversely propertional to the square root of the mass of planet. STATEMENT -2 : Self gravitational potential energy of earth is positive. STATEMENT -3 : Orbital velocity of a satellite does depend upon the mass of planet.A. T T TB. T F FC. T F TD. F F F

The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?

The values of the acceleration due to gravity on two planets are `g_(1) - g_(2)`, then the two planets must have the sameA. raduisB. massC. `(("mass")/("radius"))^(2)`D. `("mass")/(("radius")^(2))`

A body projected from the surface of the earth attains a height equal to the radius of the earth. The velocity with which the body was projected isA. `sqrt((2GM)/( R))`B. `sqrt((GM)/( R))`C. `sqrt((3GM)/( R))`D. `sqrt((5GM)/(4R))`

A body has a weight 72 N. When it is taken to a height `h=R`= radius of earth, it would weightA. 72 NB. 36 NC. 18ND. zero

Mass `M` is split into two parts `m` and `(M-m)`, which are then separated by a certain distance. What is the ratio of `(m//M)` which maximises the gravitational force between the parts ?A. `1:4`B. `1:2`C. `4:1`D. `2:1`

Mass `M` is split into two parts `m` and `(M-m)`, which are then separated by a certain distance. What is the ratio of `(m//M)` which maximises the gravitational force between the parts ?

The SI unit of gravitational constant isA. NB. JC. `m//s^(2)`D. `Nm^(2)kg^(-2)`

g is called universal constant.

Say True or False.The mass of the Earth is `6.4xx10^(6)kg`.

The gravitational force of attraction between any two objects does not depend uponA. masses of objectsB. distance between objectsC. size and shape of objectsD. all the three above

Gravitational force between two bodies existA. when they are not in contact onlyB. when they are not in contact onlyC. any of the above two casesD. can not be predicted

Gravitational force can be called asA. force without any fieldB. force at a distanceC. contact forceD. fictitious force