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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
If the earth is at one-fourth of its present distance from the sun, the duration of the year would beA. Half the present yearB. One-eighth the present yearC. One-fourth the present yearD. One-sixth the present year |
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Answer» Correct Answer - B |
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| 152. |
Venus looks brighter than other planets becauseA. It is heavier than other planetsB. It has higher density than other planetsC. It is closer to the earth than other planetsD. It has no atmosphere |
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Answer» Correct Answer - C |
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| 153. |
Distance of geostationary satellite from the surface of earth radius `(R_(e)=6400 km)` in terms of `R_(e)` isA. `13.76 R_(e)`B. `10.76 R_(e)`C. `6.56 R_(e)`D. `2.56 R_(e)` |
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Answer» Correct Answer - C |
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| 154. |
A satellite is to revolve around the earth in a circle of radius 8000 km. With what speed should this satellite be projected into orbit? What will be the time period? Take g at the surface `=9.8ms^-2` and radius of the earth =6400 km.A. 3 km /sB. 16 km /sC. 7.15 km /sD. 8 km /s |
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Answer» Correct Answer - C |
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| 155. |
The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. then, its angular momentum will beA. `sqrt(2Emr)`B. `sqrt(2Emr^(2))`C. `sqrt((E )/(mr^(2)))`D. `(E )/(2mr^(2))` |
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Answer» Correct Answer - B `E=(1)/(2)mv^(2) " " therefore 2E = mv^(2) " " therefore 2Em = m^(2)v^(2)` `therefore` Linear momentum `(p)=mv=sqrt(2Em) " "` ….(1) `therefore` Angular momentum (L) of the satellite `= mvr = sqrt(2Em). R = sqrt(2Emr^(2))` |
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| 156. |
The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. then, its angular momentum will beA. `sqrt(E )/(mr^(2))`B. `(E )/(2mr^(2))`C. `sqrt(2 Emr^(2))`D. `sqrt(2 Emr)` |
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Answer» Correct Answer - C If m is the mass and v is the orbital velocity of the satellite then its kinetic energy `E=1/2 mv^(2) rarr Em =1/2 m^(2)v^(2)` or `m^(2)v^(2)=2Em` `rarr mv=sqrt(2Em)` if r is the radius of the orbit of the satellite then its angular momentum L= mvr using Eq i we get `L=sqrt(2Em) r=sqrt(2Emr^(2))` |
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| 157. |
For a satellite moving in a circular orbit around the earth, the ratio of its potential energy to kinetic energy isA. `1`B. `-1`C. `2`D. `-2` |
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Answer» Correct Answer - D Kinetic energy of the satellite, `K=(GM_(E)m)/(2r)` Potential energy of the satellite, `U=-(GM_(E)m)/r` Their corresponding ratio is `U/K=-2` |
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| 158. |
Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?A. `S_(1)` and `S_(2)` are moving with the same speedB. The kinetic energies of the two satellites are equalC. The time period of `S_(1)` is four times that of `S_(2)`D. The potential energies of earth and satellite in the two cases are equal |
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Answer» Correct Answer - A When the two satellites move in the same orbit, their periods are equal. T is independent of mass. But the P.E. = mgh and K.E. `=(1)/(2)mv^(2)`, depend upon the mass. As their masses are different, their P.E. and K.E. will be different. But as they are orbiting in the same orbit, their orbital speeds are equal. Option (a) is correct. |
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| 159. |
Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?A. The potential energies of earth and satellite in the two cases are equalB. `S_(1)` and `S_(2)` are moving with the same periodC. The kinetic energy of two satellites are equalD. The time period of `S_(1)` is four times that `S_(2)`. |
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Answer» Correct Answer - B Both period orbital speed of satellite `v_(0)=sqrt((GM_(E))/r)` and time period of revolution of satellite, `T=[(4pi^(2)r^(3))/(GM_(E))]^(1//2)` are independent of mass of satellite. Therefore orbital speed and time period of revolution of both the satellites are same. Hence, option (b) is correct. The kinetic energy of satellite, `K=(GM_(E)m)/(2r)` and potential energy of a satellite, `U=-(GM_(E)m)/r` Both depends on mass of satellite, very with mass of satellite. |
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| 160. |
Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?A. The potential energies of earth and satellite in the two cases are equal .B. `S_(1)` and `S_(2)` are moving with the same speed.C. The kinetic energies of the two satellites are equal .D. The time period of `S_(1)` is four times that `S_(2)`. |
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Answer» Correct Answer - B (b) Both orbital speed of satellite `v_(o)=sqrt((GM_(E)m)/( r))` and time period of revolution of satellite, `T=[(4pi^(2)r^(3))/(GM_(E))]^(1//2)`are independent of mass of satellite. Therefore orbital speed and time period of revolution of both the satellites are same. Hence , option (b) is correct. The kinetic energy of a satellite, `K=-(GM_(E)m)/(r )` and potential energy of a satellite , `U=-(GM_(E)m)/( r)` Both depend on mass of satellite, vary with mass of satellite. |
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| 161. |
Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?A. The time period of `S_(1)` is four times that of `S_(2)`B. The potential energies of earth and satellite in the two casses are equalC. `S_(1)` and `S_(2)` are moving with the same speedD. The kinetic energies of the two satellite are equal |
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Answer» Correct Answer - C Speed is independent of mass of satellite |
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| 162. |
A body is projected vartically upwards from the bottom of a crater of moon of depth `( R)/(100)` where R is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body formt eh surface of the moon. |
| Answer» Correct Answer - Nearly 99.5R | |
| 163. |
Two satellites of earth `S_(1)` and `S_(2)` are moving in the same orbit. The mass of `S_(1)` is four times the mass of `S_(2)`. Which one of the following statements is true?A. The time period of `S_(1)` is four times that of `S_(2)`B. The potential energies of the earth and satellite in the two cases are equalC. `S_(1)` and `S_(2)` are moving with the same speedD. The kinetic energies of the two satellites are equal |
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Answer» Correct Answer - C |
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| 164. |
The earth is assumed to be a sphere of raduis `R`. A plateform is arranged at a height `R` from the surface of the `fv_(e)`, where `v_(e)` is its escape velocity form the surface of the earth. The value of `f` isA. `sqrt(2)`B. `(1)/(sqrt(2))`C. `(1)/(3)`D. `(1)/(2)` |
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Answer» Correct Answer - B |
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| 165. |
Calculate angular velocity of Earth so that acceleration due to gravity at 60° latitude becomes zero.(Radius of Earth = 6400 km, gravitational acceleration at poles = 10 m/s2, cos60° = 0.5)(A) 7.8 × 10-2 rad/s(B) 0.5 × 10-3 radis(C) 1 × 10-3 radis(D) 2.5 × 10-3 rad/s |
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Answer» Correct option is: (D) 2.5 × 10-3 rad/s |
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| 166. |
Calculate angular velocity of the earth so that acceleration due to gravity at `60^(@)` latitude becomes zero (radius of the earth = 6400 km, gravitational acceleration at poles = `10 m//s^(2) , cos60^(@) = 0.5`)A. `(GMm)/(11R)`B. `(GMm)/(10R)`C. `(mgR)/(11G)`D. `(10GMm)/(11R)` |
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Answer» Correct Answer - d `g_(d)=g-Romega^(2)cos^(2)phi,0=g-R omega^(2)cos^(2)phi,` `Romega^(2)cos^(2)phi=g` `omega^(2)=(g)/(Rcos^(2)phi)` `omega=sqrt((g)/(Rcos^(2)phi))` `=sqrt((10)/(6.4xx10^(6)xx(1)/(4)))=sqrt((100)/(16xx10^(6)))` `=(10)/(4)xx10^(-3)` `omega=2.5xx10^(-3)"rad/s"` |
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| 167. |
The acceleration due to gravity at the latitude `45^(@)` on the earth becomes zero if the angular velocity of rotation of the earth isA. `sqrt(2/(gR))`B. `sqrt(2gR)`C. `sqrt((2g)/R)`D. `sqrt((5R)/2)` |
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Answer» Correct Answer - C `0=g-Romega^(2)cos^(2)45^(@)rArr omega=sqrt((2g)/R)` |
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| 168. |
The radius and density of two artificial satellites are `R_(1),R_(2)` an,d ,`rho_(1),rho_(2)` respectively. The ratio of acceleration due to gravitation them will beA. `(R_(2)rho_(2))/(R_(1)rho_(1))`B. `(R_(1)rho_(2))/(R_(2)rho_(1))`C. `(R_(1)rho_(1))/(R_(2)rho_(2))`D. `(R_(2)rho_(1))/(R_(1)rho_(2))` |
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Answer» Correct Answer - C `g=4/3piGRrhorArr gpropRrho` |
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| 169. |
A tunnel has been dug into a solid sphere of non-uniform mass density as shown in the figure. As one moves from `A` to `B`, the magnitude of gravitational field intensity A. will continuously decreaseB. will decrease up to the centre of the sphere and then increaseC. may increase or decreaseD. will continuously increase |
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Answer» Correct Answer - C As the sphere is having non uniform mass density, so nothing can be predicted about the variation of gravitational field intensity. |
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| 170. |
The motion of planets in the solar system is an exmaple of the conservation ofA. Conservation of energyB. Conservation of linear momentumC. Conservation of angular momentumD. None of these |
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Answer» Correct Answer - C |
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| 171. |
Intensity of gravitational field inside the hollow spherical shell isA. maximumB. minimumC. zeroD. constant |
| Answer» Correct Answer - c | |
| 172. |
A particle of mass m is located inside a spherical shell of mass M and radius R. The gravitational force of attraction between them isA. `(GMm)/(R)`B. `(GMm)/(R^(2))`C. `(-GMm)/(R^(2))`D. 0 |
| Answer» Correct Answer - d | |
| 173. |
A spherical shell is cut into two pieces along a chord as shown in the figure. `P` is a point on the plane of the chord. The gravitational field at `P` due to the upper part is `I_(1)`, and that due to the lower part is `I_(2)`. What is the relation between them? A. `I_(1)gtI_(2)`B. `I_(1)ltI_(2)`C. `I_(1)=I_(2)`D. no definite relation |
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Answer» Correct Answer - C At point `P` we get `l_(1)-l_(2)=0` (because the gravitational field inside a shell is zero). Hence `l_(1)=l_(2)` |
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| 174. |
If mass of a satellite is doubled and time period remain constant the ratio of orbit in the two cases will beA. `1 : 2`B. `1 : 1`C. `1 : 3`D. None of these |
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Answer» Correct Answer - B |
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| 175. |
For a satellite moving in an orbit around the earth, ratio of kinetic energy to potential energy isA. 1B. `sqrt2`C. `(1)/(2)`D. `(1)/(sqrt2)` |
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Answer» Correct Answer - C For a satellite, moving in a circular orbit of radius r, orbital velocity `v_(0)=sqrt((GM)/(r ))` `therefore` Its `K.E. =(1)/(2)mv_(0)^(2)=(1)/(2)m(GM)/(r )=(1)/(2)(GMm)/(r )` and its `P.E.=(GMm)/(r )` (in magnitude) `therefore (K.E.)/(P.E.)=(GMm)/(2r)xx(r )/(GMm)=(1)/(2)` Note : We can directly use `P.E.=-2 (K.E.)` for a satellite. `therefore (K.E.)/(P.E.)=(1)/(2)` (in magnitude) |
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| 176. |
For a satellite moving in an orbit around the earth, ratio of kinetic energy to potential energy isA. 2B. `1//2`C. `1/sqrt(2)`D. `sqrt(2)` |
| Answer» Correct Answer - B | |
| 177. |
The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will beA. `1//2` yearB. `2sqrt(2)` yearsC. `4` yearsD. `8` years |
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Answer» Correct Answer - B |
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| 178. |
A particle is projected from the surface of earth of mass M and radius R with speed `v`. Suppose it travels a distance `x(lt lt R)` when its speed becomes `v` to `v//2` and `y(lt lt R)` when speed changes from `v//2` to 0. Similarly, the corresponding times are suppose `t_(1) and t_(2)`. Then `{:(,"Column-I",,"Column-II"),("(A)",x//y,"(p)",=1),("(B)",t_(1)//t_(2),"(r)",gt 1),(,,"(r)",lt 1):}` |
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Answer» Correct Answer - `(A rarr q,B rarrp)` Given, `g=(GM)/(R^(2))="constant" " "(becausex lt ltR and Y ltltR)` When speed become `v` to `(v)/(2)` `((v)/(2))=v-gxxt_(1)rArrt_(1)=(v)/(2g)` and `((v)/(2))^(2)=v^(2)-2gxrArr(3v^(2))/(8g)` Similarly, `0=(v)/(2)-gxxt_(2)rArrt_(2)=(v)/(2g)` and `0=((v)/(2))^(2)-2gyrArr y=(v^(2))/(8g)` Clearly `t_(1) = t_(2), x gt y`. |
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| 179. |
Calculate the escape speed on the surface of a planet of mass `7.5xx10^(25)g`, and radius `1.6xx10^(8)cm. G=6.67xx10^(-8)"dyne"cm^(2)g^(-2)`. |
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Answer» Here `G=7.5xx10^(25)g, R=1.6xx10^(8)cm` Escape speed `v_(e)=sqrt((2GM)/R)=sqrt(2xx6.67xx10^(-8)x7.5xx10^(25))/(1.6xx10^(8))` |
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| 180. |
The time period of revolution of moon around the earth is 28 days and radius of its orbit is `4xx10^(5)` km. If `G=6.67xx10^(-11) Nm^(2)//kg^(2)` then find the mass of the earth. |
| Answer» Correct Answer - `6.47xx10^(24)` kg | |
| 181. |
The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will beA. `4sqrt(2)` yearsB. `2sqrt(2)` yearsC. 4 yearsD. 8 years |
| Answer» Correct Answer - B | |
| 182. |
The mean distance of Mars from the sun in `1.524` times that of the Earth from the sun. Find the number of years requires for Mars make one revolution about the Sun.A. 2.35 yearsB. 1.85 yearsC. 3.65 yearsD. 2.75 years |
| Answer» Correct Answer - B | |
| 183. |
The mean radius of a planet is `6.67xx10^(3)` km. The acceleration due to gravity on its surface is `10 m//s^(2)`. If `G = 6.67xx10^(-11)Nm^(2)//kg^(2)`, then the mass of the planet will be `[R=6.67xx10^(6)m]`A. `6.67xx10^(20)kg`B. `6.67xx10^(24)kg`C. `9.8xx10^(23)kg`D. `13.34xx10^(23)kg` |
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Answer» Correct Answer - B `g=(GM)/(R^(2))` `R=6.67xx10^(3)km = 6.67xx10^(6)m` `therefore M = (gR^(2))/(G)=(10xx6.67xx10^(6)xx6.67xx10^(6))/(6.67xx10^(-11))` `= 6.67xx10^(24)kg` |
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| 184. |
A sayellite of mass m revolves in a circular orbit of radius R a round a planet of mass M. Its total energy E is :-A. `- (GMm)/(2R)`B. `+ (GMm)/(3R)`C. `- (GMm)/R`D. `+ (GMm)/R` |
| Answer» Correct Answer - A | |
| 185. |
A satellite revolves round a planet in an orbit just above the surface of planet. Taking `G=6.67xx10^(-11) Nm^(2) kg^(-2)` and the mean density of the planet `=8.0xx10^(3) kg m^(-3)`, find the period of satellite. |
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Answer» Correct Answer - 4206.7s |
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| 186. |
A planet of mass `m` revolves in elliptical orbit around the sun of mass `M` so that its maximum and minimum distance from the sun equal to `r_(a)` and `r_(p)` respectively. Find the angular momentum of this planet relative to the sun.A. `msqrt((2GMr_(p)r_(a))/((r_(p)+r_(a))))`B. `msqrt((4GMr_(p)r_(a))/((r_(p)+r_(a))))`C. `m sqrt((GMr_(p)r_(a))/((r_(p)+r_(a))))`D. `m sqrt((GMr_(p)r_(a))/(2(r_(p)+r_(a))))` |
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Answer» Correct Answer - A Using conservation of angular momentum, `mv_(p)r_(p)=mv_(a)r_(a)` As velocities are perpendicular to radius vectors at apogee and perigee `rArr v_(p)r_(p)=v_(a)r_(a)` Using conservation of energy, `-(GMm)/(r_(p))+(1)/(2)mv_(p)^(2)=(-GMm)/(r_(a))+(1)/(2)mv_(a)^(2)` By solving, the above equations, `v_(P)=sqrt((2GMr_(a))/(r_(p)(r_(p)+r_(a))))rArr L=mv_(p)r_(p)=msqrt((2GMr_(p)r_(a))/((r_(p)+r_(a))))`. |
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| 187. |
A planet revolves around the sun in an elliptical orbit. If `v_(p)` and `v_(a)` are the velocities of the planet at the perigee and apogee respectively, then the eccentricity of the elliptical orbit is given by :A. `v_(p)/v_(a)`B. `(v_(a)-v_(p))/(v_(a)+v_(p))`C. `(v_(p)+v_(a))/(v_(p)-v_(a))`D. `(v_(p)-v_(a))/(v_(p)+v_(a))` |
| Answer» Correct Answer - D | |
| 188. |
A planet of mass `m` revolves in elliptical orbit around the sun of mass `M` so that its maximum and minimum distance from the sun equal to `r_(a)` and `r_(p)` respectively. Find the angular momentum of this planet relative to the sun.A. `L=msqrt((GMr_(p)r_(a))/((r_(p)+r_(a))))`B. `L=msqrt((2GMr_(p)r_(a))/((r_(p)+r_(a))))`C. `L=Msqrt((GMr_(p)r_(a))/((r_(p)+r_(a))))`D. `L=Msqrt(((r_(p)+r_(a)))/(GMr_(p)r_(a)))` |
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Answer» Correct Answer - B From conservation of energy `-(GMm)/(r_(p))+1/2mv_(p)^(2)=(-GMm)/(r_(a))+1/2mv_(a)^(2)` `L=mv_(p)r_(p)=mv_(a)r_(a)` |
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| 189. |
The radius ofa planet is `R_1 `and a satellite revolves round it in a circle of radius `R_2`. The tiemperiod of revolution is T. find the acceleration due to the gravitational of the plane at its surface. |
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Answer» Correct Answer - A::B::C::D `T=2pi (sqrt(R_2^2)/(gR_1^2))` or, ` T^2=4pi^2 R_2^2/(gR_1^2)` or `g=(4pi^2)/T^2 R_2^2/R_1^2` `:.` Acceleration due to gravity of the planet is `=(4pi^2)/T^2=R_2^2/R_1^2` |
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| 190. |
The earth revolves around the sun in an elliptical orbit. It has a speed `v_(1)` when it is at the minimum distance `d_(1)` from the sun. When it is at the maximum distance `d_(2)` from the sun, its speed isA. `(v_(1)d_(2))/(d_(1))`B. `(v_(1)d_(1))/(d_(2))`C. `v_(1)^(2)(d_(2))/(d_(1))`D. `v_(1)((d_(1))/(d_(2)))^(2)` |
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Answer» Correct Answer - B By the principle of conservation of momentum, `I_(1)omega_(1)=I_(2)omega_(2)` `therefore md_(1)^(2)(V_(1))/(d_(1))=md_(2)^(2)(V_(2))/(d_(2))" " therefore V_(1)d_(1)=V_(2)d_(2)` |
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| 191. |
Any satellite revolving round the earth in an orbit of height 36000 km with time period of 24 hoursA. should be a geo-stationary satelliteB. is a geo-stationary satelliteC. may or may not be a geo-stationary satelliteD. is not at all a geo-stationary satellite |
| Answer» Correct Answer - c | |
| 192. |
A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit of radius `r + Deltar(Deltar lt lt r)` with time period `T + DeltaT(DeltaT lt lt T)` then.A. `(DeltaT)/(T) = 3/2 (Deltar)/(r)`B. `(DeltaT)/(T) = 2/3 (Deltar)/(r)`C. `(DeltaT)/(T) = (Deltar)/(r)`D. `(DeltaT)/(T) = -(Deltar)/(r)` |
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Answer» Correct Answer - A |
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| 193. |
An object takes 5s to reach the ground from a height of 5m on a planet. What is the value of g on the planet? |
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Answer» Given: Displacement (s) `=5m` Time `(t)=5s` Initial velocity `(u)=0m//s` To find: Gravitational acceleration `(g)=?` Formula: `s=ut+1/2"gt"^(2)` Solution: `5=0+1/2xxgxx5^(2)` `5=gxx25/2` `(5xx2)/25=g` `g=2/5` `=g=0.4m//s^(2)` Value of g on the planet is `0.4m//s^(2)` |
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| 194. |
How much below the surface of the earth does the acceleration due to gravity (i) reduced to `36%` (ii) reduces by `36% ` , of its value on the surface of the earth ? Radius of the earth = 6400 km . |
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Answer» Correct Answer - (i) 4096 km (ii) 2304 km Hint `g_(d)=g(1-(d)/(R_(e)))` |
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| 195. |
At what height the acceleration due to gravity decreases by 36% of its value on the surface of the earth ? |
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Answer» Correct Answer - 4266 km Hint : Use `g_(h)=g[(R )/(R+h)]^(2)` |
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| 196. |
Find the height above the surface of the earth where the acceleration due to gravity reduces by (a) 36% (b) 0.36% of its value on the surface ofthe earth. Radius of the earth R = 6400 km. |
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Answer» Correct Answer - zero 1600km 11.52km |
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| 197. |
Two point masses m and M are held at rest at a large distance from each other. When released, they begin moving under their mutual gravitational pull.Find their relative acceleration (a) when separation between them becomes x.Integrate the expression of a obtained above to calculate the relative velocity of the two masses when their paration is x. Write the velocity of centre of mass of the system when separation between them is x. |
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Answer» Correct Answer - `(G(M+m))/(x^(2))` `sqrt((2G(M+m))/(x))` |
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| 198. |
A rocket is launched vertically from the surface of earth with an initial velocity `v`. How far above the surface of earth it will go? Neglect the air resistance.A. `(R_(E)v^(2))/(gR_(E)-v^(2))`B. `(R_(E)v^(2))/(gR_(E)+v^(2))`C. `(R_(E)v^(2))/(2gR_(E)-v^(2))`D. `(R_(E)v^(2))/(2gR_(E)+v^(2))` |
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Answer» Correct Answer - C ( c) Let the rocket reaches a height h from the surface of earth. Total energy at the surface of the earth is `E_(s)=(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))` where m and `M_(E)` are the masses of rocket and earth respectively. At highest point, the velocity of the rocket becomes zero. `:.` Total energy at the highest point is `E_(h)=-(GM_(E)m)/((R_(E)+h))` According to law of conservation of energy, `E_(s)=E_(h)` `:.(1)/(2)mv^(2)-(GM_(E)m)/(R_(E))=-(GM_(E)m)/(R_(E)+h)` `(1)/(2)v^(2)-(GM_(E))/(R_(E))=-(GM_(E))/(R_(E)+h)` `=(gR_(E)^(2))/(R_(E))-(gR_(E)^(2))/(R_(E)+h)( :. G=(GM_(E))/(R_(E)^(2)))` `=gR_(E)(1-(R_(E))/(R_(E)+h))=gR_(E)((h)/(R_(E)+h))` `v^(2)(R_(E)+h)=2gR_(E)h` `v^(2)R_(E)=2gR_(E)h-v^(2)h` `R_(E)v^(2)=h(2gR_(E)-v^(2))` `h=(R_(E)v^(2))/(2gR_(E)-v^(2))` |
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| 199. |
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//2)`B. `R^(3//5)`C. `R^(7//2)`D. `R^(7//2)` |
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Answer» Correct Answer - D (d) According to the question, the gravitational force between the planet and the star is ` F prop (1)/(R^(5//2))` ` :. F=(GMm)/(R^(5//2))` where M and m be masses of star and planet respectively. For motion of a planet in a circular orbit, `mRomega^(2)=(GMm)/(R^(5//2))` `mR((2pi)/(T))^(2)=(GMm)/(R^(5//2))" " ( :. omega=(2pi)/(T))` `(4 pi^(2))/(T^(2))=(GM)/(R^(7//2)) implies T^(2)=(4 pi^(2))/(GM)R^(7//2)` `T^(2) prop R^(7//2) or T prop R^(7//4)` |
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| 200. |
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^(5//2)`C. `R^(3//2)`D. `R^(7//2)` |
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Answer» Correct Answer - B Gravitational force `(=(GMm)/(R^(3//2)))` providies the necessary centripetal force i.e `mRomega^(2)` so `(GMm)/(R^(3//2)=mR omega^(2)=mR(2pi)/(T)^(2)=(4pi^(2)mR)/(T^(2))` or `T^(2)=(4pi^(2)R^(5//2))/(GM), i.e T^(2) prop R^(5//2)` |
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