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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.
| 51. |
Which of the following options are correct ?A. Acceleration due to gravity decreases with increasing altitude.B. Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).C. Acceleration due to gravity increases with increasing latitude.D. Acceleration due to gravity is independent of mass of the earth. |
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Answer» Correct Answer - A::C At altitude `h`, `g_(h) = (gR^(2))/((R + h)^(2)) = (g)/((1 + h//R)^(2))` As `h` increases, `g_(h)` decreases. At depth `d`, `g_(d) = g (1- (d)/(R ))`, As `d` increases, `g_(d)` decreases At latitude `lambda`, `g_(lambda) = g - R omega^(2) cos^(2) lambda`, As `lambda` increases, the value of `cos lambda` decresas so `g_(lambda)` increases On surface of earth, `g = (GM)/(R^(2))`, i.e., `g` depends on mass `M` of the earth. |
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| 52. |
What are the main features of gravitational force? |
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Answer» Following are the main features of gravitational force : (1) It is always an attractive force. (2) It is independer of the medium between the particles. (3) It holes good over a wide range of distances (i.e., from interplanetary distance to interatomic distances) (4)The gravitational force between two particles is independent of presence or absence of other particles. (6) The total gravitational force on an particle due to number of particles is the resultant of forces of attraction exerted on the given particle due to individual particles, i.e., `vec(F) = vec(F_(1)) + vec(F_(2)) + vec(F_(3))`... it means the principle of superposition is valid. (7) It expresses the force between point masses. (8) It is a conservative force, i.e., the work done in moving a particle once around a closed path under the action of gravational force is zero. |
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| 53. |
Whenever a body is thrown vertically upwards with a certain velocity, the upwards motion is opposed by gravitational pull of earth and resisatnce of air. The velocity of the body goes on decreasing at a constant rate `(= - g)`.As soon as upward velocity of body becomes zero, it cannot rise any more. The height it has attained is the maximum height. The body then begins to fall downwards with an acceleration `= g`. Read the above passege and answer the following questions : (i) A body is thrown upwards with a velocity of `19.6 m//s`. What is the maximum height attained? (ii) With what velocity will the body hit the ground? (iii) What are the implications of this study in day to day life ? |
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Answer» (i) Here, `u= 19.6 m//s, upsilon = 0, a = - g = - 9.8 m//s^(2), s = h = ?` from `upsilon^(2) - u^(2) = 2 as` `0 - (19.6)^(2) = 2 (-9.8)h` `h = 19.6 m` (ii) If we ignore resistance due to air and other stray losses of energy, them the body will hit the ground with the same velocity `= 19.6 m//s`, with which it was thrown. (iii) In day to day life, our progress and rise in carrer is opposed by several factors influding jealousy of your friends and foes. It is the initail kick (i.e., powerful stars) and continuous motivation that lead us to our desired goel. |
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| 54. |
If the Earth were made of lead of relative density `11.4`, then find the value of acceleration due to gravity on the surface of Earth ? Radius of the Earth is `6400 km` and `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. |
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Answer» Density of Earth `rho =` relative density `xx` density of water `= 11.4 xx 10^(3) kg m^(-3)` Acceleration due to gravity on the surface of Earth is `g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho = (4 piGR rho)/(3)` `= (4 xx (3.14) xx (6.67 xx 10^(-11)) xx (6400 xx 10^(3)) xx (11.4 xx 10^(3)))/(3)` `= 20.38 ms^(-2)` |
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| 55. |
A spherical mass of `20 kg` lying on the surface of the Earth is attracted by another spherical mass of `150 kg` with a force equal to `0.23 mg f`. The centres o fthe two masses are `30 cm` apart. Calculate the mass of the Earth. Radius of the Earth is `6 xx 10^(6)m`. |
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Answer» Here, `m_(1) = 20 kg , m_(2) = 150 kg`, `r = 30 cm = 0.30 m` `F = 0.23 mg f = 0.23 xx 10^(-6) kg f` `= 0.23 xx 10^(-6) xx 9.8 N` `F = (Gm_(1)m_(2))/(r^(2))` or `G = (F r^(2))/(m_(1)m_(2)) = ((0.23 xx 10^(-6) xx 9.8) xx (0.30)^(2))/(20 xx 150)` `= 6.762 xx 10^(-11) Nm^(2) kg^(-2)` Mass of earth, `M = (gR^(2))/(G) = (9.8 xx (6 xx 10^(6))^(2))/(6.762 xx 10^(-11))` `= 5.2 xx 10^(24) kg` |
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| 56. |
A remote sensing satellite of the Earth in a circular orbit at a height of `400 km` above the surface of Earth. What is the (a) orbital speed, and (b) period of revolution of satellite ? Radius of Earth `= 6 xx 10^(6) m` and acceleration due to gravity the surface of Earth is `10 m//s^(2)`. |
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Answer» Here, `R = 6 xx 10^(6)m , g = 10 m//s^(2)`, `h = 400 xx 10^(3) m = 0.4 xx 10^(6) m` (a) Orbital speed, `upsilon = R sqrt((g)/(R + h))` ` = 6 xx 10^(6) sqrt((10)/(6 xx 10^(6) + 0.4 xx 10^(6))` `= 7.5 xx 10^(3) m//s` (b) Period of revolution, `T = (2 pi)/(R ) sqrt(((R + h)^(3))/(g)` `= (2 xx (22//7))/(6 xx 10^(6)) sqrt(((6 xx 10^(6) + 0.4 xx 10^(6))^(3))/(10))` `= 5368.5 s` |
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| 57. |
Two stars each of mass `M` and radius `R` are approaching each other for a head-on collision. They start approaching each other when their separation is `rgt gtR`. If their speed at this separation are negligible, the speed `v` with which they collide would be |
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Answer» Here, mass of each star, `M = 2 xx 10^(30) kg`. Initial distance between two stars, `r = 10^(9) km = 10^(12) m`. Initial potential energy of the systam `= (GM m)/(r )` Total K.E. of the mass `= (1)/(2) M upsilon^(2) + (1)/(2) M upsilon^(2) = M upsilon^(2)` where `upsilon` is the speed of stars with which they collide. when the stars are about to collide, the distance between their centres, `r = 2R`. `:.` Final potential energy of two stars `= - (GM M)/(2R)`. Since, gain in `K.E.` is at the cost of loss in `P.E.` `:. M upsilon^(2) = - (GM M)/(r ) - (-(GM M)/(2R)) = (-GM M)/(r) + (GM M)/(2R)` or `2 xx 10^(30) upsilon^(2) = - (6.67 xx 10^(-11) xx (2 xx 10^(30))^(2))/(10^(12)) + (6.67 xx 10^(-11) (2 xx 10^(30))^(2))/(2 xx 10^(7))` `= - 2.668 xx 10^(38) + 1.334 xx 10^(43) = 1.334 xx 10^(43) J` `:. upsilon = sqrt((1.334 xx 10^(43))/(2 xx 10^(30))) = 2.583 xx 10^(6) ms^(-1)` |
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| 58. |
There is no atomosphere on moon because |
| Answer» The value of escape speed on the surface of Moon is small (only `2.5 km s^(-1))`. It is so because the value of acceleration due to gravity on the surface of moon is small. The molecules of the atmospheric gases on the surface of the Moon have thermal speeds greater than the escape spped. due to it, all the molecules of gases have escape and there is no atmosphere. | |
| 59. |
calculate the gravitaional intensity and graviational potential at a location which is from the surface of the Earth at a height `4` times the radius of the surface. `R_(e) = 6400 km`, `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , M_(e) = 6 xx 10^(24) kg`. |
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Answer» `I = (GM)/((R + h)^(2)) = (GM)/((R + 4R)^(2)) = (GM)/(25R^(2))` `= ((6.67 xx 10^(-11)) xx (6 xx 10^(24)))/(25 xx (6.4 xx 10^(6))^(2)) = 0.39 N//kg` `V = - (GM)/((R + h)) = - (6.67 xx 10^(-11) xx 6 xx 10^(24))/(5 xx 6.4 xx 10^(6))` `= - 1.25 xx 10^(7) J kg^(-1)` |
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| 60. |
The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would beA. `2R`B. `4R`C. `R//4`D. `R//2` |
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Answer» Correct Answer - D `g_(e) = (GM_(e))/(R_(e)^(2)) = (G xx (4)/(3) pi R_(e)^(2) xx rho_(e))/(R_(e)^(2)) = (4)/(3) pi GR_(e) rho_(e)` and `g_(p) = (GM_(p))/(R_(p)^(2)) = (G xx (4)/(3) pi R_(p)^(3) xx rho_(p))/(R_(p)^(2))` `= (4)/(3) pi G R_(p) rho_(p)` As, `g_(e) = g_(p)`, so `R_(e) rho_(e) = R_(p) rho_(p)` or `R_(p) = (R_(e)rho_(e))/(rho_(p)) = (Rrho)/(2rho) = (R)/(2)` |
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| 61. |
The radii of circular orbits of two satellite `A` and `B` of the earth are `4R` and `R`, respectively. If the speed of satellite `A` is `3v`, then the speed of satellite `B` will beA. `3 V//4`B. `6 V`C. `12 V`D. `3 V//2` |
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Answer» Correct Answer - B Orbital velocity of a satellite, `upsilon = sqrt((GM)/(r ))` `:. upsilon^(2) xx r = GM =` constant `:. (3V)^(2) xx 4R = V_(B)^(2) xx R` `V_(B) = sqrt(36V^(2)) = 6V` |
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| 62. |
A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two sphere of equal radii 1 unit, with their centres at A(-2,0 ,0) and B(2,0,0) respectively, are taken out of the solid leaving behind spherical cavities as shown if fig Then: A. the gravitational force due to this object at the origin is zeroB. the gravitational force at the point `B(2, 0, 0)` is zeroC. the gravitational potential is the same at all points of the circly `y^(2) + z^(2) = 36`D. the gravitaitonal potential is the same at all points on the circle `y^(2) + z^(2) = 4`. |
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Answer» Correct Answer - A::B::D The mass of the sphere with cavities is symmetrically situsted w.r.t. the origin. The circle `y^(2) + z(2) = 36` has a radius `6` and all the points on it are at a distance `6` from the centre of sphere where the whole mass of the sphere can be supposed to be concentrated. the circle is outside the sphere. the sinilar axplanation is valid for `y^(2) + z^(2) = 4`. The gravitational pull at the centre of sphere is zero. |
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| 63. |
Statement-1 : A body released from a height equal to the radius `(R )` of the earth. The velocity of the body when it strikes the surface of the earth will be `sqrt(2 g R)`. Statement -2 : As `upsilon^(2) = u^(2) + 2` as.A. Statement -1 is true , Statement -2 is true , Statement-2 is a correct explanation of Statement -1.B. Statement -1 is true , Statement-2 is true , Statement-2 is a correct explanation of Statement-1.C. Statement-1 is true , Statement-2 is false.D. Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - D Here, statement-1 is wrong and statement-2 is correct. At height `R`, velocity of the body, `upsilon = 0`. `K.E.` of body `= 0`. `P.E.` of body `= - (Gm)/((R + R)) = - (GMm)/(2R)` Total energy `= K.E. + P.E. = - (GMm)/(2R)` On the surface of earth, let `upsilon` be the velocity of the body `K.E.` of body `= (1)/(2) m upsilon^(2)`, `P.E.` of body `= - (GMm)/(R )` Total energy `= (1)/(2) m upsilon^(2) - (GMm)/(R )`. According to law of conservation of energy `(1)/(2) m upsilon^(2) - (GMm)/(R ) = - (GMm)/(2R)` or `(1)/(2) upsilon^(2) = (GM)/(R ) = - (GM)/(2R) = (gR^(2))/(2R)` or `upsilon = sqrt(gR)` |
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| 64. |
Two equal masses of `6.40 kg` are separted by a distance of `0.16 m`. A small body is released from a point `P`, equidistant from the two masses and at a distance of `0.06 m` from the line joining them. Fig. (a) Calculate the velocity of this body when it passes through `Q`. (b) Calculate the acceleration of this body at `P` and `Q` if its mass is `0.1 kg`. Use `G = 6.67 xx 10^(-11) N m^(2) kg^(-2)` |
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Answer» In figure, `AP = BP = sqrt((0.08)^(2) + (0.06)^(2)) = 0.1 m` Here, `m_(1) = 6.4 kg = m_(2)` and `m_(3) = 0.1 kg` (a) When body falls from `P` to `Q`, then gain in `KE` of body is equal to loss in `PE` of the system. `:. (1)/(2) m_(3) upsilon^(2) = - [(Gm_(1)m_(3))/((0.1)) + (Gm_(2) m_(3))/(0.1)]` `+ [(Gm_(1)m_(3))/(0.08) + (Gm_(2) m_(3))/(0.08)]` or `(1)/(2) upsilon^(2) = - [(Gm_(1))/(0.1) + (Gm_(2) )/(0.1)] + [(Gm_(1))/(0.08) + (Gm_(2) )/(0.08)]` or `upsilon^(2) = - 2 xx G [((6.4)/(0.1) + (6.4)/(0.1)) - ((6.4)/(0.08) + (6.4)/(0.08))]` `= - 2 xx (6.67 xx 10^(-11)) xx (-32)` `= 64 xx 6.67 xx 10^(-11)` or `upsilon = (64 xx 6.67 xx 10^(-11))^(1//2) = 6.53 xx 10^(-5) ms^(-1)` |
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| 65. |
If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what percent? |
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Answer» We know that, `g = (GM)/(R^(2))` or `g prop (1)/(R^(2))` So `(g_(2))/(g_(1)) = ((R_(1))/(R_(2)))^(2)` If `R_(1) = R`, then `R_(2) = R +- n% R = (1 +- n%)R` `:. (g_(2))/(g_(1)) = [(R)/((1 +- n%)R)]^(2) = (1)/((1 +- n%)^(2))` `= (1 +- n%)^(-2)` `%` increase or decrease in the value of `g` is `= ((g_(2) - g_(1))/(g_(1))) xx 100 = ((g_(2))/(g_(1)) - 1) xx 100` `= [(1 +- n%)^(-2) - 1] xx 100` |
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| 66. |
Calculate (i) kinetic energy (ii) potential energy and (iii) total energy of a satellite of mass `200 kg` orbiting around the earth in an orbit of height `100 km` from the surface of earth. Given, mass of earth `= 10^(25) kg`,radius of earth `= 6.4 xx 10^(6) m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. |
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Answer» Here, `m = 200 kg , M = 10^(25) kg`, `R = 6.4 xx 10^(6) m` `h = 100 km = 0.1 xx 10^(6) m` Orbital velocity of satellite is `upsilon_(0) = sqrt((GM)/(R + h))` (i) `KE` of satellite, `K = (1)/(2) m upsilon_(0)^(2) = (1)/(2) m (GM)/(R + h)` `= (1)/(2) xx (200 xx(6.67 xx 10^(-11)) xx 10^(25))/((6.4 xx 10^(6) + 0.1 xx 10^(6))` `= 10 xx 10^(9) J = 10^(10) J` (ii) `PE` of satllite, `U = - (GM m)/(R + h) = - 2 xx K` `= - 2 xx 10^(10) J` (iii) Total energy of satllite `= K + U` `= 10^(10) + (-2 xx 10^(10))` `= - 10^(10) J` |
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| 67. |
Statement-1 : An artificial satellie moving in a circular orbit around the earth has a total energy (i.e., sum of potential energy and kinetic energy) `E_(0)`. Its potential energy is `-E_(0)`. Statement-2 : Potential energy of the body at a point in a gravitaitonal field of earth is `-(GMm)/(R )`.A. Statement -1 is true , Statement -2 is true , Statement-2 is a correct explanation of Statement -1.B. Statement -1 is true , Statement-2 is true , Statement-2 is a correct explanation of Statement-1.C. Statement-1 is true , Statement-2 is false.D. Statement-1 is false , Statement-2 is true. |
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Answer» Correct Answer - D Here, statement-1 is wrong but statement-2 is correct. As for a satellite `K.E. = (GMm)/(2R)` and `P.E. = - (GMm)/(R )` Toatal energy, `E_(0) = K.E. + P.E.` `= (GMm)/(2R) - (GMm)/(R )` `= - (GMm)/(2R) = (PE)/(2)` `PE = 2 E_(0)`. |
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| 68. |
A body weighs `54 kg f` on the surface of Earth. How much will it weigh on the surface of mers whose mass is `1//9` and the redius is `1//2` of that of earth? |
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Answer» `g_(p) = g_(e) .(M_(p))/(M_(e)) xx ((R_(e))/(R_(p)))^(2) = g_(e) xx (1)/(9) xx (3)^(2) = g_(e)` Weight of body on planet `= mg_(p) = mg_(e) = 36 xx 9.8 N = 36 kg wt` |
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| 69. |
Moelcules in air in the atmosphere are attracted by gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree. |
| Answer» Air molecules in the atmosphere experience vertically downward force due to gravity, just like an apple falling vertically from a tree. Due to temperature, the air molecules have random motion. Due to it, the velocity of air molecules is not exactly in the vertical downward direction. the downward gravity pull on ait molecules increase the density of air in the atmosphere close to earth. It is due to this reason, the density of air decrease as we go up. | |
| 70. |
Will `1 kg` sugar be more at poles or at the equator? |
| Answer» The value of `g` larger at the poles than at the equator. If the suger is weighed in a physical balance then there will be no difference. If it is weighed by a spring balance, calibrated at the equator, then `1 kg` of suger will have a lesser amount at poles. | |
| 71. |
Since the Moon is gravitational attracted to the Earth, why does it not simply crash into the Earth? |
| Answer» The Moon is orbiting around the Earth in a certain orbit with a certain period. The centripetal force required for the orbital motion is provided to the Moon by the gravitational pull of Earth. The Moon can crash into the Earth if its tangential velocity is reduced to zero. As Moon has tangential velocity while orbiting around Earth, it simply falls around Earth rather than into it and hence cannot crash into the Earth. | |
| 72. |
What is the angle between the equatorial plane and the orbital plane of (a) Polar satellitee ? (b) Geostationary satellite? |
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Answer» (a) Angle between the equatorial plane and orbital plane of polar satellite is `90^(@)`. (b) Angle between equatorial plane and orbital plane of the geostationary satellite is `0^(@)`. |
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| 73. |
A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water. Is poured into one side until it stands at a distance of `10 mm` above the water level on the other side. Meanwhile the water rises by `65mm` from its original level (see diagram). The density of the oil is: A. `650 kg m^(-3)`B. `425 kg m^(-3)`C. `800 kg m^(-3)`D. `928 kg m^(-3)` |
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Answer» Correct Answer - D Pressure at `B =` presure at `C` `(65 + 65 + 10) rho_(oil) g = (65 + 65) rho_(w) g` or `140 rho_(oil) = 130 xx rho_(w)` or `rho_(oil) = (13)/(14) rho_(w) = (13)/(14) xx 10^(3) kg//m^(3)` `=928 kg//m^(3)` |
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| 74. |
If the law of gravitational, instead of being inverse-square law, becomes an inverse-cube lawA. planets will not have elliptic orbits.B. circular orbits of planets is not possible.C. projectile motion of a stone thrown by hand on the surface of the earth will be appoximately perabolic.D. there will be no gravitational force inside a spherical shell of uniform density. |
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Answer» Correct Answer - A::B::C If the law of gravitational becomes an inverse cube law then `F = (GMm)/(r^(3)) = (m upsilon^(2))/(r )` or `upsilon = (sqrt(GM))/(r )` `:.` Time period of revolution of a planet, `T = (2 pi r)/(upsilon) = (2 pi r^(2))/(sqrt(GM))` or `T^(2) prop r^(4)`. It mean a planet will not have an elliptical orbit. The circular orbit of a planet may not be possible as the gravitational attractive force obays inverse square law and not inverse cube law. A stone thrown by hand on the surface of the earth will follow nearly parabolic path, under the gravitational force. There will be some gravitational force inside a spherical shell of uniform density. |
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| 75. |
There have been suggestions that the value of the gravitational constant `G` becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earthA. nothing will change.B. we will become hotter after billions of years.C. we will be going around but not strictly in closed orbits.D. after sufficiently long time we will leave the solar system. |
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Answer» Correct Answer - C::D When `G` decreases with time, the gravitational force `F = (GM m)/(r^(2))` will become weaker with time. Then `r` increases with time. Due to it, earth willl be going around the sun not strictly in closed orbit and after long time it will leave the solar system. |
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| 76. |
The masses and radii of the Earth and the Moon are `M_1, R_1 and M_2,R_2` respectively. Their centres are at a distance d apart. The minimum speed with which a particel of mass m should be projected from a point midway between the two centres so as to escape to infinity is ........ |
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Answer» Gravitational potential energy of the particle of mass `m` at a distance `r//2` from the centre of the Earth `= - (GM_(1) m)/((r//2)) = - (2GM_(1) m)/(r )` Gravitational potential energy of the particle of mass `m` at a distance `r//2` from the centre of the Moon `= - (GM_(2) m)/((r//2)) = - (2 Gm_(2) m)/(r )` Total potential energy of the particle, `U = - (2 Gm_(1) m)/(r ) - (2 Gm_(2) m)/(r )` `= - (2Gm)/(r )(M_(1) + M_(2))` Since, the P.E. at infinity is zero, so work required so shift the mass from the given position to infinity is, `W = 0 - U = (2Gm)/(r ) (M_(1) + M_(2))` or `upsilon sqrt(4G (M_(1) + M_(2))//r)` |
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| 77. |
Two bodies of masses `m_1` and `m_2` are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance `r` between them is. |
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Answer» Let `upsilon_(r)` be the relative velocity of approach of two bodies at distance `r` apart. The reduced mass of the system of two particles is, `mu = (m_(1) m_(2))/(m_(1) + m_(2))`. According to law of conservation of mechincal energy. Decrease in potential energy = increase in K.E. `0 - (-(Gm_(1) m_(2))/(r)) = (1)/(2) mu upsilon_(r)^(2)` or `(Gm_(1) m_(2))/(r) = (1)/(2) ((m_(1) m_(2))/(m_(1) + m_(2))) upsilon_(r)^(2)` or `upsilon_(r) = sqrt((2G (m_(1) + m_(2)))/(r))` |
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| 78. |
Both earth and moon are subjected to the gravitational force of the sun. as observed from the sun, the orbit of the moonA. will be elliptical.B. will not be strictly elliptical the total gravitational force on it is central.C. os not elliptical but will necessarily be a closed curve.D. deviates considerably from being elliptical due to influence of plants other than earth. |
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Answer» Correct Answer - B Moon is revolving around earth in almost circular orbit. Sun exerts gravitational pull on both, earth and moon. When observed from sun, the orbit of the moon will not be stricly elliptical because the total gravitational force (i.e., force due to earth on moon and force due to sun on moon) is not central. |
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| 79. |
If the radius of the Earth shrinks by `2%`, mass remaing same, then how would the have of acceleration due to gravity change? |
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Answer» Let `M,R` be tha mass and redius of the Earth. The acceleration due to gravity on the surface of Earth is given by `g = GM//R^(2)` ..(i) Differentiating it, we have, `dg = - (2GM)/(R^(3))dR` ..(ii) Dividing (iii) by (i), we have, `(dg)/(g) = - (2dR)/(R )` ..(iii) When redius of earth shrinks by `2%` then, `(dR)/(R ) = - (2)/(100)` ..(iv) From (iii) & (iv), the `%` increase in the value of `g` is `= (dg)/(g) xx 100 = - (2dR)/(R ) xx 100 = - [(-2)/(100)] xx 100` `= 4 %` |
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| 80. |
A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours. |
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Answer» As `T^(2)= k r^(3)` or `T prop r^(3//2)` `:. (T_(1))/(T_(2)) = ((r_(2))/(r_(1)))^(3//2)` or `T_(2) = T_(1) ((r_(2))/(r_(1)))^(3//2) = 24 ((2.5 R + R)/(6R + R))^(3//2)` `= 24 ((1)/(2))^(3//2) = 6 sqrt(2) hour` |
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| 81. |
A geostationary satellite is orbiting the Earth at a height of `6 R` above the surface of Earth, where `R` is the radius of the Earth. The time period of another satellites is `6 sqrt(2) h`. Find its height from the surface of Earth. |
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Answer» Here, `r_(1) = 6R + R = 7 R, T_(1) = 24 h, T_(2) = 6 sqrt(2) h`, `r_(2) = r_(1) ((r_(2))/(r_(1)))^(3//2) = 7R((6sqrt(2))/(24))^(2//3) = 7R((1)/(2)) = 3.5 R` `:.` Height of satellite from the surface of earth `= 3.5 R - R` `= 2.5 R` |
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| 82. |
The orbit of a geostationary satellite is concentric and coplanar with the equator of Earth and rotates along the direction of rotation of Earth. Calculate the height and speed. Take mass of Earth `= 6 xx 10^(-11) Nm^(2) kg^(-2)`. Given `pi^(2) = 10`. |
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Answer» Here, `T = 24 h = 24 xx 3600 s, M = 6 xx 10^(24) kg`, `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2), R = 6.4 xx 10^(6) m` `h = [(T^(2) GM)/(4pi^(2))]^(1//3) - R` `= [((24 xx 3600)^(2) xx (6.67 xx 10^(-11)) xx 6 xx 10^(24))/(4 xx 10)]^(1//2)` `- 6.4 xx 10^(6)` `= 42.11 xx 10^(6) - 6.4 xx 10^(6)` `= 35.71 xx 10^(6) m`. Orbital velocity, `upsilon_(0) = (2pi (R + h))/(T)` `= (2 xx 3.14 xx 42.11 xx 10^(6))/(24 xx 3600) = 3.062 xx 10^(3) ms^(-1)` |
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| 83. |
Two indentical geostationary satellite each of mass `m` are moving with equal speed `upsilon` in the same orbit but their sense of rotation brings them on a collision course. What will happen to the debris? |
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Answer» The collision of two satellite will be inclastic collision and they stick together after collision. Let `V` be the velocity of two satellies after collision. According to law of conservation of linear momentum `m upsilon + (-m upsilon) = (m + m) V` or `V = 0` The kinetic energy of the debris in its orbit `= (1)/(2) (m + m)V^(2) = (1)/(2) xx 2 m xx 0 = 0` Due to gravity, the debris will fall towards Earth. During its fall, it meets with air resisatnce and may get burnt up. |
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| 84. |
A person brings a mass of 1 kg from infinity to a point . Initally the mass was at rest but it moves at a speed of 2 `ms^-1` as it reaches A. The work done by the person on the mass is -3J. The potential at A isA. `-2 J//kg`B. `-3 J//kg`C. `-5 J//kg`D. `-7 J//kg` |
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Answer» Correct Answer - C Work done `= P.E. + K.E., - 3 = PE + (1)/(2) xx 1xx 2^(2)` or `P.E. = - 3 - 2 = - 5J` ot Gravitational potential `= P.E.//mass = - 5//1 = - 5J//kg` |
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| 85. |
One of the satellite of jupiter, has an orbital period of `1.769` days and the radius of the orbit is `4.22 xx 10^(8) m`. Show that mass of jupiter is about one thousandth times that of the radius of the sun. (Take `1` year `= 365.15` mean solar day). |
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Answer» For a satellite of jupiter period, `T_(1) = 1.769 days = 1.769 xx 24 xx 60 xx 60 s` Radius of the orbit of satellite, `r_(1) = 4.22 xx 10^(8) m` Mass of the jupiter, `M_(1)` is given by `M_(1) = (4 pi^(2) r_(1)^(3))/(GT_(1)^(2)) = (4pi^(2) xx (4.22 xx 10^(8))^(3))/(G xx (1.769 xx 24 xx 60 xx 60)^(2))` ..(i) We know that the orbital period of Earth around the sun, `T = 1 year = 365.25 xx 24 xx 60 xx 60 s`, Orbitial radius, `r = 1 A.U. = 1.496 xx 10^(11) m` Mass of the sun is given by `M = ( 4 pi^(2) r^(3))/(GT^(2)) = (4pi^(2) xx (1.496 xx 10^(11))^(3))/(G xx (365.25 xx 24 xx 60 xx 60)^(2))` ..(ii) Dividing (2) by (1), we get `(M)/(M_(1)) = (4pi^(2) xx (1.496 xx 10^(11))^(3))/(G xx (365.25 xx 24 xx 60 xx 60)^(2))xx (G xx (1.769 xx 24 xx 60 xx 60)^(2))/(4pi^(2) xx (4.22 xx 10^(8))^(3)) = 1046` or `(M_(1))/(M) = (1)/(6400) ~~(1)/(1000)` or `M_(1) ~~ (1)/(1000) M`, which was to be proved. |
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| 86. |
The mass of planet Jupiter is `1.9xx10^(7)kg` and that of the Sun is `1.9x10^(30)kg`. The mean distance of Jupiter from the Sun is `7.8xx10^(11)`m. Calculate te gravitational force which Sun exerts on Jupiter. Assuming that Jupiter moves in circular orbit around the Sun, also calculate the speed of Jupiter `G=6.67xx10^(-11)Nm^(2)kg^(-2)`. |
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Answer» Here, `M_(J) = 1.9 xx 10^(27) kg`, `M_(S) = 1.99 xx 10^(30) kg, r = 7.8 xx 10^(11) m`, `g = 6.67 xx 10^(-11) Nm^(2) kg^(-2), F= ?` Now `F = (GM_(J) M_(S))/(r^(2))` `= (6.67 xx 10^(-11) xx 1.9 xx 10^(27) xx 1.99 xx 10^(30))/((7.8 xx 10^(11))^(2))` `= 4.15 xx 10^(23)N` Since, the gravitational pull of sun to jupiter provides the required centreipetal force to the jupiter, therefore, the velocity of the jupter `upsilon` can be given `F = (M_(J) upsilon^(2))/(r )` or `upsilon = sqrt((F r)/(M_(J))) = sqrt((4.15 xx10^(23) xx 7.8 xx 10^(11))/(1.9 xx 10^(27)))` `= 1.3 xx 10^(4) ms^(-1)` |
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| 87. |
The gravitational field due to a mass distribution is given by `E=K/x^3` in X-direction. Taking the gravitational potential to be zero at infinity, find its value at a distance x.A. `K//x`B. `K//2x`C. `K//x^(2)`D. `K//2x^(2)` |
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Answer» Correct Answer - D Since, `I = - dV//dr` or `dV = - I dr` So, `V = underset(0) overset(x)int - I dr = underset(0) overset(x)int - K r^(-3) dr` `= - K((r^(-3+ 1))/(-3 + 1))_(0)^(x) = (K)/(2x^(2))` |
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| 88. |
In a certain region of space gravitational field is given by `I = - (k//r)`. Taking the reference point to be at `r = r_(0)`, with gravitational potential `V = V_(0)`, find the gravitational potential at distance `r`. |
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Answer» We know that, the garvitational intensity is equal to the negative of gradient of potential i.e., `I = - (dV)/(dr)`. Here, `I = - (k)/(r )`, so `(dV)/(dr) = (k)/(r )` or `int_(V_(0))^(V) dV = int_(r_(0))^(r) (k)/(r )` or `V - V_(0) = k log_(e) .(r )/(r_(0))` or `V = V_(0) + k log_(e).(r )/(r_(0))` |
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| 89. |
How far from Earth must a body be along a line joining the sun to the earth so that resultage gravitational pull on the body due to Earth and sun is zero ? Distance between sun and the Earth is `1.5 xx 10^(8) km`. Mass of sun `= 3.25 xx 10^(5)` times mass of Earth. |
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Answer» Let `x` be the distance of body of mass `m` from Earth, where the resultant gravitational pull on the body due to Earth and sun is zero. Then `(GM_(E)m)/(x^(2)) = (GM_(S)m)/((r - x)^(2))` or `((r - x)^(2))/(x^(2)) = (M_(S))/(M_(E)) = 3.25 xx 10^(5)` or `(r - x)/(x) = sqrt(3.25 xx 10^(5)) = 570` or `x = (r)/(570 + 1) = (1.5 xx 10^(8))/(570) = 2.63 xx 10^(5) km` |
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| 90. |
Two bodies of masses `100 kg` and `10,000 kg` are at a distance `1 m` apart. At which point on the line joining them will the resultant gravitational field intensity is zero? What is the gravitational potential at that point ? `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. |
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Answer» Let `x` be the distance of the point `P` from `100 kg` body where the resultant gravitational intensity is zero. `:.` Gravitational intensity at `P` due to body of mass `100 kg` is equal and opposite to that due to body of mass `10,000 kg`. Hence, `(G xx 100)/(x^(2)) = (G xx 10,000)/((1 - x)^(2))` or `100 x^(2) = (1 - x)^(2)` or `10 x = 1 - x` or `11 x = 1` or `x = 1//11 m` Gravitational potential at `P` `= - (Gm_(1))/(r_(1)) + (-(Gm_(2))/(r_(2))) = - G[(m_(1))/(r_(1)) + (m_(2))/(r_(2))]` `= - 6.67 xx 10^(-11) [(100)/(1//11) + (10,000)/(10//11)]` `~~ - 8 xx 10^(-7) J//kg` |
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