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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.
| 1. |
Three equal masses of `m kg` each are fixed at the vertices of an equilateral triangle `ABC`. a. What is the force acting on a mass `2m` placed at the centroid `G` of the triangle? b. What is the force if the mass at the vertex `A` is doubled? Take `AG=BG=CG=1m` |
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Answer» (a) Refer to Figure , `/_CBC = 30^(@) = /_CDX` `AD = (2)/(3) AF = (2)/(3) xx a sin 60^(@)` `= (2)/(3) xx a (sqrt(3))/(2) = (a)/(sqrt(3)) = BD = CD` Force acting on mass at `D` due to mass at `A` is `vec(F_(DA)) = (Gm xx 2m)/((a//sqrt(3))^(2)) hatj = (6Gm^(2))/(a^(2))hatj` Force on mass at `D` due to mass at `B` is `vec(F_(DB)) = (Gm xx 2m)/((a//sqrt(3))^(2)) (-hati cos 30^(@) - hatj sin 30^(@))` Force on mass at `D` due to mass at `C` is `vec(F_(DC)) = (Gm xx 2m)/((a//sqrt(3))^(2)) (-hati cos 30^(@) - hatj sin 30^(@))` Resulting force on a mass at `D` is `vec(F_(0)) = vec(F_(DA)) = vec(F_(DB)) = vec(F_(DC))` `= (6 Gm^(2))/(a^(2)) hatj + (6m^(2))/(a^(2)) (-hati cos 30^ (@) - hatj sin 30^(@))` `+ (6m^(2))/(a^(2)) (-hati cos 30^(@) - hatj sin 30^(@)) = 0` (b) If mass at `A` is doubled, then finding `vec(F_(DA)), vec(F_(DB))` and `vec(F_(DC))` in terms of rectagular components along `X`-axis and `Y`-axis, we note that the componet force along `X`-axis cancel out but the component force along `Y`-axis will survive. Then `vec(F_(R)) = (12 Gm^(2))/(a^(2)) hatj - (6Gm^(2))/(a^(2)) hatj = (6Gm^(2))/(a^(2)) hatj` |
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| 2. |
If the distance between the earth and the sun were half its present value, the number of days in a year would have beenA. `64.5`B. `70.24`C. `182.5`D. `730` |
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Answer» Correct Answer - B Here `T_(1) = 365 days, r_(2) = r//3, r_(1) = r` `:. T_(2) = T_(1) ((r_(2))/(r_(1)))^(3//2) = 365 ((r//3)/(r ))^(3//2)` `= 365 xx (1)/(3sqrt(3)) = 70.24 days` |
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| 3. |
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 a correct explanation of the Asseration.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
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Answer» Correct Answer - A Here both assertion and reason are true and reason is correct explanation of assertion. In case of planet orbiting the sun, `tau = 0` as `tau = (dvecL)/(dt), (dvecL)/(dt) = 0` or `L` is constant. |
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| 4. |
A satellite revolves in an orbit close to the surface of a planet of mean density `5.51 xx 10^(3) kg m^(-3)`. Calculate the time period of satellite. Given `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. |
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Answer» `T = (2pi R)/(upsilon_(0)) = (2pi R)/(sqrt(GM//R)) = (2pi R)/(sqrt((G)/(R ) xx (4)/(3) pi R^(2) rho))` `= 2 sqrt((3pi)/(4G rho))` `:. T = 2 sqrt((3 xx 22)/(7 xx 4 xx 6.67 xx 10^(-11) xx 5.5 xx 10^(3)))` `= 5065 s` |
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| 5. |
A particle is projected vertivally upwards from the surface of earth `(radius R_e)` with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is …… |
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Answer» Let `upsilon_(e)` be the escape velocity of Earth. A particle can escape when its total energy `(KE + PE)` becomes zero at infinity. So `(1)/(2) m upsilon_(e)^(2) + (-(GMm)/(R )) = 0` or `(1)/(2) m upsilon_(e)^(2) = (GMm)/(R )` Supplied `KE = (1)/(2) xx (1)/(2) m upsilon_(e)^(2) = (GMm)/(2R)` Let the particle rises to a height `h`, then `(1)/(2) xx (1)/(2) m upsilon_(e)^(2) = (GMm)(R + h)` `(GMm)/(2R) = (GMm)/(R + h)` or `h = R` |
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| 6. |
A body is projected vertically upwards from the surface of the Earth so as to reach a height equal to the radius of the Earth. Neglecting resistance due to it, calculate the initial speed which should be imparted to the body. Mass of Earth `= 5.98 xx 10^(24) kg`, Radius of Earth `= 6400 km`, `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`. |
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Answer» According to law of conservation of mechanical energy, `-(GM m)/(R ) + (1)/(2) m upsilon^(2) = -(GM m)/((R + R)) = - (GM m)/(2R)` `(1)/(2) m upsilon^(2) = (GM m)/(R ) [1 - (1)/(2)] = (GM m)/(2R)` or `upsilon = sqrt((GM)/(R ))` |
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| 7. |
A body is projected vertically upwards from the surface of a planet of radius `R` with a velocity equal to hall the escape velocity for that planet. The maximum height attained by the body isA. `R//2`B. `R//3`C. `R//5`D. `R//4` |
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Answer» Correct Answer - B Using law of conservation of energy, we have `(1)/(2)m((upsilon_(e))/(2))^(2) - (GM m)/(R ) = - (GM m)/((R + h))` or `(1)/(2)m[(1)/(4)((2GM)/(R ))] - (GM m)/(R ) = (-GM)/((R + h))` On solving, we get, `h = (R )/(3)` |
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| 8. |
Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet isA. `-5000 J`B. `-1000 J`C. `-2400 J`D. `5000 J` |
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Answer» Correct Answer - A `upsilon_(e) = sqrt((2GM)/(R )) = 100 ms^(-1)` or `(GM)/(R ) = ((100)^(2))/(2) = 5000` `P.E., U = - (GM)/(R ) = - 5000 J` |
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| 9. |
The ratio of escape velocity at earth `(v_(e))` to the escape velocity at a planet `(v_(y))` whose radius and density are twiceA. `1 : 2`B. `1 : 2sqrt(2)`C. `1 : 4`D. `1 : sqrt(2)` |
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Answer» Correct Answer - B On earth, `g_(e) = (GM)/(R_(e)^(2)) = (G)/(R_(e)^(2)) xx (4)/(3) piR_(e)^(2)rho_(e)` `= (4)/(3) pi GR_(e) rho_(e)` Escape velocity, `upsilon_(e) = sqrt(2g_(e)R_(e)) = R_(e) sqrt((8)/(3) piG rho_(e))` On planet, `upsilon_(p) = R_(p)sqrt((8)/(3) pi G rho_(p))` `:. (upsilon_(e))/(upsilon_(p)) = (R_(e))/(R_(p)) sqrt((rho_(e))/(rho_(e))) = (R_(e))/(2R_(e)) sqrt((rho_(e))/(2rho_(e))) = (1)/(2sqrt(2))` |
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| 10. |
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. `1//2`B. `sqrt(2)`C. `1//sqrt(2)`D. `1//3` |
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Answer» Correct Answer - C As `f upsilon` is the escape velocity of the body from the platform, then using the law of conservation of total energy or `-(GM m)/((R + R)) + (1)/(2)m(f upsilon)^(2) = 0` or `f upsilon = sqrt((GM)/(R )) = sqrt((g R^(2))/(R )) = sqrt(g R)`. The value of escape velocity from the surface of earth `upsilon = sqrt(2g R)` or `sqrt(g R) = upsilon//sqrt(2)` `:. f upsilon = upsilon//sqrt(2)` or `f = 1//sqrt(2)` |
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| 11. |
Statement-1 : Two soild sphere of radius `r` and `2r`, made of same material, are kept in contact. The mutual grvitational force to attraction between them is proportional to `1//r^(4)`. Statement-2 : Gravitational attraction between two point mass bodies varies inversely as the square of the distance between them.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. Let `rho` be the density of the material of each sphere. Then, `M_(1) = (4)/(3) pi r^(3) rho` and `M_(2) = (4)/(3) pi (2r^(3)) rho` Disatnce between their centres `= r + 2r = 3r` Now `F = (GM_(1)M_(2))/((3r)^(2))` `= ((G(4pi)/(3) r^(3) rho)((4)/(3) pi 8 r^(3) rho))/(9r^(2))` This gives `F prop r^(4)` |
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| 12. |
Assertion: Generally the path of projectile form the earth is parabolic but it is elliptical for projection going to a very large height. Reason: The path of projectile is independent of the gravitational force of earth. |
| Answer» Upto ordinary heights, the change in the distance of a projectile from the centre of the Earth is negligible small, as compared to the radius of the Earth. Due to it, the projectile moves under a uniform gravitational force and its path is parabolic, the gravitational force decrease, which is inversely proportinal to the square of the distance of the projectile from the centre of the Earth. Under such a variable force, the path of projectile is elliptical. | |
| 13. |
Is the potential energy of a system of bodies positive or negative ? Give reason in support of your answer. |
| Answer» Since, the gravitational force between the different bodies of a system are attractive in nature, so the potential energy of the system is nagative. | |
| 14. |
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 change in direction, but its magnitude remains constantC. the total mechanical energy of `S` varies periodically with timeD. the linear momentum of `S` remains constant is magnitude |
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Answer» Correct Answer - A When a satellite is moving in an elliptical orbit around the earth, the gravitational force on the satellite by the earth is acting towards the centre of the earth. Therefore, the acceleration of the satellite will also be directed towards the centre of the earth |
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| 15. |
A satellite of mass `m` is in a circular orbit of radius `2R_(E)` about the earth. The energy required to transfer it to a circular orbit of radius `4R_(E)` is (where `M_(E)` and `R_(E)` is the mass and radius of the earth respectively)A. `1.65 xx 10^(9)J`B. `3.13 xx 10^(9)J`C. `6.26 xx 10^(9)J`D. `4.80 xx 10^(9)J` |
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Answer» Correct Answer - B Total energy of satellite in circular orbit of radius `2R` `E_(1) = PE + KE = - (GM m)/(2R) + (1)/(2)m [sqrt((GM)/(2R))]^(2)` `= -(GM m)/(4R)` Total energy of satellite in circular orbit of radius `4R` `E_(2) = -(GM m)/(4R) + (1)/(2)m [sqrt((GM)/(4R))]^(2) = -(GM m)/(8R)` Energy spent `= E_(2) -E_(1) = -(GM m)/(8R) + (GM m)/(4R)` `=(GM m)/(8R)=(gR^(2) m)/(8R) = (mg R)/(8)` `= (400 xx 9.80 xx (6.4 xx 10^(6)))/(8) = 3.13 xx 10^(9)J` |
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| 16. |
What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?A. `(5Gm M)/(6R)`B. `(2Gm M)/(3R)`C. `(Gm M)/(2R)`D. `(Gm M)/(3R)` |
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Answer» Correct Answer - A Let `K` be the `KE` given to the satellite on the surface if earth to take it to the desired location. Following the law of conservation of total energy, we have `-(G mM)/(R ) + K = - (G mM)/((2R +R)) + (1)/(2)m((GM)/(3R))` `= - (GM m)/(3R) + (GM m)/(6R) = - (GM m)/(6R)` or `K = (GM m)/(R ) - (GM m)/(6R) = (5GM m)/(6R)` |
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| 17. |
does speed of satellite remain constant in an orbit ? Explain. |
| Answer» If the orbital path of a satellite is circular, then its speed is constant and if the orbital path of a satellite is elliptical one, then its speed in its orbit is not constant. In that case its areal velocity is constant. | |
| 18. |
Which of the following is true for a satellite in an orbitA. it is a freely falling bodyB. it suffers an accelerationC. it does not require energy for its motion in the orbitD. its speed is constant |
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Answer» Correct Answer - A::C::D A satellite in an orbit is a freely falling body. It does not require any energy for its motion in the orbit and its speed is constant. |
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| 19. |
it possible to put an artifical1 satellite into orbit in such a way that it will always remain directly over New Delhi. |
| Answer» Not possible, because a satellite will appear stationary only when ot revolves in an orbit concentric and coplanar with the equatorial plane, having time period of revolution `24` hours and it should have the sense of revolution from west to east that of Earth. As New Delhi is not in the equatorial plane, hence it will not be possible to put a geostationary satellite over New Delhi. | |
| 20. |
Assertion: The time period of revolution of a satellite close to surface of earth is smaller then that revolving away from surface of earth. Reason: The square of time period of revolution of a satellite is directely proportioanl to cube of its orbital radius.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 a correct explanation of the Asseration.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
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Answer» Correct Answer - A Both the Assertion and the Reason are correct and Reason is correct explanation of Assertion. |
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| 21. |
A satellite of mass `m` is in a circular orbit of radius `r` round the Earth. Calculate its angular momentum with respect to the centre of the orbit in terms of the mass `M` of the Earth and `G`. |
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Answer» Angular momentum of satellite, `L = m upsilon r = m r sqrt(GM//r) = (m^(2) GM r)^(1//2)` |
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| 22. |
A satellite with kinetic energy `E_(k)` is revolving round the earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer space |
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Answer» Let `upsilon_(0), upsilon_(e)` be the orbital and escape speeds of the satellite, then `upsilon_(e) = sqrt(2) upsilon_(0)` Energy in the given orbit, `E_(1) = (1)/(2) m upsilon_(0)^(2) = E` ..(i) Energy for the escape speed, `E_(2) = (1)/(2) m upsilon_(e)^(2) = (1)/(2) m (sqrt(2) upsilon_(0))^(2) = 2E` `:.` Energy required to be supplied `= E_(2) - E_(1) = E` |
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| 23. |
Two objects of masses `m` and `4m` are at rest at an infinite separation. They move towards each other under mutual gravitational attraction. If `G` is the universal gravitaitonal constant, then at separation `r`A. the total energy of the two objects is zeroB. net angular momentum of both the objects is zero about any pointC. the total `K.E.` of the objects is `4Gm^(2)//r`D. their relative velocity of approach is `((8Gm)/(r ))^(1//2)` |
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Answer» Correct Answer - A::B::C Initially both the objects are at rest at an infinite separation, hence their initial `PE` and `KE` are zero. Therefore, following law of conservation of machanical energy, total energy of the two objects at separtion `r` will be zero. There will be two equal and opposite forces acting on two objects. The net torque on two onjects is zero. So, the angular momentum will remain conserved. Initially, both the objects were at rest, therefore, the angular momentum about any point is zero. The reduced mass of the two objects, `mu = (m xx 4m)/(m + 4m) = (4m)/(5)` If `upsilon_(r )` is the relative velocity of approach and `r` is the distance between the two objects, then total `KE` of objects = loss in `PE` of objects `= (Gm xx 4m)/(r ) = (4Gm^(2))/(r )` As total `KE` of objects = loss in `PE` of objects, so `(1)/(2) mu upsilon_(r )^(2) = (4Gm^(2))/(r )` or `(1)/(2) xx (4m)/(5) upsilon_(r )^(2) = (4Gm^(2))/(r )` or `upsilon_(r ) = sqrt((10Gm)/(r ))` |
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| 24. |
A solid sphere of uniform density and radius `R` applies a gravitational force of attraction equal to `F_(1)` on a particle placed at `P`, distance `2R` from the centre `O` of the sphere. A spherical cavity of radius `R//2` is now made in the sphere as shown in figure. The particle with cavity now applies a gravitational force `F_(2)` on same particle placed at `P`. The radio `F_(2)//F_(1)` will be |
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Answer» If `M` and `m` are the masses of solid sphere and particle respectively and `R` is the radius of the sphere, then gravitational force of attraction acting on the particle at `P` is, `F_(1) = GM m//4 R^(2)`. Mass of sphere of radius `R//2 = (4)/(3) pi (R//2)^(3) rho = (M)/(8)` The gravitational force of attraction `(F_(2))` acting on the particle due to sphere with cavity = force due to soild sphere - force due to the sphere creating the cavity assumed to be present at that position alone So, `F_(2) = (GM m)/((2R)^(2)) - (G(M//8)m)/((3R//2)^(2)) = (7)/(36) (GM m)/(R^(2))` Hence, `(F_(2))/(F_(1)) = (7)/(36) (GM m)/(R^(2))((GM m)/(4R^(2))) = (7)/(9)` |
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| 25. |
A mass `M` is broken into two parts of masses `m_(1)` and `m_(2)`. How are `m_(1)` and `m_(2)` related so that force of gravitational attraction between the two parts is maximum? |
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Answer» Let `m_(1) = m`, then `m_(2) = M - m`. Gravitational force of attraction between them when placed distance `r` apart will be `F = (Gm(M - m))/(r^(2))`. Differentiating it w.r.t.`m`, we get `(dF)/(dm) = (G)/(r^(2)) [m(d)/(dm) (m - m) + (M - m) (dm)/(dm)]` `= (G)/(r^(2))[m (-1) + M - m] = (G)/r^(2)(M - 2m)` If `F` is maximum, then `(dF)/(dm) = 0`, `:. (G)/(r^(2)) (m - 2m) = 0` or `M = 2m` or `m = (M)/(2)` |
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| 26. |
A rocket of mass `m` is field vertically from the surface of Mars of mass `M`, radius `R` with a sped `upsilon`. If `20%` of its initial energy is lost due to Martain atmosheric resistance, how far will the rocket go from the surface of Mars before returing to it? Let `G` be the gravational constant. |
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Answer» Total energy of rocket is `E = KE + PE = (1)/(2) m upsilon^(2) + (-(Gmm)/(R ))` Since `20%` energyis lost, hence energy remained with rocket `= 80%` of `E` `= (80)/(100) E = (3)/(5) E = (4)/(5) [(1)/(2) m upsilon^(2) - (GMm)/(R )]` Let `h` be the heighest distance of rocket from the surfcace of Mars upto which rocket can go. At this location the total energy will its potential energy. Hence `(4)/(5) [(1)/(2) m upsilon^(2) - (GMm)/(R )] = - (GMm)/((R + h))` or `(2)/(5) [upsilon^(2) - (GMm)/(R )] = - (GMm)/((R + h))` or `(2)/(5) [(upsilon^(2) R - 2 GM)/(R )] = - (GM)/(R + h)` or `(R + h) = (5 R GM)/(2 (2 GM - upsilon^(2) R))` or `h = (5R GM)/(2(2 GM - upsilon^(2)R)) - R` |
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| 27. |
Assertion : The gravitational attraction of moon is much less than that of earth. Reason : Gravitational force of a given mass `(M)` depends upon `M//r^(2)`, which is smaller for moon.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 a correct explanation of the Asseration.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
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Answer» Correct Answer - A Both the Assertion and the Reason are correct and Reason is correct explanation of Assertion. |
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| 28. |
A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`. Answer the following questions: Time period of revollution of satellite around the earth isA. `3550 s`B. `7100 s`C. `5330 s`D. `8880 s` |
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Answer» Correct Answer - B Time period of revolution of satellite, `T = (2pi(R + h))/(upsilon_(0))` `= (2 xx (22//7) xx (6.4 xx 10^(6) + 1.6 xx 10^(6)))/(7.1 xx 10^(3))` `= 7082 ~~ 7100 s` |
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| 29. |
A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`. Answer the following questions: The value of `h` isA. `1.5 xx 10^(5) m`B. `3.2 xx 10^(5)m`C. `3.2 xx 10^(6)m`D. `1.6 xx 10^(6)m` |
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Answer» Correct Answer - D According to law of conservation of total mechanical energy, total energy of rocket at the surface of earth = total energy of rocket at the highest point or `(1)/(2) m upsilon^(2) + ((-GMm)/(R )) = 0 + ((-GMm)/(R + h))` or `(upsilon^(2))/(2) = (GM)/(R ) - (GM)/((R + h)) = (gR^(2))/(R ) - (gR^(2))/((R + h))` `= gR (1-(R )/(R + h)) = gR ((h)/(R + h))` or `upsilon^(2) (R + h) = 2g Rh` or `R upsilon^(2) = 2g R h - upsilon^(2)h` `= (2g R - upsilon^(2))h` or `h = (R upsilon^(2))/((2gR - upsilon^(2)))` `= (6.4 xx 10^(6) xx (5 xx 10^(3))^(2))/((2 xx 9.8 xx 6.4 xx 10^(6)) - (5 xx 10^(3))^(2))` `= 1.6 xx 10^(6)m` |
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| 30. |
A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`. Answer the following questions: The value of `h` is |
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Answer» Let the rocket be fired with velocity `upsilon` from the surface of Earth and it reaches a height `h` from the surface of Earth where its velocity becomes zero. Total energy of rocket at the surface of Earth `= K.E. + P.E. = (1)/(2) m upsilon^(2) + ((-GM m)/(R ))` At the highest point, `upsilon = 0, K.E. = 0` and `P.E. = - (GM m)/((R + h))` Total energy `= K.E. + P.E. = 0 + ((-GM m)/(R + h)) = - (GM m)/(R + h)`. According to law of caonservation of energy. `(1)/(2)m upsilon^(2) - (GM m)/(R ) = - (GM m)/((R + h))` or `(1)/(2) upsilon^(2) = (GM)/(R ) - (GM)/((R + h)) = (g R^(2))/(R ) - (gR^(2))/((R + h)) = gR (1-(R)/(R + h)) = gR ((h)/(R + h))` or `upsilon^(2) (R + h) = 2 g Rh` or `R upsilon^(2) = 2 gRh - upsilon^(2) h = (2gR - upsilon^(2))h` or `h = (R upsilon^(2))/(2g R - upsilon^(2)) = ((6.4 xx 10^(6) )xx (5 xx 10^(3))^(2))/(2 xx 9.8 xx(6.4 xx 10^(6)) - (5 xx 10^(3))^(2)) = 1.6 xx 10^(6) m` |
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| 31. |
What are the conditions under which a rocket, fired from the earth, launches an artificial satellite of the earth? |
| Answer» Following are the basic conditions, (i) The rocket must take the satellite to a suitable height above surface of Earth. (ii) From the desired height, the satellite must be projected with a suitable speed, called the orbital speed. (iii) In the orbital path of satellite, the air resistance should be negligible so that its speed does not decrease and it does not burn due to the heat produced. | |
| 32. |
Why is the weight of an object on the Moon `(1//6)`th its weight on the Earth?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 a correct explanation of the Asseration.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
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Answer» Correct Answer - A Here, both assertion and reason are true and reason is correct explanation of assertion. Acceleration due to gravity on the moon is `g//6` and there is no resisatnce to the motion of ballon due to absence of atmosphere on the moon. |
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| 33. |
(1) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts, (2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius, (3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be cosidered to be concentrated at its C.G.., (4) The radius of gyration of any body rotating about ab axis is the length of the perpendicular dropped from thr C.G. the body to the axis. which one of the following paries of statements is correct ?A. `(4) and (1)`B. `(1) and (2)`C. `(2) and (3)`D. `(3) and (4)` |
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Answer» Correct Answer - A The statement (1) and (4) are true. Thus option (a) is correct. |
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| 34. |
What is the tota, energy of a satellite revolving around Earth ? |
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Answer» The satellite revolving around the Earth has two types of erergies. (i) Gravitational potential energy `U` due to position of the satellite in the orbit. where, `U = - GMm//r` (ii) Kinetic energy `K` due to orbital speed of satellite. Where, `K = (1)/(2) m upsilon^(2) = (1)/(2) m ((GM)/(r )) = GM (m)/(2r)` Total energy `= U = K = (-GM m)/(r ) + (GM m)/(2r)` `= (-GMm)/(2r)` |
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| 35. |
Assertion: A person sitting in an artificial satellite revolving around the earth feels weightless. Reason: There is no gravitational force on the satellite. |
| Answer» No, because both the body and the arficial satellite area in a sate of free fall and body inside the satellite is in a weightlessness state. | |
| 36. |
A rocket is fired the Earth towards the Moon. At what distance from the Moon is the gravitational force on the rocket is zero. Mass of Earth is `6 xx 10^(24) kg` , mass of moon is ` 7.4xx 10^(22) kg` and distance between moon and earth is `3.8xx 10^(8)m`. Neglect the effect of the sun and other planes. |
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Answer» Let `x` be the distance between rocket and Moon, when the net gravitational force on roccket is zero, i.e., gravitational force on rocket due to Earth = gravitational force on rocket due to Moon. Then, `(GM_(e)m)/((r - x)^(2)) = (GM_(m)m)/(x^(2))` `(r - x)/(x) = sqrt((M_(e))/(M_(m))) = sqrt((6 xx 10^(24))/(7.4 xx 10^(22))) = 9.0` Using `r = 3.8 xx 10^(8) m` and solving the above relation, we get `x = 3.8 xx 10^(7) m` |
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| 37. |
What are the two factors which determine why some bodies in solar system have atmosphere and others do not? |
| Answer» the ability of a body (planet) to hold the atmosphere depends on : (i) acceleration due to gravity and (ii) surface temperature. | |
| 38. |
Why does a body lose weight at the centre of the earth? |
| Answer» The weight of the body is the force with which the body is attracted by the Earth towards its centre. Quantitatively, the weight of body of mass `m` is equal to `mg`, where` g` is the acceleration due to gravity. At the centre of Earth, `g = 0`, so weight of body is zero at the centre of Earth. | |
| 39. |
Should the speed of two artificial satellites of the earth having the different masses but the same orbital radius be the same? |
| Answer» Yes, it is so because the orbital speed of a satllite is independent of the mass of a satellite. Therefore, the speeds of the arficial satellies of different masses but of the same orbital radius will be the same. | |
| 40. |
A point mass `m` is a distance `x` from the centre of mass `M` and radius `R` on its axis. Find the gravitational force between the two. What will this force be if `x gt gt R` and `x lt lt R`? For what value of `x` is the force maximum? |
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Answer» In case of sperical shell, the gravitational intensity at a point inside the shell, `I = 0`. So force on particle of mass `m` inside the shell, `F = mI = 0` |
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| 41. |
A satellite revolving around Earth loses height. How will its time period be changed? |
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Answer» Time period of satellite is given by , `T = 2 pi sqrt(((R + h)^(3))/(GM))`. Therefore, `T` will decrease, when `h` decreases. |
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| 42. |
Determine the escape speed of Moon. Given, the radius of Moon is `1.74 xx 10^(6) m`, its mass is `7.36 xx 10^(22) kg`. Does your answer throw light on why the moon has no atmosphere? `G = 6.67 xx 10^(-11) n m^(2) kg^(-2)`. |
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Answer» Here, `R = 1.74 xx 10^(6) m`, `m = 7.36 xx 10^(22) kg`, , `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2), upsilon_(e) = ?` `upsilon_(e) = sqrt((2GM)/(R )) = sqrt((2 xx 6.67 xx 10^(-11) xx 7.36 xx 10^(22))/(1.74 xx 10^(6)))` `= 2375 ms^(-1) = 2.38 km s^(-1)` Since, the average thermal speed of the gas molecules like oxygen, hydrogen etc. on the surface of Moon is greater than the escape speed of moon `(= 2.38 km s^(-1))`, hence these gases have escaped from the surface of Moon and Moon has no atmosphere. |
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| 43. |
According to kepler, the line joining a planet to the sun sweeps out……..in equal intervals of time. |
| Answer» Correct Answer - equal areas | |
| 44. |
Give one example each of central and non-central force. |
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Answer» Examples of central force. Gravitationaal force of a point mass and electric force due to a point change. Examples of non-central force. Nuclear forces which hold the nucleons together in the nucleus and magnetic forces between two current carrying loops. |
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| 45. |
The linear speed of a planet around the sun is not constant in its orbit. Comment. |
| Answer» The areal speed of the planet around the sun is constant but not the linear speed of the planet is large while passing close to the sun and is small while passing away from the sun. | |
| 46. |
A planet of radius `R=(1)/(10)xx(radius of Earth)` has the same mass density as Earth. Scientists dig a well of depth`(R )/(5)` on it and lower a wire of the same length and a linear mass density `10^(-3) kg m(_1)` into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it inplace is (take the radius of Earth`=6xx10^6m` and the acceleration due to gravity on Earth is `10ms^(-2)`A. `96 N`B. `108 N`C. `120 N`D. `150 N` |
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Answer» Correct Answer - B Let `R_(p)` be the radius of the planet. Then `R_(p) = (R_(e))/(10) = (6 xx 10^(6))/(10) = 6 xx 10^(5) m` As `g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho = (4)/(3) pi G rho R` So `g prop R` `:. (g_(P))/(g_(e)) = (R_(P))/(R_(e)) = (1)/(10)` or `g_(P) = (g_(e))/(10) = (10)/(10) = 1 ms^(-2)` Acceleration due to gravity at depth `x` on a planet is `g_(x) = (1-(x)/(R )) g_(P)` Let `lambda` be the mass per unit lenght of wire. Given, `lambda = 10^(-3) kg m^(-1)`. Force on a small segment of wire of lenght `dx` at a depth `x`from surface is `dF = (lambda dx)g_(x) = lambda dx xx (1- (x)/(R )) g_(P)` `= lambda (1- (x)/(R )) g_(P) dx` Total force on wire is `:. F = underset(0) overset(R//5) int lambda (1- (x)/(R )) g_(P) dx = lambda [x - (x^(2))/(2R)]_(0)^(R//5) g_(P)` `= lambda [(R)/(5) - (R )/(50)] g_(P) = 10^(-3) [(9R)/(50)] g_(P)` `= 10^(-3) xx (9)/(50) xx 6 xx 10^(5) xx 1= 108 N` |
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| 47. |
The escape velocity of a body form the earth depends on (i) the mass of the body. (ii) the location from where it is projected. (iii) the direction of projection. (iv) the height of the location form where the body is launched. |
| Answer» The escape velocity is independent of the mass of body and the direction of projection. It depends upon the gravitational potentail at a point from where the body is launched. Since, this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors. | |
| 48. |
Assertion: if an earth satellite moves to a lower orbit, there is some dissipation of energy but the satellite speed increases. Reason: The speed of satellite is a constant quantity. |
| Answer» The orbiting satellite loses kinetic energy due to atmospheric friction. Therefore, in a particular orbit, the gravitational attraction on the satellite becomes greater than the force required to keep the satellite in than the force required to keep the satellite in the orbit. Due to it, the satellite moves down towards the Earth into a lower orbit. In the lower orbit the potential energy decrease (become more negative) so that the K.E. correcspondingly increases. that is why, the satellite describes a smaller orbit with increased speed. Infact, due to atmospheric frinction, the satellite spirals down towards the Earth with increasing speed and ultimately burns out in the lower denser atmosphere. | |
| 49. |
The change in the gravitational potential energy when a body of a mass `m` is raised to a height `nR` above the surface of the earth is (here `R` is the radius of the earth) |
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Answer» Change in `P.E. = P.E.` on the height `n R` `- P.E.` on the surface of Earth `= - (GMm)/((R + nR)) - (GMm)/(R )` `= ((n)/(n + 1)) (GM m)/(R ) = ((n)/(n + 1)) mg R` |
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| 50. |
The change in the gravitational potential energy when a body of a mass `m` is raised to a height `nR` above the surface of the earth is (here `R` is the radius of the earth)A. `m g R((n)/(n - 1))`B. `n m g R`C. `mgR((n^(2))/(n^(2) + 1))`D. `mgR((n)/(n + 1))` |
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Answer» Correct Answer - D Change in potential energy, `DeltaE = GMn((1)/(R ) - (1)/(R + nR)) = (GM m)/(R(R + nR)) xx nR` `= (GMm)/(R ) ((n)/(1 + n)) = gRm ((n)/(1 + n))` |
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