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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.
| 1651. |
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions `A,B` and `C` are `K_(A),K_(B)` and `K_(C)` respectively. AC is the major axis and `SB` is perpendicular to `AC` at the position of the sun as shown in the figure. Then A. `K_(B) gt K_(A) gt K_(C)`B. `K_(A) lt K_(B) lt K_(C)`C. `K_(B) lt K_(A) lt K_(C)`D. `K_(A) gt K_(B) gt K_(C)` |
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Answer» Correct Answer - D When a planet orbit about the sum, the angular momentum of the planet about the sun remains conserved. `mvr` =constant or `v.r`=constant it means `v prop1/r` At position `A`, the distance of the planet from the sun is least and at position `C`, it is largest Hence speed of the planet is greatest at `A` and least at `C` hence `K_(A)gtK_(B)gtK_(C)` |
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| 1652. |
A planet moves aruond the sun in an elliptical orbit such that its kinetic energy is `K_(1)` and `K_(2)` when it is nearest to the sun and farthest from the sun respectively. The smallest distance and the largest distance between the planet and the sun are `r_(1)` and `r_(2)` respectively.A. If total energy of the planet is `E` then `(r_(2))/(r_(1))=(E-K_(1))/(K_(2)-E)`B. If the total energy of the planet is `E`, then `(r_(2))/(r_(1))=(E-K_(2))/(E-K_(1))`C. If `r_(2)=2r_(1)`, the total energy of the planet energy of the planet in terms of `K_(1)` and `K_(2)` is `(2K_(1)-K_(2))`D. If `r_(2)=2r_(1)`, the total energy of the plenet energy of the planet in terms of `K_(1)` and `K_(2)` is `(2K_(2)-K_(1))` |
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Answer» Correct Answer - D `Fpropr^(-5/2)` `rArr F=(GMm)/(r^(5//2))` `rArr (mV_(0)^(2))/r=(GMm)/(r^(5//2))` `=V_(0)^(2)=(GM)/(r^(5//2))xxr` `rArr V_(0)^(2)=(GM)/(r^(3//2))` `:. V_(0)=(sqrt(GM))/(r^(3//4))` take logrithm on both sides `logV_(0)=lgosqrt(GM)-logr^(3//4)` `log_(e)V_(0)=-3/4log_(e)r+log_(e)sqrt(GM)` `y=mx+c` slope of `log_(e)V_(0)` versus `log_(e)r` is `m=-3/4` |
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| 1653. |
If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube law- (a) 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 approximately parabolic. (d) there will be no gravitational force inside a spherical shell of uniform density. |
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Answer» (a) planets will not have elliptic orbits. (c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic. |
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| 1654. |
If the sun and the planets carried huge amounts of opposite charges, (a) all three of Kepler’s laws would still be valid. (b) only the third law will be valid. (c) the second law will not change. (d) the first law will still be valid. |
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Answer» (c) the second law will not change. (d) the first law will still be valid. |
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| 1655. |
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 the happens for our earth.A. nothing will changeB. we will becomes hotter after billions of years.C. we will be going around but not strictly in losed orbitsD. after sufficiently long time we will leave the solar system |
| Answer» Correct Answer - C::D | |
| 1656. |
If the law of gravitational, instead of being inverse-square law, becomes an inverse-cube lawA. planets will not have elliptical orbitsB. circular orbits of planets is not possibleC. projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic.D. There will be gravitational force inside a spherical shell of uniform density. |
| Answer» Correct Answer - A::C | |
| 1657. |
Choose the correct statement. In planetary motionA. the speed along the orbit remains constantB. the angular speed remains constantC. the total angular momentum remains constantD. the radius of the orbit remains constant |
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Answer» Correct Answer - c The orbits of planets are elliptical. The speed of the planet and its angular speed (or angular frequency) keep changing. Since no net torque acts on the planet, its angular momentum remains constant. |
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| 1658. |
A satellite is orbiting the earth in a circular orbit of radius `r`. ItsA. Knetic enegy varies as rB. angular momentum varies as `(1)/(sqrt(r ))`C. linear momentum varies as `(1)/( r)`D. frequency of revolution varies as `(1)/(r^(3//2))` |
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Answer» Correct Answer - D `K prop (1)/( r)` Hence, option (a) is incorrect. Angular momentum `L=mvr=msqrt((GM_(E))/( r))r=msqrt(GM_(E)r) :. L prop sqrt(r )` Hence, option ( c) is incorrect. Frequency of revolution, `v=(1)/(T)=(1)/(2pi)sqrt((GM_(E))/(r^(3))) :. v prop (1)/(r^(3//2)` |
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| 1659. |
Two satellites A and B move round the eaerth in the same orbit. The mass of B is twice the mass of A.A. Speeds of A and B are equalB. The potential energy of earth +A is same is that of earth +BC. The kinetic energy of A and B are equalD. The total energy of earth +A is same as that of earth +B |
| Answer» Correct Answer - A | |
| 1660. |
The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statements about the curves `A,B` and `C` A. A shows the kinetic energy, `B` the total energy and `C` the potential energy of the system.B. `C` shows the total energy, `B` the kinetic energy and `A` the potential energy of the system.C. `C` and `A` are kinetic and potential energies and `C` is the total energy of the system.D. A shows the kinetic energy, `B` the potential energy and `C` the total energy of the system. |
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Answer» Correct Answer - d `T.E.=P.E. +K.E.impliesP.E.(-)` `K.E.(+)impliesT.E.(-)` |
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| 1661. |
The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statements about the curves `A,B` and `C` A. A shows the kinetic energy `B` the total energy and `C` the potential energy of the systemB. C shows the total energy B the kinetic energy and A the potential energy of the systemC. C and A are kinetic and potential energies respectively and B is total energy of the systemD. A and B are the kinetic and potential energies respectively and C is the total energy of the system. |
| Answer» Correct Answer - D | |
| 1662. |
The figure shows the variation of energy with the orbit radius of a body in circular planetary motion. Find the correct statements about the curves `A,B` and `C` A. C shows the total energy, B the kinetic energy and A the potential energy of the satelliteB. A shows the kinetic energy, B the total energy and C the potential energy of the satelliteC. A and B are the kinetic and potential energies and C the total energy of the satelliteD. C and A are the kinetic potential energies respectively and B the total energy of the satellite |
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Answer» Correct Answer - C |
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| 1663. |
A planet is moving in an elliptic orbit. If `T,V,E` and `L` stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the centre of force, thenA. `T` is conservedB. `V` is always positiveC. `E` is always negativeD. magnitude of `L` is conserved but its direction changes continuously |
| Answer» Correct Answer - C | |
| 1664. |
The first Indian satellite wasA. AppleB. BhaskaraC. RohiniD. Aryabhatta |
| Answer» Correct Answer - d | |
| 1665. |
The escape velocity for a planet is `v_(e)`. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will beA. `v_(e )`B. `(v_(e ))/(sqrt(2))`C. `(v_(e ))/(2)`D. zero |
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Answer» Correct Answer - B From energy conservation Potential inside sphere `V=-(GM)/(2R^(3))(3R^(2)-r^(2))` `V_(s)=-(GM)/(R ), V_(0)=-(3GM)/(2R)` Loss of potential energy of partical = Gain of kinetic energy `m(V_(s)-V_(0))=(1)/(2)mv^(2)` `v^(2)=2[-(GM)/(R )(-(3GM)/(R ))]rArr v= sqrt((GM)/(R )) rArr v = sqrt((2GM)/(2R)) rArr v = (V_(e ))/(sqrt(2))` |
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| 1666. |
Statement-1: Escape velocity from surface of a particular planet is `V`. If a tunnel is made inside the surface, the escape velocity from a point inside the tunnel must be greater than `V`. Statement-2: Gravitational force is a conservative force.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. Statement-1 is true, Statement-2 is true, statement-2 is Not a correct explanation for statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false Statement-2 is True. |
| Answer» Correct Answer - b | |
| 1667. |
The escape velocity for a planet is `v_(e)`. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will beA. `v_(e)`B. `(v_(e))/(sqrt(2))`C. `(v_(e))/2`D. zero |
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Answer» Correct Answer - b From energy conservation Potential inside sphere `V=-(GM)/(2R^(3))(3R^(2)-r^(2))` `V_(S)=-(GM)/R,V_(0)=-(3GM)/(2R)` Loss of potential energy of particle =Gain of kinetic energy `m(V_(S)-V_(0))=1/2mv^(2)` `v^(2)=2[-(GM)/R-(-(3GM)/(2R))]impliesv=sqrt((GM)/R)` `v=sqrt((2GM)/R)impliesv=(V_(e))/(sqrt(2))` |
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| 1668. |
The escape velocity of a body from the earth is `11.2 km//s`. If a body is projected with a velocity twice its escape velocity, then the velocity of the body at infinity is (in `km//s`)A. `11.2 " kms"^(-1)`B. `22.4 " kms"^(-1)`C. `19.4 " kms"^(-1)`D. `15.2 " kms"^(-1)` |
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Answer» Correct Answer - C We know that escape velocity, `v_(e)=sqrt((2GM_(e))/(R_(e)))`. Substituting the values, we get `v_(e) = 112 " kms"^(-1)`. But if the projected velocity is double then the speed become `sqrt(3)` times and having the value `112 xx sqrt(3)=19.4 " kms"^(-1)`. |
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| 1669. |
There is a concentric hole of radius R in a solid sphere of radius 2R. Mass of the remaining portional is M. What is the gravitational potential at centre?A. `-(5G)/(7R)`B. `-(5G)/(14R)`C. `-(3G)/(7R)`D. `-(9G)/(14R)` |
| Answer» Correct Answer - D | |
| 1670. |
Four similar particles of mass m are orbiting in a circle of radius r in the same direction and same speed because of their mutual gravitational attractive force as shown in the figure . Speed of a particle is given by A. `sqrt((Gm)/(4r) ( 1+ 2sqrt2))`B. `sqrt((Gm)/(r) ( 1+ sqrt2))`C. `sqrt((Gm)/(r) ( 1+ 2sqrt2))`D. `sqrt((Gm)/(4r) ( 2 + sqrt2))` |
| Answer» Correct Answer - A | |
| 1671. |
The K.E. of a satellite moving in a circular orbit around a planet is `1.5xx 10^(10)J`. What is its potential energy?A. `0.75xx10^(10)`JB. `-3xx10^(10)`JC. `3xx10^(10)J`D. `6xx10^(9)`J |
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Answer» Correct Answer - B `P.E.=-2(K.E.)=-3xx10^(10)J` |
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| 1672. |
The radius of the earth is reduced by `1%` with mass remaining the same. The escape velocity form the earthA. increases by `0.5%`B. decreases by `11%`C. remains the sameD. decreases by `0.5%` |
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Answer» Correct Answer - a `(Deltav_(e))/(v_(e))=-(1)/(2)(DeltaR)/(R)` `=-(1)/(2)xx1%=-0.5" increases"` `(Deltav_(e))/(v_(e))=-0.5%" increase."` |
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| 1673. |
If M be the mass of the earth, R its radius (assumed spherical) and G gravitational constant, then the amount of work that must be done on a body of mass m, so that it completely escapes from the gravity of the earth of the earth is given byA. `(GmM)/(R)`B. `(GmM)/(2R)`C. `(3GmM)/(2R)`D. `(3GmM)/(4R)` |
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Answer» Correct Answer - a `"P.E. "=("GMm")/(R)` |
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| 1674. |
The escape velocity of a body from the earth is `11.2 km//s`. If a body is projected with a velocity twice its escape velocity, then the velocity of the body at infinity is (in `km//s`)A. `19.4`B. `194`C. `1.94`D. `0.194` |
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Answer» Correct Answer - A Velocity at infinity `v_(oo)=sqrt(v^(2)-v_(e)^(2))=sqrt([2(11.2)]^(2)-(11.2)^(2))` |
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| 1675. |
There is no atomosphere on moon becauseA. it is closer to the earthB. is revolves round the earthC. it gets light from the sunD. the escape velocity of gas molecules is lesser than their root mean square velocity. |
| Answer» Correct Answer - d | |
| 1676. |
Given mass of the moon is 1/81 of the mass of the earth and corresponding radius is 1/4 of the earth. If escape velocity on the earth surface is 11.2 km / s , the value of same on the surface of the moon isA. `1.25km//s`B. `2.5m//s`C. `5m//s`D. `10m//s` |
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Answer» Correct Answer - B `v_(e)=sqrt((2GM)/(R )),v_("e(moon)")=sqrt((2GM xx 4)/(81xx R))` `= sqrt((4)/(81)((2GM)/(R )))=(2)/(9)xx v_(e )=2.5 m//s` |
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| 1677. |
An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth of radius R. The height of the satellite above the surface of the earth isA. RB. R/2C. 3 RD. 6 R |
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Answer» Correct Answer - a `v_(c)=(1)/(2)v_(e)` `sqrt((GM)/(R+h))=(1)/(2)sqrt((2GM)/(R))` `(GM)/(R+h)=(1)/(4)xx(2GM)/(R)` `R+h=2R` `h=2R-R=R.` |
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| 1678. |
A body is projected vertically upwards from the surface of the earth with a velocity equal to half of escape velocity of the earth. If `R` is radius of the earth, maximum height attained by the body from the surface of the earth isA. `(R)/(6)`B. `(R)/(3)`C. `(2R)/(3)`D. R |
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Answer» Correct Answer - b `(h)/(R)=(u^(2))/(v_(e)^(2)-u^(2))=((1)/(4)v_(e)^(2))/(v_(e)^(2)-(1)/(4)v_(e)^(2))=((1)/(4)v_(e)^(2))/((3)/(4)v_(e)^(2))=(1)/(3)` `h=(R)/(3)` |
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| 1679. |
A body is projected vertically upwards from the surface of the earth. If its kinetic energy is equal to half of its minimum value required to escape from the gravitational influece, then the height upto which it rises isA. 4RB. 3RC. 2RD. R |
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Answer» Correct Answer - D Given K.E. `=(1)/(2)` (K.E.) escape `=(1)/(2)((1)/(2)mv_(e)^(2))=(1)/(4)xx m((2GM)/(R ))=(GMm)/(2R)` For upward motion, loss in K.E. = gain in P.E. `therefore (GMm)/(2R)=-((GMm)/(R+h))-(-(GMm)/(R ))` `therefore (1)/(2R)=(1)//(R )-(1)/(R+h)` `therefore (1)/(R+h)=(1)/(2R)" " therefore h = R` |
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| 1680. |
A body is projected vertically upwards from the surface of the earth with a velocity equal to half of escape velocity of the earth. If `R` is radius of the earth, maximum height attained by the body from the surface of the earth isA. `(R )/(6)`B. `(R )/(3)`C. `(2R)/(3)`D. R |
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Answer» Correct Answer - B `1/2m(v_(e))/(2)^(2)=(mgh)/(R ) rarr 1/2m (2gR)/(4)=(mgR)/(R )` `rarr 1/4=(1)/(1+(R )/(h)rarr 1+(R )/(h)=4 rarr(R )/(h)=3` `therefore h=(R )/(3)` |
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| 1681. |
A point `P` lies on the axis of a fixed ring of mass `M` and radius `a`, at a distance `a` from its centre `C`. A small particle starts from `P` and reaches `C` under gravitational attraction only. Its speed at `C` will be. |
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Answer» Correct Answer - `sqrt((2GM)/(a)(1-(1)/(sqrt(2))))` Potential at the axis of ring `V_(p) = -(GM)/(sqrt(R^(2)+x^(2)))` `V_(p)=-(GM)/(sqrt(a^(2)+a^(2))) rArr V_(p) = -(GM)/(sqrt(2)a) rArr V_(0) = -(GM)/(a)` From energy conservation Loss of potential of particle = Gain of kinetic energy. `U_(i)-U_(f)=K_(f)-K_(i)` `-(GMm)/(sqrt(2)a)-(-(GMm)/(a))-(1)/(2)mv^(2) rArr (GM)/(a) (1-(1)/(sqrt(2)))=(v^(2))/(2)` `v^(2) = (2GM)/(a) (1-(1)/(sqrt(2))) rArr v= sqrt((2GM)/(a)(1-(1)/(sqrt(2))))` |
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| 1682. |
Find the centre of gravity of a uniform vertical rod height = R/3, where R is the radius of Earth. The lower end of the rod is touching the surface of the Earth. |
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Answer» Correct Answer - `y=(int_(0)^((R )/(3))(yg_(0)R^(2)lambda dy)/((y+R)^(2)))/(int_(0)^((R )/(3))(g_(0)R^(2)lambda dy)/((y+R)^(2)))=4R ln((4)/(3))-R` from surface `y=(int_(0)^((R )/(3))(yg_(0)R^(2)lambda dy)/((y+R)^(2)))/(int_(0)^((R )/(3))(g_(0)R^(2)lambda dy)/((y+R)^(2)))=4R ln((4)/(3))-R` from surface |
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| 1683. |
Find the binding energy of a satellite of mass `m` in orbit of radius `r`, (R = radius of earth, g = acceleration due to gravity) |
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Answer» The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite. It is equal to negative of total mechanical energy of satellite in its orbit. Thus, binding energy `= -E = (GMm)/(2r)` but, `g = (GM)/(R^(2)) rArr GM = gR^(2)` `:.` Binding energy `= |E|=(gmR^(2))/(2r)`. |
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| 1684. |
Statemet-1: Two satellites A & B are in the same orbit around the earth B being behind A. B cannot overtake A by increasing its speed. Statement-2: it will then go into a different orbit.A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
| Answer» Correct Answer - A | |
| 1685. |
At which depth from Earth surface, acceleration due to gravity is decreased by 1 % ? |
| Answer» `(Delta g_(d))/g=d/R_(e) implies 1/100=d/6400 :. d=64` km. | |
| 1686. |
Two solid sphere of same size of a metal are placed in contact by touching each other prove that the gravitational force acting between then is directly proportional to the fourth power of their radius. |
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Answer» The weight of the spheres may be assumed to be concentrated at their centres. So `F=(G[4/3 pi R^(3) rho]xx[4/3 pi R^(3) rho])/((2R)^(2))=4/9 (Gpi^(2) rho^(2))R^(4)` `:. F prop R^(4)` |
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| 1687. |
Two solid sphere of same size of a metal are placed in contact by touching each other prove that the gravitational force acting between then is directly proportional to the fourth power of their radius. |
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Answer» The weights of the spheres may be assumed to be concentrated at their centres. So, `F=(G[(4)/(3)piR^(3)rho])xx([(4)/(3)piR^(3)rho])/((2R)^(2))=(4)/(9)[Gpi^(2)rho^(2))R^(4)thereforeFpropR^(4)` |
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| 1688. |
The intensity of gravitational field at a point situated at a distance of 8000km form the centre of the earth is `6N//kg`. The gravitational potential at that point is -(in joule /kg)A. 6B. `4.8xx10^(7)`C. `8xx10^(5)`D. `4.8xx10^(2)` |
| Answer» Correct Answer - B | |
| 1689. |
Find the height from the surface of earth at which weight of a body of mass m reduced to 36% of its weight on the surface. (Re = 6400 km.) |
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Answer» h = ?, R = 6400 km g' = g\(\big(1-\frac{2h}{R}\big)\)= g - \(\frac{2gh}{R}\) or g - g' = \(\frac{2gh}{R}\) Percentage decrease in weight = \(\frac{mg-mg'}{mg}\) x 100 = \(\frac{g-g'}{g}\) x 100 \(\frac{g-g'}{g}\) x 100 = \(\frac{2gh}{gR}\) x 100 = \(\frac{2h}{R}\times100\) 36 = \(\frac{2\times h}{6400}\times100\) or h = 1.152 km |
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| 1690. |
If R is radius of the earth, the height above the surface of the earth where the weight of a body is `36%` less than its weight on the surface of the earth isA. 4 R/5B. R/5C. R/6D. R/4 |
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Answer» Correct Answer - d `mgg=mg((R)/(R+h))^(2) therefore (R)/(R+h)=(8)/(10)` `10R=8R+8h" "therefore h=(R)/(4)` |
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| 1691. |
The acceleration due to gravity at the poles is `10ms^(-2)` and equitorial radius is `6400 km` for the earth. Then the angular velocity of rotaiton of the earth about its axis so that the weight of a body at the equator reduces to `75%` isA. `(1)/(1600)"rad/s"`B. `(1)/(800)"rad/s"`C. `(1)/(400)"rad/s"`D. `(1)/(200)"rad/s"` |
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Answer» Correct Answer - a `g_(phi)=g-Romega^(2)cos^(2)phi` `7.5=10-6.4xx10^(6)xx omega^(2)xx"1 (at equator)"` `6.4xx10^(6)omega^(2)=2.5` `omega^(2)=sqrt((25)/(64xx10^(6)))=(5)/(8)xx10^(-3)` `=8.25xx10^(-4)=(1)/(1600)"rad/s."` |
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| 1692. |
The value of acceleration due to gravity at the height h from the ground is______________.A. `g=(GM)/(R+h)`B. `g=(GM)/(sqrt(R+h))`C. `g=(GM)/((R+h)^(2))`D. `g=GM(R+h)^(2)` |
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Answer» Correct Answer - A `g=(GM)/((R+h)^(2))` |
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| 1693. |
The value of acceleration due to gravity at poles is ________________ `m//s^(2)`A. 9.72m/sB. `9.83m//s^(2)`C. `9.83 m//s`D. `9.72m//s^(2)` |
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Answer» Correct Answer - B `9.83m//s^(2)` |
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| 1694. |
The value of g at the centre of the Earth is zero. Explain?A. Centre of EarthB. PolesC. Infinite distanceD. Both a and b |
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Answer» Correct Answer - D Both a and c |
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| 1695. |
In above problem find the gravitational potential at a point whose co-ordinates are `(5,4)` in (J//kg) .A. `-180`B. `180`C. `-90`D. zero |
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Answer» Correct Answer - A `V = 20 xx 5 + 20 xx 4 = 180J//kg` . |
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| 1696. |
In the above problem find the work done is shifting of mass `1kg` from origin (0,0) to a point (5 ,4) (In J) .A. `-180`B. `180`C. `-90`D. zero |
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Answer» Correct Answer - B `W = m (V_(f) -V_(i)) = 1 (180 -0) = 180 J` . |
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| 1697. |
`v = 2x^(2) + 3y^(2) + zx` Find gravitational at a point (x ,y ,z) . |
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Answer» `vecg=-((delV_g)/(delx)hati+(delV_g)/(dely)hatj+(delV_g)/(delz)hatk)` ` vecg=-[(x+z)hati+(6y)hatj+(x)hatk]` |
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| 1698. |
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 beA. `v=sqrt(GM((1)/( R)-(1)/(r )))`B. `v=sqrt(GM((1)/( 2R)-(1)/(r )))`C. `v=sqrt(GM((1)/( R)+(1)/(r )))`D. `v=sqrt(GM((1)/( 2R)+(1)/(r )))` |
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Answer» Correct Answer - B (b) Since the speeds of the stars are negligible when they are at a distance r, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the systme is `E_(i)=KE+PE=0+(-(GMM)/( r))=-(GM^(2))/( r)` where M represent the mass of each star and r is initial seperation between them. When two stars collide their centres will be at a distance twice the radius of a star i.e. 2R. Let v be the speed with which two stars collide. Then total energy of the system at the instant of their collision is given by. `E_(i)=KE+PE=0+(-(GMM)/(r ))=-(GM^(2))/(r )` where M represents the mass of each star and r is initial separation between them. When two stars collide their centres will be at a distancee twice the radius of a star i.e., 2R Let v be the speed with which two stars collide. Then total energy of the system at the instant of their collision is given by `E_(f)=2xx((1)/(2)Mv^(2))+(-(GMM)/(2R))=Mv^(2)-(GM^(2))/(2R)` According to law of conservation of mechanical energy `E_(f)=E_(i)` `Mv^(2)-(GM^(2))/(2R)=-(GM^(2))/(r ) or v^(2)=GM((1)/(2R)-(1)/( r)) or v=sqrt(GM((1)/(2R)-(1)/( r)))` |
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| 1699. |
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 beA. `v=sqrt(GM(1/R-1/r))`B. `v=sqrt(GM(1/(2R)-1/r))`C. `v=sqrt(GM(1/R+1/r))`D. `v=sqrt(GM(1/(2R)+1/r))` |
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Answer» Correct Answer - B Since the speeds of the stars are negligible when they are at distance `r`, hence the initial kinetic energy of the system is zero. Therefore, the initial total energy of the system is `E_(i)=KE+PE=0+(-(GMM)/r)=-(GM^(2))/r` Where `M` represents the mass of each star and `r` is initial separation between them, twice the radius of a star i.e. `2R`. Let `v` be the speed with which two stars collides. Then total energy of the system at the instant of their collision is given by `E_(f)=2xx(1/2 Mv^(2))+(-(GM_(E)m)/(2R))=Mv^(2)-(GM^(2))/(2R)` According to law of conservation of mechanical energy `E_(f)-E_(i)` `Mv^(2)-(GM^(2))/(2R)=-(GM^(2))/r` or `v^(2)=GM(1/(2R)-1/r)` or `v=sqrt(GM(1/(2R)-1/r))` |
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| 1700. |
Define Binding Energy. |
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Answer» The energy required to remove the satellite its orbit to infinity is called Binding Energy of the system, i.e., Binding Energy (B.E.) = -E = GMm/2r |
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