2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 6 InterviewSolution

251.	Gravitation on moon is `(1)/(6)` th of that on earth. When a balloon filled with hydrogen is released on moon then thisA. fall with acceleration less than g/6B. fall with acceleration g/6C. rise with acceleration g/6D. rise with uniform velocity
Answer» Correct Answer - b

Discussion

252.	Gravitation on moon is `(1)/(6)` th of that on earth. When a balloon filled with hydrogen is released on moon then thisA. Will rise with an acceleration less then `(g/6)`B. Will rise with acceleration `(g/6)`C. Will fall down with an acceleration less than `((5g)/6)`D. Will fall down with acceleration `(g/6)`
Answer» Correct Answer - D

Discussion

253.	Gravitation on moon is `(1)/(6)` th of that on earth. When a balloon filled with hydrogen is released on moon then thisA. will rise with an acceleration less then `(g)/(6)`B. will rise with acceleration `(g)/(6)`C. will fail down with an acceleration less than `(5g)/(6)`D. will fall down with acceleration `(g)/(6)`
Answer» Correct Answer - D There will be no buoyant force on the moon.

Discussion

254.	Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:
Answer» Correct Answer - `2J`

Discussion

255.	Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:A. 2JB. 20 JC. 40 JD. 10 J
Answer» Correct Answer - A `DeltaV=-E_(g).dr` Because field is uniform `therefore2=-E_(g).20impliesE=-(1)/(10),DeltaV=+(1)/(10)[4]=(2)/(5)` work done in taking a 5 kg body to height 4 m =m (change in gravitational potential) `=5[(2)/(5)]implies2J`

Discussion

256.	Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:A. 2 JB. 20 JC. 40 JD. 10 J
Answer» Correct Answer - A

Discussion

257.	An athlete complete one round of a circular track of diameter 200m in 40s. What will be the distance covered and the displacement at the end of 2 minutes ?
Answer» Number of rounds complered in 2 minutes `n=(2xx60s)/(40s)=3`. `:.` Distance covered=`nxx2pir=nxxpiD=3xx3.14xx200=1884` m on completing 3 roounds, final position of athlete coincides with his initial position.therefore,displacement = zero.

Discussion

258.	Write true or false for the following statements:All objects attract each other along the line joining their centre of mass
Answer» True Explanation: Yes, all objects attract each other along the line joining their centre of mass. The force of attraction is the gravitational force of attraction and is given as, F = G\(\frac{m_1\times m_2}{R^2}\) Where, G = Universal gravitational Constant m₁, m₂= Masses of bodies R = Distance between two bodies Gravitational force acts only along the line joining the two masses.

258.

Write true or false for the following statements:All objects attract each other along the line joining their centre of mass

Answer»

True

Explanation: Yes, all objects attract each other along the line joining their centre of mass.

The force of attraction is the gravitational force of attraction and is given as,

F = G\(\frac{m_1\times m_2}{R^2}\)

Where, G = Universal gravitational Constant

m₁, m₂= Masses of bodies

R = Distance between two bodies

Gravitational force acts only along the line joining the two masses.

Discussion

259.	What name has been given to the force with which two objects lying apart attract each other?
Answer» Correct Answer - Gravitational force

Discussion

260.	If the gravitational force between two bodies of masses 10 kg and 100 kg separated by a distance 10 m is `6.67 xx 10^(-10)` N, then the force between the given masses would be ______ N if they are placed in a geostationary satellite without change in the distance between them.A. `3.335 xx 10^(-10)`B. `3.335 xx 10^(-11)`C. `6.67 xx 10^(-10)`D. `6.67 xx 10^(-11)`
Answer» Correct Answer - C

Discussion

261.	If the distacne between the earth and the sun gets doubled then what would be the duration of the yearA. `730 sqrt2` daysB. `91sqrt2` daysC. 365 daysD. 730 days
Answer» Correct Answer - a `(T_(2))/(T_(1))=n^(3//2)=(2)^(3//2)` `T_(2)=2sqrt2T_(1)` `T_(2)=365xx2sqrt2=730sqrt2" days."`

Discussion

262.	The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`A. `3.80xx10^(8)` metersB. `3.37xx10^(6)` metersC. `7.60xx10^(4)` metersD. `1.90xx10^(2)` meters
Answer» Correct Answer - A

Discussion

263.	The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`A. `3.8xx10^(8)m`B. `3.37xx10^(6)m`C. `7.6xx10^(4)m`D. `1.9xx10^(2)m`
Answer» Correct Answer - A Here, Potential energy of system, `U=-7.79xx10^(28)J` Mass of the earth, `M_(E)=6xx10^(24)kg` Mass of the moon, `M_(M)=7.4xx10^(22)kg` The potential energy of the earth-moon system is `U=-(GM_(E)M_(M))/r` Where `r` is the mean distance between earth and moon. `:. r=-(GM_(E)M_(M))/U` `=-(6.67xx10^(-11)xx6xx10^(24)xx7.4xx10^(22))/(-7.79xx10^(28))` `=3.8xx10^(8)m`

Discussion

264.	The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`A. `3.8xx10^(8) m`B. `3.37xx10^(6)` mC. `7.60xx10^(4)` mD. `1.9xx10^(2)` m
Answer» Correct Answer - A (a) Here, Potential energy of system, `U= -7.79xx10^(28)J` Mass of the earth, `M_(E)=6xx10^(24)kg` Mass of the moon, `M_(M)=7.4xx10^(22) kg` The potential energy of the earth-moon system is `U=-(GM_(E)M_(M))/( r)` where r is the mean distance between earth and moon. `:. r= - (GM_(E)M_(M))/(U)=-(6.67x10^(-11)xx6xx10^(24)xx7.4 xx 10^(22))/(-7.79xx10^(28))` `=3.8xx10^(8) m`

Discussion

265.	The mass of the Earth and Moon are `6xx10^(24)` kg and `7.4xx10^(22)` kg respectively. The distance between them is `3.84xx10^(5)`km. Calculate the gravitational force of attraction between the two? `G=6.7xx10^(-11)Nm^(2)//kg^(2)`
Answer» Given: `m_(1)=6x10^(24)kg` `m_(2)=7.4xx10^(22)kg` `r=3.84xx10^(5)km=3.84xx10^(8)m` `G=6.7xx10^(-11)Nm^(2)//kg^(2)` To find: `F=?` Formula `F=(Gm_(1)m_(2))/(r^(2))` Solution: `F=(6.7xx10^(-11)xx6x10^(24)xx7.4xx10^(22))/((3.84xx10^(8))^(2))` `=(297.48xx10^(-11)xx10^(24)xx10^(22))/(14.74xx10^(16))` `=(300xx10^(19))/15` `=20xx10^(19)` `F=2xx10^(20)N` The gravitational force of attraction between Earth and Moon is `2xx10^(20)N`

Discussion

266.	If the orbital radius of the moon is `3.84xx10^(8)m` and period, is 27 dyas, then the orbital radius of communication satellite placed in orbit above the equator will beA. `4.26xx10^(7)m`B. `5.25xx10^(7)m`C. `3.26xx10^(7)m`D. `2.26xx10^(7)m`
Answer» Correct Answer - a `(T_(2))/(T_(1))=((r_(2))/(r_(1)))^(3//2)` `((1)/(27))^(2//3)=(r_(2))/(r_(1))` `(1)/(9)=(r_(2))/(r_(1))` `r_(2)=(r_(1))/(9)=(3.84xx10^(8))/(9)=4.26xx10^(7)m`.

Discussion

267.	The mass of the earth is `6 × 10^(24)` kg and that of the moon is` 7.4 xx 10^(22)` kg. If the distance between the earth and the moon is `3.84xx10^(5)` km, calculate the force exerted by the earth on the moon. (Take G `= 6.7 xx 10^(–11) N m^(2) kg^(-2)`)
Answer» Correct Answer - `2.02 xx 10^(20)N` Here, mass of Earth, `m_(1)=6xx10^(24)` kg , mass of Moon, `m_(2)=7.4xx10^(22)`kg distance between earth and moon, `r=3.84xx10^(5) km =3.84xx10^(8)m` force exerted by earth on moon, `F=? G=6.7XX10^(-11) Nm^(2)//kg^(2)` Thus, `F=(Gm_(1)m_(2))/r^(2)=((6.7xx10^(-11)xx6xx10^(24))xx(7.4xx10^(22)))/(3.84xx10^(8))^(2)=2.02xx10^(20)N`

Discussion

268.	The square of time period of revolution around the Sun is directly proportional to the _______ of the planet from the Sun.A. mean distanceB. square of the distanceC. cube to the distanceD. cube of the mean distance
Answer» Correct Answer - D cube of the mean distance

Discussion

269.	With what velocity would a shell have to be projected horizontally so that it might describle a circular orbit round the earth ? Acceleration due to gravity = ` 9.81 m s^(-2)` and radius of the earth = ` 6.4 xx 10 ^(6) m.`
Answer» Correct Answer - `7.9 xx 10^(3) ms^(-1)`

Discussion

270.	The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earthA. Is the sameB. Is smallerC. Is greaterD. Varies with its phase
Answer» Correct Answer - A

Discussion

271.	What do we call the gravitational force between the earth and an object?
Answer» The gravitational force between the earth and an object is called force of gravity.

Discussion

272.	What is Mass?
Answer» Mass is the measurement of inertia and inertia is the property of any object which opposes the change in state of the object. It is inertia because of which an object in rest has tendency to remain in rest and an object in motion has tendency to remain in motion. Inertia depends upon the mass of an object. Object having greater mass has greater inertia and vice versa. Mass of an object remains constant everywhere, i.e. mass will remain same whether that object is at the moon, at the earth or anywhere in the universe.

272.

What is Mass?

Answer»

Mass is the measurement of inertia and inertia is the property of any object which opposes the change in state of the object. It is inertia because of which an object in rest has tendency to remain in rest and an object in motion has tendency to remain in motion.

Inertia depends upon the mass of an object. Object having greater mass has greater inertia and vice versa. Mass of an object remains constant everywhere, i.e. mass will remain same whether that object is at the moon, at the earth or anywhere in the universe.

Discussion

273.	What is the acceleration of free fall?
Answer» The acceleration produced in the motion of an object when it falls freely towards the earth is termed acceleration of free fall. It is also called acceleration due to gravity. Its value on earth’s surface is 9.8 ms^-2.

Discussion

274.	What is the importance of universal law of gravitation?
Answer» The importance of universal law of gravitation can be understood from the following points: a) All the planets revolve around the sun due to gravitational force between the sun and planets. Hence gravitational force is responsible for the existence of solar system. b) Tides in oceans are formed due to the gravitational force between the moon and water in oceans. c) The atmosphere of the earth is attached with it due to gravitational force of earth. d) Artificial satellites revolve around the earth due to gravitational force of earth. e) We can stand on the surface of earth due to gravitational force of earth.

Discussion

275.	Define Free fall.
Answer» When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration. This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.

275.

Define Free fall.

Answer»

When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration.

This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.

Discussion

276.	What is the importance of the universal law of Gravitation?
Answer» The universal law of gravitation successfully explained several phenomena which were believed to be unconnected: (i) the force that binds us to the earth; (ii) the motion of the moon around the earth; (iii) the motion of planets around the Sun; and (iv) the tides due to the moon and the Sun.

276.

What is the importance of the universal law of Gravitation?

Answer»

The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:

(i) the force that binds us to the earth;

(ii) the motion of the moon around the earth;

(iii) the motion of planets around the Sun; and

(iv) the tides due to the moon and the Sun.

Discussion

277.	Average density of the earthA. does not depent on gB. is a complex function of gC. is directly proportional to gD. is inversely proportional to g
Answer» Correct Answer - A

Discussion

278.	From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure Taking gravitational potential `V =0at r = oo,` the potential at (G = gravitational constatn) A. `(2-GM)/(3R)`B. `(-2GM)/(R)`C. `(-GM)/(2R)`D. `(-GM)/(R)`
Answer» Correct Answer - D

Discussion

279.	The dependence of acceleration due to gravity `g` on the distance `r` from the centre of the earth, assumed to be a sphere of radius `R` of uniform density is as shown in Fig. below: The correct figure isA. B. C. D.
Answer» Correct Answer - B

Discussion

280.	From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure Taking gravitational potential `V =0at r = oo,` the potential at (G = gravitational constatn) A. `(-GM)/(2R)`B. `(-GM)/(R)`C. `(-2GM)/(3R)`D. `(-2GM)/(R)`
Answer» Correct Answer - B

Discussion

281.	The dependence of acceleration due to gravity `g` on the distance `r` from the centre of the earth, assumed to be a sphere of radius `R` of uniform density is as shown in Fig. below: The correct figure isA. B. C. D.
Answer» Correct Answer - A

Discussion

282.	A satellite revolves around the earth at a height of 1000 km. The radius of the earth is `6.38 xx 10^(3)`km. Mass of the earth is `6xx10^(24)kg and G=6.67xx10^(-14)"N-m"^(2)kg^(-2)`. Determine its orbital velocity and period of revolution.
Answer» Given, `h= 1000 km = 10^(6) m` `R=6.38xx10^(3)"km"=6.38xx10^(6)"m"` `rArr h+R=7.38xx10^(6) "m",M=6xx10^(24) kg` `:.` Orbital velocity, `v_(0)=sqrt((GM)/(R+h))` `=sqrt((6.67xx10^(-11)xx6xx10^(24))/(7.38xx10^(6)))` `=7364 "ms"^(-1)` `:.` Period of revolution, `T=sqrt((2pi(R+h))/(v_(0)))=(2pixx7.38xx10^(6))/(7364)` `rArr T=6297 s`.

Discussion

283.	State the value of universal gravitational constant. What was its value back in 1947?
Answer» The value of the universal gravitational constant in 1947 was the same as that of now. G = 6.673 ×1 0^-11 N m² kg^-2

283.

State the value of universal gravitational constant. What was its value back in 1947?

Answer»

The value of the universal gravitational constant in 1947 was the same as that of now.

G = 6.673 ×1 0^-11 N m² kg^-2

Discussion

284.	A satellite of mass `m` is circulating around the earth with constant angular velocity. If the radius is `R_(0)` and mass of earth is M, then the angular momentum about the centre of the earth isA. `m sqrt(GM//R_(0))`B. `Msqrt(GmR_(0))`C. `m sqrt(GMR_(0))`D. `M sqrt(GM//R_(0))`
Answer» Correct Answer - C Angular momentum = linear momentum `xx` perpendicular distance from the axis of rotation = mass `xx` orbital velocity `xx` radius `=mxxsqrt((GM)/(R_(0)))xxR_(0)=msqrt(GMR_(0))`.

Discussion

285.	Two identical thin ring each of radius `R` are co-axially placed at a distance `R`. If the ring have a uniform mass distribution and each has mass `m_(1)` and `m_(2)` respectively, then the work done in moving a mass `m` from the centre of one ring to that of the other is :A. `(Gm)/(m_(2)R)(sqrt(2)+1)m`B. `(Gm(m_(1)-m_(2)))/(sqrt(2)R)(sqrt(2)-1)`C. `(Gmsqrt(2))/(R)(m_(1)+m_(2))`D. zero
Answer» Correct Answer - B `V_(B)=` Potential at B due to A + Potential at A due to B `V_(B)=(-Gm_(2))/(R)-(-Gm_(1))/(sqrt(2)R)rArrV_(A)=(-Gm_(1))/(R)-(-Gm_(2))/(sqrt(2)R)` `W_(ArarrB)=m(V_(B)-V_(A))=(Gm(m_(1)-m_(2)))/(sqrt(2)R)(sqrt(2)-1)`.

Discussion

286.	The gravitational field in a region is given by `vecE=(10Nkg^-1)(veci+vecj)`. Find the work done by an external agent to slowly shift a particle of mass 2 kg from the point (0,0) to a point (5m,4m)`
Answer» `vecE=10hati+10hatj` `vecF=mvecE=20 hati+20hatj` `W=intvecF.dvecr=int_(x_(1))^(x_(2))F_(x)dx+int_(y_(1))^(y_(2))F_(y)dy` `=int_(0)^(5)20dx+int_(0)^(4) 20dy` `=20xx5+20xx4=180 J` Work done by external agent `=-180 J` Since particle moves slowly, initial `K.E.=0` Final `K.E.=0` total work done `=0` Work done by field `=180 J` Work done by external agent =`-180 J`

Discussion

287.	The gravitational force in a region is given by, `vec(F)=ayhati+axhatj`. The work done by gravitational force to shift a point mass `m` from `(0,0,0)` is `(x_(0),y_(0),z_(0))` isA. ma`x_(0) y_(0)z_(0)`B. ma`x_(0)y_(0)`C. -ma`x_(0)y_(0)`D. `0`
Answer» Correct Answer - B `W=intvec(F).vec(dr)=maint_(0)^(x_(0)y_(0)z_(0))(yhati+xhatj)(dxhati+dyhatj)` `=maintd(xy)=ma(xy)`

Discussion

288.	Two identical thin ring each of radius `R` are co-axially placed at a distance `R`. If the ring have a uniform mass distribution and each has mass `m_(1)` and `m_(2)` respectively, then the work done in moving a mass `m` from the centre of one ring to that of the other is :A. zeroB. `(Gm(m_(1)-m_(2))(sqrt(2)-1))/(sqrt(2)R)`C. `(Gmsqrt(2)(m_(1)+m_(2)))/R`D. `(Gm_(1)m(sqrt(2)+1))/(m_(2)R)`
Answer» Correct Answer - B `V_(1)=(-Gm_(1))/R-(Gm_(2))/(sqrt(2)R)` and `V_(2)=(-Gm_(2))/R-(Gm_(1))/(sqrt(2)R)` `DeltaV=V_(2)-V_(1)=(-Gm_(2))/R-(Gm_(1))/(sqrt(2)R)+(-Gm_(1))/R-(Gm_(2))/(sqrt(2)R)` `=G(m_(1)-m_(2))(1/R-1/(sqrt(2)R))` Hence `W=m(DeltaV)=(mG(m_(1)-m_(2))(sqrt(2)-1))/(sqrt(2)R)`

Discussion

289.	Potential (V) at a point in space is given by `v=x^(2)+y^(2)+z^(2)`. Gravitational field at a point (x,y,z) isA. `-2x hat(i)-2y hat(j)-2z hat(k)`B. `2x hat(i)+2y hat(j)+2z hat(k)`C. `x hat(i)+y hat(j)+ z hat(k)`D. `-x hat(i)-y hat(j)-z hat(k)`
Answer» Correct Answer - A

Discussion

290.	Is there any meaning of "Weight of the earth"?
Answer» No. Because the weight is the force by which the earth attracts a body. So the weight of the earth itself has no meaning.

Discussion

291.	Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particles be? Can one of the three particles be?
Answer» If stationary, the net force on each of the two particles will not be zero, and they will exert the gravitational force on each other. So they will move towards each other and not be in equilibrium. If moving around their center of mass, they may maintain the relative distance as the gravitational force on each of the particles is balanced by the centrifugal force and the net force is zero. Again for the same reason, three particles cannot be in equilibrium if stationary. If moving, they may maintain their relative distances for some configurations, again the gravitational force is balanced by centrifugal force. One of the three particles can be in equilibrium even without moving. This situation can happen if two of the particles are moving around the center of mass and the third particle is stationary at the center of mass.

291.

Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particles be? Can one of the three particles be?

Answer»

If stationary, the net force on each of the two particles will not be zero, and they will exert the gravitational force on each other. So they will move towards each other and not be in equilibrium. If moving around their center of mass, they may maintain the relative distance as the gravitational force on each of the particles is balanced by the centrifugal force and the net force is zero.

Again for the same reason, three particles cannot be in equilibrium if stationary. If moving, they may maintain their relative distances for some configurations, again the gravitational force is balanced by centrifugal force.

One of the three particles can be in equilibrium even without moving. This situation can happen if two of the particles are moving around the center of mass and the third particle is stationary at the center of mass.

Discussion

292.	Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. `(GMm)/(12R^2)`B. `(GMm)/(3R^2)`C. `(GMm)/(8R)`D. `(GMm)/(6R)`
Answer» Correct Answer - D (d) Energy required = (Potential energy of the Earth - mass system when mass is at distance 3R) -(Potentail energy of the Earth- mass system when mass is at distacnce 2R) `=(-GMm)/(3R) -((-GMm)/(2R)) = (GMm)/(3R) + (GMm)/(2R)` `=(-2GMm+3GMm)/(6R) = (GMm)/(6R)`

Discussion

293.	The escape velocity of a body depeds upon mass asA. `m^(0)`B. `m^(1)`C. `m^(3)`D. `m^(2)`
Answer» Correct Answer - A

Discussion

294.	The escape velocity of a body depeds upon mass asA. `m^0`B. `m^1`C. `m^2`D. `m^3`
Answer» Correct Answer - A (a) Escape velcoity, `v_c = sqrt(2gR) = sqrt(2GM)/(R ) rArr V_e propm^0` Where M, R are the mass and radius of the planet respectively. In this experssion the mass of the body (m) is not present showing that the escape velocity is independent of the mass.

Discussion

295.	The escape velocity of a projectile from the earth is approximatelyA. 11.2 m / secB. 112 km / secC. 11.2 km / secD. 11200 km / sec
Answer» Correct Answer - C

Discussion

296.	The escape velocity of a body depeds upon mass asA. `m^(2)`B. `m`C. `m^(0)`D. `m^(-1)`
Answer» Correct Answer - C

Discussion

297.	Ravi can throw a ball at a speed on the earth which can cross a river of width `10 m`. Ravi reaches on an imaginary planet whose mean density is twice that of the earth. Find out the maximum possible radius of the planet so that if Ravi throws the ball at the same speed it may escape from the planet. Given radius of the earth `= 6.4 xx 10^(6) m`.
Answer» Correct Answer - 4 km Range of throw is =10m `(u^(2))/(g)=10impliesu^(2)=100impliesu=10m//s` `v_(e)=sqrt((2GM)/(R))implies(v_(e))/(v_(ep))=sqrt((M_(e))/(R_(e))xx(R_(p))/(M_(p)))` `(v_(e))/(v_(p))=sqrt(((S_(e))/(S_(p)))(R_(e))/(R_(p)))^(2)` (S=density) `implies(11.2)/(10)=(R_(e))/(R_(p))xx(1)/(sqrt(2))impliesR_(p)=(10)/(11.2xxsqrt(2))xxR_(e)` `implies(10xx6.4xx10^(6))/(11.2xxsqrt(2))=40.42km`

Discussion

298.	Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to reach centre of earth if it is projected into the tunnel from surface of earth with speed needed for it to escape the gravitational field to earth.
Answer» Correct Answer - `t=sin^(-1)((1)/(sqrt(3)))sqrt((R_(e))/(g))` let x= distance of the particle form the surface acceleration `(vdv)/(dx)=(GM)/(R^(3))(R-x)` `implies_(v_(e))^(v)vdv=int_(0)^(x)g(1-(x)/(R))dx` `implies v=sqrt(v_(e)^(2)+2g(x-(x^(2))/(2R)))=(dx)/(dt)` `impliesint_(0)^(t)dt=int_(R)^(0)(dx)/(sqrt(2g[R+x-(x^(2))/(2R)]))` `impliest=int_(R)^(0)(dx)/(sqrt((g)/(R)[3R^(2)-(x-R)^(2)]))` let x-R=`sqrt(3)Rsintheta` `dx=sqrt(3)Rcosthetad theta` `thereforet=sqrt((R)/(g))intd theta=sqrt((R)/(g))sin^(-1)((x-R)/(sqrt(3)R))_(0)^(R)` `impliest=sqrt((R)/(g))sin^(-1)((1)/(sqrt(3)))`

Discussion

299.	Define Interatomic Forces.
Answer» The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces.

Discussion

300.	Find the gravitatinal force of between a point mass m placed at a disance x on the prolongation of a thin rod of mass M andlength l from near end .
Answer» Correct Answer - `(GMm)/(x(x+1))`

Discussion

Explore topic-wise InterviewSolutions in .

Gravitation on moon is `(1)/(6)` th of that on earth. When a balloon filled with hydrogen is released on moon then thisA. fall with acceleration less than g/6B. fall with acceleration g/6C. rise with acceleration g/6D. rise with uniform velocity

Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:

Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:A. 2JB. 20 JC. 40 JD. 10 J

Gravitational potential difference between surface of a planet and a point situated at a height of 20 m above its surface is 2joule/kg. if gravitational field is uniform, then the work done in taking a 5kg body of height 4 meter above surface will be-:A. 2 JB. 20 JC. 40 JD. 10 J

An athlete complete one round of a circular track of diameter 200m in 40s. What will be the distance covered and the displacement at the end of 2 minutes ?

Write true or false for the following statements:All objects attract each other along the line joining their centre of mass

What name has been given to the force with which two objects lying apart attract each other?

If the distacne between the earth and the sun gets doubled then what would be the duration of the yearA. `730 sqrt2` daysB. `91sqrt2` daysC. 365 daysD. 730 days

The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`A. `3.8xx10^(8)m`B. `3.37xx10^(6)m`C. `7.6xx10^(4)m`D. `1.9xx10^(2)m`

The mass of the earth is `6xx10^(24)kg` and that of the moon is `7.4xx10^(22)kg`. The potential energy of the system is `-7.79xx10^(28)J`. The mean distance between the earth and moon is `(G=6.67xx10^(-11)Nm^(2)kg^(-2))`A. `3.8xx10^(8) m`B. `3.37xx10^(6)` mC. `7.60xx10^(4)` mD. `1.9xx10^(2)` m

The mass of the Earth and Moon are `6xx10^(24)` kg and `7.4xx10^(22)` kg respectively. The distance between them is `3.84xx10^(5)`km. Calculate the gravitational force of attraction between the two? `G=6.7xx10^(-11)Nm^(2)//kg^(2)`

If the orbital radius of the moon is `3.84xx10^(8)m` and period, is 27 dyas, then the orbital radius of communication satellite placed in orbit above the equator will beA. `4.26xx10^(7)m`B. `5.25xx10^(7)m`C. `3.26xx10^(7)m`D. `2.26xx10^(7)m`

The mass of the earth is `6 × 10^(24)` kg and that of the moon is` 7.4 xx 10^(22)` kg. If the distance between the earth and the moon is `3.84xx10^(5)` km, calculate the force exerted by the earth on the moon. (Take G `= 6.7 xx 10^(–11) N m^(2) kg^(-2)`)

The square of time period of revolution around the Sun is directly proportional to the _______ of the planet from the Sun.A. mean distanceB. square of the distanceC. cube to the distanceD. cube of the mean distance

With what velocity would a shell have to be projected horizontally so that it might describle a circular orbit round the earth ? Acceleration due to gravity = ` 9.81 m s^(-2)` and radius of the earth = ` 6.4 xx 10 ^(6) m.`

The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earthA. Is the sameB. Is smallerC. Is greaterD. Varies with its phase

What do we call the gravitational force between the earth and an object?

What is Mass?

What is the acceleration of free fall?

What is the importance of universal law of gravitation?

Define Free fall.

What is the importance of the universal law of Gravitation?

Average density of the earthA. does not depent on gB. is a complex function of gC. is directly proportional to gD. is inversely proportional to g

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure Taking gravitational potential `V =0at r = oo,` the potential at (G = gravitational constatn) A. `(2-GM)/(3R)`B. `(-2GM)/(R)`C. `(-GM)/(2R)`D. `(-GM)/(R)`

The dependence of acceleration due to gravity `g` on the distance `r` from the centre of the earth, assumed to be a sphere of radius `R` of uniform density is as shown in Fig. below: The correct figure isA. B. C. D.

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure Taking gravitational potential `V =0at r = oo,` the potential at (G = gravitational constatn) A. `(-GM)/(2R)`B. `(-GM)/(R)`C. `(-2GM)/(3R)`D. `(-2GM)/(R)`

The dependence of acceleration due to gravity `g` on the distance `r` from the centre of the earth, assumed to be a sphere of radius `R` of uniform density is as shown in Fig. below: The correct figure isA. B. C. D.

A satellite revolves around the earth at a height of 1000 km. The radius of the earth is `6.38 xx 10^(3)`km. Mass of the earth is `6xx10^(24)kg and G=6.67xx10^(-14)"N-m"^(2)kg^(-2)`. Determine its orbital velocity and period of revolution.

State the value of universal gravitational constant. What was its value back in 1947?

A satellite of mass `m` is circulating around the earth with constant angular velocity. If the radius is `R_(0)` and mass of earth is M, then the angular momentum about the centre of the earth isA. `m sqrt(GM//R_(0))`B. `Msqrt(GmR_(0))`C. `m sqrt(GMR_(0))`D. `M sqrt(GM//R_(0))`

The gravitational field in a region is given by `vecE=(10Nkg^-1)(veci+vecj)`. Find the work done by an external agent to slowly shift a particle of mass 2 kg from the point (0,0) to a point (5m,4m)`

The gravitational force in a region is given by, `vec(F)=ayhati+axhatj`. The work done by gravitational force to shift a point mass `m` from `(0,0,0)` is `(x_(0),y_(0),z_(0))` isA. ma`x_(0) y_(0)z_(0)`B. ma`x_(0)y_(0)`C. -ma`x_(0)y_(0)`D. `0`

Potential (V) at a point in space is given by `v=x^(2)+y^(2)+z^(2)`. Gravitational field at a point (x,y,z) isA. `-2x hat(i)-2y hat(j)-2z hat(k)`B. `2x hat(i)+2y hat(j)+2z hat(k)`C. `x hat(i)+y hat(j)+ z hat(k)`D. `-x hat(i)-y hat(j)-z hat(k)`

Is there any meaning of "Weight of the earth"?

Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particles be? Can one of the three particles be?

Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. `(GMm)/(12R^2)`B. `(GMm)/(3R^2)`C. `(GMm)/(8R)`D. `(GMm)/(6R)`

The escape velocity of a body depeds upon mass asA. `m^(0)`B. `m^(1)`C. `m^(3)`D. `m^(2)`

The escape velocity of a body depeds upon mass asA. `m^0`B. `m^1`C. `m^2`D. `m^3`

The escape velocity of a projectile from the earth is approximatelyA. 11.2 m / secB. 112 km / secC. 11.2 km / secD. 11200 km / sec

The escape velocity of a body depeds upon mass asA. `m^(2)`B. `m`C. `m^(0)`D. `m^(-1)`

Assume that a tunnel is dug across the earth (radius=R) passing through its centre. Find the time a particle takes to reach centre of earth if it is projected into the tunnel from surface of earth with speed needed for it to escape the gravitational field to earth.

Define Interatomic Forces.

Find the gravitatinal force of between a point mass m placed at a disance x on the prolongation of a thin rod of mass M andlength l from near end .