2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 28 InterviewSolution

1351.	For a body to escape from earth, angle from horizontal at which it should be fired isA. `45^(@)`B. `0^(@)`C. `90^(@)`D. any angle
Answer» Correct Answer - D

Discussion

1352.	One projectile after deviating from itspath starts movnig round the earth in a circular path of radius equal to nine times the radius of earth R.A. `2 pi sqrt(R/g)`B. `27xx2pi sqrt(R/g)`C. `pi sqrt(R/g)`D. `0.8xx3pi sqrt(R/g)`
Answer» Correct Answer - B

Discussion

1353.	A planet is revolving round the sun in an elliptical orbit. Of the following the property which is a constant during the motion of the planet is :A. The force of attraction between the planet and sun.B. The total energy of the `"planet plus sun"` system.C. The linear momentum of the planet.D. The kinetic energy of the planet about the sun.
Answer» Correct Answer - B

Discussion

1354.	A cone and a cylinder having same base area and height are placed on a horizontal surface. What is the ratio of the heights of centre of gravity of the cone and the cylinder from the surface?A. 5 : 3B. 3 : 5C. 3 : 2D. 1 : 2
Answer» Correct Answer - D

Discussion

1355.	If a graph is plotted between `T^(2) and r^(3)` for a planet, then its slope will be be (where `M_(S)` is the mass of the sun)A. `(4pi^(2))/(GM_(S))`B. `(GM_(S))/(4pi)`C. `4piGM_(S)`D. `GM_(S)`
Answer» Correct Answer - A (a) Time period of a revolution of a planet, `T=( 2pi r)/(v)=(2pi r)/(sqrt((GM_(S))/(r )))=(2 pi r^(3//2))/(sqrt(GM_(S))` where `M_(S)` is the mass of the sun Squaring both sides, we get `T^(2)=(4 pi^(2) r^(3))/(GM_(S))` The graph between `T^(2)` and `r^(3)` is a straight line whose slope is is `(4 pi^(2))/(GM_(S))`.

Discussion

1356.	If a graph is plotted between `T^(2) and r^(3)` for a planet, then its slope will be be (where `M_(S)` is the mass of the sun)A. `(4pi^(2))/(GM)`B. `(GM)/(4pi^(2))`C. `4pi GM`D. Zero
Answer» Correct Answer - A

Discussion

1357.	For a planet, the graph of `T^(2)` against `r^(3)` is plotted. The slope of the graph isA. `4pi GM`B. `(4pi^(2))/(GM)`C. `(GM)/(4pi^(2))`D. `(GM)/(4pi)`
Answer» Correct Answer - B The graph of `T^(2)` against `r^(3)` is a straightline. `because T^(2)=(4pi^(2)r^(3))/(GM) therefore (T^(2))/(r^(3))=" Slope " = (4 pi^(2))/(GM)`

Discussion

1358.	A planet of mass m is revolving round the sun (of mass `M_(s`) in an elliptical orbit. If `vecv` is the velocity of the planet when its position vector from the sun is `vecr`, then areal velocity of the positon vector of the planet is :A. `vecv + vecr`B. `vecr xx vecv`C. `1/2(vecv xx vecr)`D. `1/2(vecr xx vecv)`
Answer» Correct Answer - D

Discussion

1359.	One revolution of a given planet around the sun is 1000 days. If the distance between the planet and the sun is made `(1)/(4)`th of original value, then how many days will make one year?A. 180 daysB. 400 daysC. 125 daysD. 250 days
Answer» Correct Answer - C

Discussion

1360.	If the earth stops rotating about its axis , then what will be the change in the value of g at a place in the equitorial plane ? Radius of the earth = 6400 km.
Answer» Correct Answer - `3.4 cms^(-2)`

Discussion

1361.	In the earth-moon system, if `T_(1)` and `T_(2)` are period of revolution of earth and moon respectively about the centre of mass of the system themA. `T_(1) gt T_(2)`B. `T_(1) = T_(2)`C. `T_(1) lt T_(2)`D. Insufficient data
Answer» Correct Answer - B

Discussion

1362.	Acceleration due to gravity at the centre of the earth is :-A. `g`B. `g/2`C. zeroD. infinite
Answer» Correct Answer - C

Discussion

1363.	If the earth stops rotating sudenly, the value of g at a place other than poles would :-A. DecreaseB. Remain constantC. IncreaseD. Increase or decrease depending on the position of earth in the orbit round the sun
Answer» Correct Answer - c

Discussion

1364.	What is the relation between the period of rotation `(R_(T))` and periond of revolution `(R_(V))` of moon?A. `R_(T) = R_(V)`B. `R_(V) gt R_(T)`C. `R_(V) lt R_(T)`D. No relation exists
Answer» Correct Answer - A

Discussion

1365.	The acceleration due to gravity on a planet is `1.96 ms^(-1)`. If it is safe to jump from a height of 2 m on the earth, what will be the corresponding safe height on the planet?A. 5mB. 2mC. 10mD. 20m
Answer» Correct Answer - C

Discussion

1366.	If the acceleration due to gravity on a planet is `6.67 ms^(-2)` and its radius is `4 xx 10^(6)` m, then the mass of the planet is _____ .A. `16 xx 10^(23) kg`B. `726 xx 10^(23) kg`C. `16 xx 10^(24) kg`D. `26 xx 10^(24) kg`
Answer» Correct Answer - A

Discussion

1367.	Average density of the earthA. `g=(4)/(3)pi rhoG`B. `g=(3)/(4)piR rho G`C. `g=(4)/(3pi)R rhoG`D. `g=(4)/(3)pi R rho G`
Answer» Correct Answer - d `g=(GM)/(R^(2))=(Grho)/(R^(2))(4pi)/(3)R^(3)` `g=(4)/(3)pi rho GR.`

Discussion

1368.	The acceleration due to gravity on a planet of mass `10^(25)` Kg and radius 2580 Km in `ms^(-2)` isA. 1B. 10C. 20D. 100
Answer» Correct Answer - d `g=(GM)/(R^(2))=(6.67xx10^(-11)xx10^(25))/((2580xx10^(3))^(2))` `=(6.67xx10^(14))/(6.66xx10^(12))=10^(+2)=100ms^(-1)`

Discussion

1369.	A satellite `P` is revolving around the earth at a height `h`=radius of earth `(R)` above equator. Another satellite `Q` is at a height `2h` revolving in oppisite direction. At an instant the two are at same vertical line passing through centres of sphere. Find the least time of after which again they are in this situation.
Answer» Correct Answer - `(2piR^(3//2)(6sqrt(6)))/(sqrt(GM)(2sqrt(2)+3sqrt(3)))` `omega = sqrt((Gm)/(r^(3))) rArr omega_(1) = sqrt((Gm)/((R+R)^(3))) rArr omega_(1) = sqrt((Gm)/((2R)^(3)))` `omega_(2) = sqrt((Gm)/((R+2R)^(3))) rArr omega_(2) = sqrt((Gm)/((3R)^(3))) rArr omega_("rel")=omega_(1)+omega_(2)` `omega_(2) = (sqrt(Gm))/(R^(3//2))[(1)/(2sqrt(2))+(1)/(3sqrt(3))] rArr omega_("rel") = (sqrt(Gm))/(R^(3//2))xx((3sqrt(3)+2sqrt(2)))/(6sqrt(6))` `T = (2pi)/(omega_("rel")) rArr T = (2piR^(3//2)xx6sqrt(6))/(sqrt(Gm)(3sqrt(3)+2sqrt(2)))`

Discussion

1370.	A satellite `P` is revolving around the earth at a height `h`=radius of earth `(R)` above equator. Another satellite `Q` is at a height `2h` revolving in oppisite direction. At an instant the two are at same vertical line passing through centres of sphere. Find the least time of after which again they are in this situation.
Answer» Correct Answer - `(2piR^(3//2)(6sqrt(6)))/(sqrt(GM)(2sqrt(2)+3sqrt(3)))` `omega=sqrt((Gm)/(r^(3))impliesW_(1)=sqrt((Gm)/((R+R)^(3)))` `omega_(1)=sqrt((Gm)/((2R)^(3)))` `omega_(2)=sqrt((Gm)/((R+2R)^(3)))impliesomega_(2)=sqrt((Gm)/((3R)^(2)))` `omega_(rel)=omega_(1)+omega_(2)` `omega_(2)=(sqrt(Gm))/(R^(3//2))[1/(2sqrt(2))+1/(3sqrt(3))]` `omega_(rel)=(sqrt(Gm))/(R^(3//2))xx((3sqrt(3)+2sqrt(2)))/(6sqrt(6))` `T=(2pi)/(omega_(rel))impliesT=(2piR^(3//2)xx6sqrt(6))/(sqrt(Gm)(3sqrt(3)+2sqrt(2)))`

Discussion

1371.	The identical metal spheres of radius 10 cm are in contact with each other. If density of the metal is 10 g/cc, the graviational force between the spheres is `(pi^(2)=10)`A. `2.96xx10^(-4)N`B. `2.96xx10^(-6)N`C. `1.58xx10^(-4)N`D. `1.584xx10^(-6)N`
Answer» Correct Answer - b `F=Gxx(((4)/(3)pir^(3)rho)^(2))/((2r)^(2))=(4)/(9)pi^(2)Grho^(2)r^(4)` `=29.6xx10^(-11)(10xx10^(3))^(2)xx(10xx10^(-2))^(4)` `=29.6xx10^(-7)` `=2.96xx10^(-6)N`

Discussion

1372.	If the radius of the earth is 6400 km, the height above the surface of the earth where the value of acceleration due to gravity will be `4%` of its value on the surface of the earth isA. 6400 kmB. 64 kmC. 57600 kmD. 25600 km
Answer» Correct Answer - d `(g_(h))/(g)=((R)/(R+h))^(2)` `(4)/(100)=((R)/(R+h))^(2)` `(1)/(5)=(R)/(R+h)` `R+h=5R` `h=4R=4xx6400=25600km`.

Discussion

1373.	What will be velocity of a satellite revolving around the earth at a height h above surface of earth if radius of earth is R :-A. `R^(2) sqrt(g/(R+h))`B. `R g/((R+h)^(2))`C. `R sqrt(g/(R+h))`D. `R sqrt((R+h)/g)`
Answer» Correct Answer - C

Discussion

1374.	The mass of lift is 500 kg. When it ascends with an acceleration of `2m//s^(2)`,the tension in the cable will be `[g=10m//s^(2)]`A. 6000 NB. 4000 NC. 5000 ND. 50 N
Answer» Correct Answer - a `T=m(g+a)` `=500(10+2)=500xx12=6000N.`

Discussion

1375.	A point mass `m` is released from rest at a distance of `3R` from the centre of a thin-walled hollow sphere of radius `R` and mass `M` as shown. The hollow sphere is fixed in position and the only force on the point mass is the gravitational attraction of the hollow sphere. There is a very small hole in the hollow sphere through which the point mass falls as shown. The velocity of a point mass when it passes through point `P` at a distance `R//2` from the centre of the sphere is A. `sqrt((2GM)/(3R))`B. `sqrt((5GM)/(3R))`C. `sqrt((25GM)/(24R))`D. none of these
Answer» Correct Answer - D Inside the sphereicla shell, `V` is constant, so from energy conservation. `(-GMm)/(3R)=(mv^(2))/2-(GMm)/R` `(v^(2))/2=(GM)/R[1-1/3]=(GM)/Rxx2/3` or `v=sqrt((4GM)/(34R))`

Discussion

1376.	An artificial satellite in the presence of frictional forces will move into an orbit closer to the earth and may have increased kinetic energy. Explain this.
Answer» `E` (total Energy)`=1/2mv^(2)+(-(GMm)/r)` where `m` is the mass of the satellite `M=`mass of the earth and `r=` radius of the orbit. Being the dynamics of circlar motion, `(GMm)/(r_(2))=(mv^(2))/r` or `v^(2)=(GM)/r` Hence `E=(-1)/2(GM)/r m=-1/2 mv^(2)=-k` (kinetic energy) `E_(i)` (initial energy) `=-1/2(GM)/(r_(0))` `E_(f)` (final energy) `=-1/2(GM)/r` By the principle of conservation of energy `E_(i)=E_(f)+W`( work done by frictional force) `implies-1/2(GM)/(r_(0))=1/2(GM)/r+W` or `1/2 GM(1/r-1/(r_(0)))=W ` (a positive quantity) `:. 1/rgt1/(r_(0))` or `rltr_(0)` Also, `K_(0)=-K+W` or `K-K_(0)=W` (a positive quantity) `:. KltK_(0)`

Discussion

1377.	Work done by an external agent in bringing three particles each of mass `m` at the vertices of an equilateral triangle of length `l` isA. `(3Gm^(2))/(l)`B. `-(3Gm^(2))/(l)`C. `(3Gm^(2))/(2l)`D. none of these
Answer» Correct Answer - B `W=U=-(3Gmm)/l=-(3Gm^(2))/l`

Discussion

1378.	Find the work done in bringing three particles each having a mass of 100 g, from large distances to the vertices of an equilateral triangle of side 20 cm.A. `5.0 xx 10^(-12) J`B. `2.25 xx 10^(-10)J`C. `4.0 xx 10^(-11) J`D. `6.0 xx 10^(-15) J`
Answer» Correct Answer - A

Discussion

1379.	The ratio of KE of a planet at the points `1` and `2` is : A. `((r_(1))/(r_(2)))^(2)`B. `((r_(2))/(r_(1)))^(2)`C. `(r_(1))/(r_(2))`D. `(r_(2))/(r_(1))`
Answer» Correct Answer - B `v_(1)r_(1)=v_(2)r_(2)` `((KE)_(1))/((K.E.)_(2))=(1/2mv_(1)^(2))/(1/2mv_(2)^(2))=((v_(1))/(v_(2)))^(2)=((r_(2))/(r_(1)))^(2)`

Discussion

1380.	If `M` is the mass of the earth and `R` its radius, the ratio of the gravitational acceleration and the gravitational constant isA. `R^(2)/M`B. `M/R^(2)`C. `MR^(2)`D. `M/R`
Answer» Correct Answer - B

Discussion

1381.	In a gravitational field, at a point where the gravitational potential is zeroA. The gravitational field is necessarily zeroB. The gravitational field is not necessarily zeroC. Nothing can be said definitely about the gravitational fieldD. None of these
Answer» Correct Answer - A

Discussion

1382.	The mass of moon 1% of mass of earth. The ratio of gravitational pull of earth on moon and that of moon on earth will beA. `1:1`B. `1:10`C. `1:100`D. `2:1`
Answer» Correct Answer - A (a) The gravitational forces as mutually equal and opposite.

Discussion

1383.	A planet has radius `((1)/(36))` th of the radius of the earth. The escape velocity on the surface of the planet was found to be `(1)/(sqrt(6))` times the escape velocity from the surface of e earth. The planet is surrounded by a thin layer of atmosphere having thickness h (ltlt radius of the planet). The average density of the atmosphere on the planet is d and cceleration due to gravity on the surface of the earth is `g_(e)`. Find the value of atmospheric pressure on the surface of the planet.
Answer» Correct Answer - 6 dgh

Discussion

1384.	Using a telescope for several nights, you found a celestial body at a distance of`2xx10^(11)m` m from the sun travelling at a speed of `60kms^(-1)` Knowing that mass of the sun is`2xx10^(30)kg,` calculate after how many years you expect to see the body again at the same location.
Answer» Correct Answer - The body will never return to the same location

Discussion

1385.	Two identical thin ring, each of radius R meters, are coaxially placed a distance R metres apart. If `Q_1` coulomb, and `Q_2` coulomb, are repectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other isA. zeroB. `(Gm(m_(1) - m_(2))(sqrt(2) - 1))/(sqrt(2)R)`C. `(Gmsqrt(2)(m_(1) + m_(2)))/(R)`D. `(Gmm_(1)(sqrt(2)-1))/(m_(2)R)`
Answer» Correct Answer - B

Discussion

1386.	Choose the wrong option.A. Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass. .B. That the gravitational mass and inertial mass are equal is an experimental result.C. That the acceleration due to the gravity on Earth is the same for all bodies and is due to the quality of gravitational mass and inertial mass.D. Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot .
Answer» Correct Answer - D (d) According to the principle of equivalence, the gravitational mass of a body is equal to its inertial mass.

Discussion

1387.	in the previous question, the change in potential energy.A. `(GM_(E)m)/(2R_(E))`B. `(GM_(E)m)/(4R_(E))`C. `(GM_(E)m)/(8R_(E))`D. `(GM_(E)m)/(R_(E))s`
Answer» Correct Answer - B The change in potential energy is twice the change in the total energy. `:. DeltaU=2DeltaE=2((GM_(E)m)/(8R_(E)))=(GM_(E)m)/(4R_(E))`

Discussion

1388.	In previous question calculate orbital speed of satellite.A. `6.335 km//sec`B. `7.335 km//sec`C. `8.335 km//sec`D. `9.335 km//sec`
Answer» Correct Answer - A

Discussion

1389.	in the previous question, the change in potential energy.A. `(GM_(E)m)/(2R_(E))`B. `(GM_(E)m)/(4R_(E))`C. `(GM_(E)m)/(8R_(E))`D. `(GM_(E)m)/(16R_(E))`
Answer» Correct Answer - B (b) The change in potential energy is twice the change in the total energy. `:. DeltaU=2DeltaE=2((GM_(E)m)/(8R_(E)))=(GM_(E)m)/(4R_(E))`

Discussion

1390.	On what factors does the value of 'g' depends?
Answer» The value of 'g' depends on, (a) shape of the earth (b) latitude (c) altitude (d) depth

1390.

On what factors does the value of 'g' depends?

Answer»

The value of 'g' depends on,

(a) shape of the earth

(b) latitude

(c) altitude

(d) depth

Discussion

1391.	An elephant and an ant are to be projected out of earth into space. Do we need different velocities to do so?
Answer» The escape velocity, v_e = √(2gR), does not depend upon the mass of the projected body. Thus, we need the same velocity to project an elephant and an ant into space.

Discussion

1392.	The weight of a body is 20 N. What is the gravitational pull of the body on the earth?
Answer» The gravitational pull of the body on the earth is 20 N.

Discussion

1393.	A particle of mass m is moving in a horizontal circle of radius R under a centripetal force equal to `-A/r^(2)` (A = constant). The total energy of the particle is :- (Potential energy at very large distance is zero)A. `A/R`B. `- A/R`C. `A/(2R)`D. `- A/(2R)`
Answer» Correct Answer - D

Discussion

1394.	An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-E_(0)`B. `E_(0)`C. `-2E_(0)`D. `2E_(0)`
Answer» Correct Answer - D

Discussion

1395.	An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-E_(0)`B. `1.5 E_(0)`C. `2 E_(0)`D. `E_(0)`
Answer» Correct Answer - C

Discussion

1396.	An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-2E`B. 2EC. `(2E)/(3)`D. `-(2E)/(3)`
Answer» Correct Answer - A We know that, the potential energy of the satellite `U=-(GM_(e)m)/(R_(e))` The kinetic energy of the satellite `K = (1)/(2)(GM_(e)m)/(R_(e))` The total energy, `E=U+K=-(GM_(e)m)/(R_(e))+(1)/(2)(GM_(e)m)/(R_(e))=-(1)/(2)(GM_(e)m)/(R_(e))` `rArr2E=-(GM_(e)m)/(R_(2))rArr -2E =U` So, potential energy = -2 (total energy) `rArr PE = -2E`.

Discussion

1397.	The Earth is rotating with angular velocity ω about its own axis. R is the radius of the Earth.If Rω2 = 0.03386 m/s2, calculate the weight of a body of mass 100 gram at latitude 25°.(g = 9.8 m/s2)
Answer» Given: Rω²= 0.03386 m/s², θ = 25°, m = 0.1 kg, g = 9.8 m/s² To find: Weight (W) Formulae: i) g’ = g – Rω² cos² θ ii) W = mg Calculation: From formula (i), g’ = 9.8 – [0.03386 – cos² (25°)] ∴ g’ = 9.8 – [0.03386 × (0.9063)²] ∴ g’ = 9.8 – 0.02781 ∴ g’ = 9.772 m/s² From formula (ii), W = 0.1 × 9.772 ∴ W = 0.9772 N

1397.

The Earth is rotating with angular velocity ω about its own axis. R is the radius of the Earth.If Rω2 = 0.03386 m/s2, calculate the weight of a body of mass 100 gram at latitude 25°.(g = 9.8 m/s2)

Answer»

Given: Rω²= 0.03386 m/s², θ = 25°,

m = 0.1 kg, g = 9.8 m/s²

To find: Weight (W)

Formulae:

i) g’ = g – Rω² cos² θ

ii) W = mg

Calculation: From formula (i),

g’ = 9.8 – [0.03386 – cos² (25°)]

∴ g’ = 9.8 – [0.03386 × (0.9063)²]

∴ g’ = 9.8 – 0.02781

∴ g’ = 9.772 m/s²

From formula (ii),

W = 0.1 × 9.772

∴ W = 0.9772 N

Discussion

1398.	If the angular speed of the Earth is 7.26 × 10-5 rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.
Answer» Given: R = 6.4 × 10⁶ m, ω = 7.26 × 10^-5 rad/s To find: Change in weight (∆W) Formulae: (i) ∆g = g_p – g_eq= Rω² (ii) ∆W = m∆g Calculation: From formula (i) and (ii), ∆W = m(Rω²) = 1 × 6.4 × 10⁶(7.26 × 10^-5)² = 3373 × 10^-5N

1398.

If the angular speed of the Earth is 7.26 × 10-5 rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.

Answer»

Given: R = 6.4 × 10⁶ m, ω = 7.26 × 10^-5 rad/s

To find: Change in weight (∆W)

Formulae:

(i) ∆g = g_p – g_eq= Rω²

(ii) ∆W = m∆g

Calculation: From formula (i) and (ii),

∆W = m(Rω²)

= 1 × 6.4 × 10⁶(7.26 × 10^-5)²

= 3373 × 10^-5N

Discussion

1399.	Define potential energy.
Answer» Potential energy is the work done against conservative force (or forces) in achieving a certain position or configuration of a given system.

Discussion

1400.	Explain with examples the universal law which states that “Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy”.
Answer» Example 1: 1. A spring in its natural state, possesses minimum potential energy. Whenever we stretch it or compress it, we perform work against the conservative force. 2. Due to this work, the relative distances between the particles of the system change i.e., configuration changes and potential energy of the spring increases. 3. The spring finally regains its original configuration of minimum potential energy on removal of the applied force. This explains that the spring always try to rearrange itself in order to attain minimum potential energy. Example 2: 1. When an object is lying on the surface of the Earth, the system of that object and the Earth has minimum potential energy. 2. This is the gravitational potential energy of the system as these two are bound by the gravitational force. While lifting the object to some height, we do work against the conservative gravitational force. 3. In its new position, the object is at rest due to balanced forces. However, now, the object has a capacity to acquire kinetic energy, when allowed to fall. 4. This increase in the capacity is the potential energy gained by the system. The object falls on the Earth to achieve the configuration of minimum potential energy on dropping it from the new position.

1400.

Explain with examples the universal law which states that “Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy”.

Answer»

Example 1:

1. A spring in its natural state, possesses minimum potential energy. Whenever we stretch it or compress it, we perform work against the conservative force.

2. Due to this work, the relative distances between the particles of the system change i.e., configuration changes and potential energy of the spring increases.

3. The spring finally regains its original configuration of minimum potential energy on removal of the applied force. This explains that the spring always try to rearrange itself in order to attain minimum potential energy.

Example 2:

1. When an object is lying on the surface of the Earth, the system of that object and the Earth has minimum potential energy.

2. This is the gravitational potential energy of the system as these two are bound by the gravitational force. While lifting the object to some height, we do work against the conservative gravitational force.

3. In its new position, the object is at rest due to balanced forces. However, now, the object has a capacity to acquire kinetic energy, when allowed to fall.

4. This increase in the capacity is the potential energy gained by the system.

The object falls on the Earth to achieve the configuration of minimum potential energy on dropping it from the new position.

Discussion

Explore topic-wise InterviewSolutions in .

For a body to escape from earth, angle from horizontal at which it should be fired isA. `45^(@)`B. `0^(@)`C. `90^(@)`D. any angle

One projectile after deviating from itspath starts movnig round the earth in a circular path of radius equal to nine times the radius of earth R.A. `2 pi sqrt(R/g)`B. `27xx2pi sqrt(R/g)`C. `pi sqrt(R/g)`D. `0.8xx3pi sqrt(R/g)`

A cone and a cylinder having same base area and height are placed on a horizontal surface. What is the ratio of the heights of centre of gravity of the cone and the cylinder from the surface?A. 5 : 3B. 3 : 5C. 3 : 2D. 1 : 2

If a graph is plotted between `T^(2) and r^(3)` for a planet, then its slope will be be (where `M_(S)` is the mass of the sun)A. `(4pi^(2))/(GM_(S))`B. `(GM_(S))/(4pi)`C. `4piGM_(S)`D. `GM_(S)`

If a graph is plotted between `T^(2) and r^(3)` for a planet, then its slope will be be (where `M_(S)` is the mass of the sun)A. `(4pi^(2))/(GM)`B. `(GM)/(4pi^(2))`C. `4pi GM`D. Zero

For a planet, the graph of `T^(2)` against `r^(3)` is plotted. The slope of the graph isA. `4pi GM`B. `(4pi^(2))/(GM)`C. `(GM)/(4pi^(2))`D. `(GM)/(4pi)`

One revolution of a given planet around the sun is 1000 days. If the distance between the planet and the sun is made `(1)/(4)`th of original value, then how many days will make one year?A. 180 daysB. 400 daysC. 125 daysD. 250 days

If the earth stops rotating about its axis , then what will be the change in the value of g at a place in the equitorial plane ? Radius of the earth = 6400 km.

In the earth-moon system, if `T_(1)` and `T_(2)` are period of revolution of earth and moon respectively about the centre of mass of the system themA. `T_(1) gt T_(2)`B. `T_(1) = T_(2)`C. `T_(1) lt T_(2)`D. Insufficient data

Acceleration due to gravity at the centre of the earth is :-A. `g`B. `g/2`C. zeroD. infinite

If the earth stops rotating sudenly, the value of g at a place other than poles would :-A. DecreaseB. Remain constantC. IncreaseD. Increase or decrease depending on the position of earth in the orbit round the sun

What is the relation between the period of rotation `(R_(T))` and periond of revolution `(R_(V))` of moon?A. `R_(T) = R_(V)`B. `R_(V) gt R_(T)`C. `R_(V) lt R_(T)`D. No relation exists

The acceleration due to gravity on a planet is `1.96 ms^(-1)`. If it is safe to jump from a height of 2 m on the earth, what will be the corresponding safe height on the planet?A. 5mB. 2mC. 10mD. 20m

If the acceleration due to gravity on a planet is `6.67 ms^(-2)` and its radius is `4 xx 10^(6)` m, then the mass of the planet is _____ .A. `16 xx 10^(23) kg`B. `726 xx 10^(23) kg`C. `16 xx 10^(24) kg`D. `26 xx 10^(24) kg`

Average density of the earthA. `g=(4)/(3)pi rhoG`B. `g=(3)/(4)piR rho G`C. `g=(4)/(3pi)R rhoG`D. `g=(4)/(3)pi R rho G`

The acceleration due to gravity on a planet of mass `10^(25)` Kg and radius 2580 Km in `ms^(-2)` isA. 1B. 10C. 20D. 100

The identical metal spheres of radius 10 cm are in contact with each other. If density of the metal is 10 g/cc, the graviational force between the spheres is `(pi^(2)=10)`A. `2.96xx10^(-4)N`B. `2.96xx10^(-6)N`C. `1.58xx10^(-4)N`D. `1.584xx10^(-6)N`

If the radius of the earth is 6400 km, the height above the surface of the earth where the value of acceleration due to gravity will be `4%` of its value on the surface of the earth isA. 6400 kmB. 64 kmC. 57600 kmD. 25600 km

What will be velocity of a satellite revolving around the earth at a height h above surface of earth if radius of earth is R :-A. `R^(2) sqrt(g/(R+h))`B. `R g/((R+h)^(2))`C. `R sqrt(g/(R+h))`D. `R sqrt((R+h)/g)`

The mass of lift is 500 kg. When it ascends with an acceleration of `2m//s^(2)`,the tension in the cable will be `[g=10m//s^(2)]`A. 6000 NB. 4000 NC. 5000 ND. 50 N

An artificial satellite in the presence of frictional forces will move into an orbit closer to the earth and may have increased kinetic energy. Explain this.

Work done by an external agent in bringing three particles each of mass `m` at the vertices of an equilateral triangle of length `l` isA. `(3Gm^(2))/(l)`B. `-(3Gm^(2))/(l)`C. `(3Gm^(2))/(2l)`D. none of these

Find the work done in bringing three particles each having a mass of 100 g, from large distances to the vertices of an equilateral triangle of side 20 cm.A. `5.0 xx 10^(-12) J`B. `2.25 xx 10^(-10)J`C. `4.0 xx 10^(-11) J`D. `6.0 xx 10^(-15) J`

The ratio of KE of a planet at the points `1` and `2` is : A. `((r_(1))/(r_(2)))^(2)`B. `((r_(2))/(r_(1)))^(2)`C. `(r_(1))/(r_(2))`D. `(r_(2))/(r_(1))`

If `M` is the mass of the earth and `R` its radius, the ratio of the gravitational acceleration and the gravitational constant isA. `R^(2)/M`B. `M/R^(2)`C. `MR^(2)`D. `M/R`

In a gravitational field, at a point where the gravitational potential is zeroA. The gravitational field is necessarily zeroB. The gravitational field is not necessarily zeroC. Nothing can be said definitely about the gravitational fieldD. None of these

The mass of moon 1% of mass of earth. The ratio of gravitational pull of earth on moon and that of moon on earth will beA. `1:1`B. `1:10`C. `1:100`D. `2:1`

Using a telescope for several nights, you found a celestial body at a distance of`2xx10^(11)m` m from the sun travelling at a speed of `60kms^(-1)` Knowing that mass of the sun is`2xx10^(30)kg,` calculate after how many years you expect to see the body again at the same location.

in the previous question, the change in potential energy.A. `(GM_(E)m)/(2R_(E))`B. `(GM_(E)m)/(4R_(E))`C. `(GM_(E)m)/(8R_(E))`D. `(GM_(E)m)/(R_(E))s`

In previous question calculate orbital speed of satellite.A. `6.335 km//sec`B. `7.335 km//sec`C. `8.335 km//sec`D. `9.335 km//sec`

in the previous question, the change in potential energy.A. `(GM_(E)m)/(2R_(E))`B. `(GM_(E)m)/(4R_(E))`C. `(GM_(E)m)/(8R_(E))`D. `(GM_(E)m)/(16R_(E))`

On what factors does the value of 'g' depends?

An elephant and an ant are to be projected out of earth into space. Do we need different velocities to do so?

The weight of a body is 20 N. What is the gravitational pull of the body on the earth?

A particle of mass m is moving in a horizontal circle of radius R under a centripetal force equal to `-A/r^(2)` (A = constant). The total energy of the particle is :- (Potential energy at very large distance is zero)A. `A/R`B. `- A/R`C. `A/(2R)`D. `- A/(2R)`

An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-E_(0)`B. `E_(0)`C. `-2E_(0)`D. `2E_(0)`

An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-E_(0)`B. `1.5 E_(0)`C. `2 E_(0)`D. `E_(0)`

An artificial satellite moving in a circular orbit around the earth has a total energy `E_(0)`. Its potential energy isA. `-2E`B. 2EC. `(2E)/(3)`D. `-(2E)/(3)`

The Earth is rotating with angular velocity ω about its own axis. R is the radius of the Earth.If Rω2 = 0.03386 m/s2, calculate the weight of a body of mass 100 gram at latitude 25°.(g = 9.8 m/s2)

If the angular speed of the Earth is 7.26 × 10-5 rad/s and radius of the Earth is 6,400 km, calculate the change in weight of 1 kg of mass taken from equator to pole.

Define potential energy.

Explain with examples the universal law which states that “Every system always configures itself in order to have minimum potential energy or every system tries to minimize its potential energy”.