2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 37 InterviewSolution

1801.	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 NB. 108 NC. 120 ND. 150 N
Answer» Correct Answer - B `E_(G)=(4piGrrho)/(3)` `dF=E_(G)lamdadr` `F=underset((4R)/(5))overset(R)int(4piGrholamda)/(3)rdr=(4piGrholamda)/(3)[(r^(2))/(2)]_((4R)/(5))^(R)` `=(4piGrholamda)/(3xx2)[R^(2)-(16R^(2))/(25)]=(4pi)/(6)Grholamdaxx(9)/(25)R^(2)` `F=(4pi)/(6)Gxx(M)/((4pi)/(3)R_(e)^(3))xxlamdaxx(9)/(25)xx(R_(e)^(2))/(100)` After solving [`Fimplies108N]`

Discussion

1802.	If the radius of a planet is R and its density is `rho` , the escape velocity from its surface will beA. `v_(e) prop R sqrt(rho)`B. `v_(e) prop (1)/(sqrt(rhoR))`C. `V_(e) prop rhoR`D. `V_(e) prop (sqrt(rho))/( R)`
Answer» Correct Answer - A `v_(e)=sqrt((2GM)/(R ))=sqrt((2G.(4)/(3)pi R^(3)rho)/(R ))` `= sqrt((8)/(3)G pi rho R^(2))=2R sqrt((2)/(3)G pi rho)` Thus `v_(e) prop R sqrt(rho)` `because` The remaining quantities are constant.

Discussion

1803.	The escape velocity from the surface of the earth of radius `R` and density `rho`A. `sqrt(2pi g rhoR)`B. `2Rsqrt((2Gpirho)/(3))`C. `sqrt(4piG rho R)`D. `sqrt((4)/(3)piGrhoR)`
Answer» Correct Answer - b `v_(e)=sqrt((2GM)/(R))=sqrt((2Ge)/(R)(4pi)/(3)R^(3))=2Rsqrt((2Gepi)/(3)).`

Discussion

1804.	What is the radius of a planet of density `rho` if at its surface escape velocity of a body is `v`.
Answer» Correct Answer - `vsqrt((3)/(8piGp))`

Discussion

1805.	If the radius of a planet is R and its density is `rho` , the escape velocity from its surface will beA. `v_(e) prop rho R`B. `v_(e) prop sqrt(rho)R`C. `v_(e) prop sqrt(rho)/R`D. `v_(e) prop 1/(sqrt(rho) R)`
Answer» Correct Answer - B

Discussion

1806.	Escape velocity on the earthA. Is less than that on the moonB. Depends upon the mass of the bodyC. Depends upon the direction of projectionD. Depends upon the height from which it is projected
Answer» Correct Answer - D

Discussion

1807.	The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would beA. `2 R`B. `4 R`C. `1/4 R`D. `1/2 R`
Answer» Correct Answer - D

Discussion

1808.	The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would beA. `2R`B. `4R`C. `(1)/(4)R`D. `(1)/(2)R`
Answer» Correct Answer - D

Discussion

1809.	Say True or False.Aceleration is a scalar quantity.
Answer» Acceleration is a vector quantity

Discussion

1810.	Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.A. `(3Gm^(2))/(a)`B. `(3Gm^(2))/(2a)`C. `(Gm^(2))/(2a)`D. `(Gm^(2))/(a)`
Answer» Correct Answer - B

Discussion

1811.	Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.A. `(3Gm^(2))/(a)`B. `(3Gm^(2))/(2a)`C. `(4Gm)^(2)/(3a)`D. `(Gm^(2))/(a)`
Answer» Correct Answer - B

Discussion

1812.	A body starts from rest from a point distant `r_(0)` from the centre of the earth. It reaches the surface of the earth whose radius is `R`. The velocity acquired by the body isA. `2GMsqrt(1/R-1/(r_(0)))`B. `sqrt(2GM(1/R-1/(r_(0)))`C. `GMsqrt(1/R-1/(r_(0)))`D. `sqrt(GM(1/R-1/(r_(0)))`
Answer» Correct Answer - B `-(GMm)/(r_(0))=1/2mv^(2)-(GMm)/R` or `1/2mv^(2)=(GMm)/R-(GMm)/(r_(0))` or `(v^(2))/2=(GM)/R-(GM)/(r_(0))` or `v^(2)=2GM[1/R-1/(r_(0))]` or `v=sqrt(2GM(1/R-1/(r_(0))))`

Discussion

1813.	A body starts from rest from a point distant `r_(0)` from the centre of the earth. It reaches the surface of the earth whose radius is `R`. The velocity acquired by the body isA. `GM((1)/(R)-(1)/(R_(0)))`B. `2GM((1)/(R)-(1)/(R_(0)))`C. `sqrt(2GM((1)/(R)-(1)/(R_(0))))`D. `2GMsqrt(((1)/(2)-(1)/(R_(0)))`
Answer» Correct Answer - C

Discussion

1814.	A body starts from rest from a point distant `r_(0)` from the centre of the earth. It reaches the surface of the earth whose radius is `R`. The velocity acquired by the body isA. `GM((1)/(R)-(1)/(R_(0)))`B. `2GM((1)/(R)-(1)/(R_(0)))`C. `sqrt(2GM((1)/(R)-(1)/(R_(0))))`D. `2GMsqrt(((1)/(R)-(1)/(R_(0))))`
Answer» Correct Answer - C

Discussion

1815.	An infinite number of particles each of mass `m` are placed on the positive X-axis of `1m, 2m, 4m, 8m,...` from the origin. Find the magnitude of the resultant gravitational force on mass `m` kept at the origin.A. `-8Gm`B. `-3Gm`C. `-4Gm`D. `-2Gm`
Answer» Correct Answer - D

Discussion

1816.	State whether the following statements are True or False : The mass of a body is the amount of matter present in it.
Answer» Answer is True

Discussion

1817.	State any one characteristic of gravitational force.
Answer» Gravitational force between two particles does not depend on the nature of the medium between them.

Discussion

1818.	State the SI and CGS units of G.
Answer» The SI unit of G is N.m²/kg² and CGS unit is the dyne.cm²/g².

Discussion

1819.	The Earth is acted upon by the gravitational force of attraction due to the sun. They why does the Earth not fall towards sun?
Answer» The earth does not fall into sun due to gravitational pull of sun on earth. This is because earth is not stationary.It is revolving around the sun in a particular orbit. The centripetal force requared by the earth for revolution around the sun is provided by gravitational pull of sun on the earth, In other words, gravitation pull of sun on earth up in providing necessary centripetal force required by the earth for revolution around the sun.Hence the earth does not fall into the sun.

Discussion

1820.	A ball is thrown up with a speed of 15 m/s . How high wiil it go before it begins to fall ? `(g=9.8 m//s^(2)`
Answer» Please note that here the ball is going up against the gravity ,so the value of g is to be taken as negative . Here Initial speed of ball u=15 /s Final speed of ball ,v=0 (The ball stops ) Acceleration due to gravity `g= -9.8 m//s^(2)` (Retardation ) And Height h=? (to be calculated ) Now ,Putting al these values in the formula `v^(2)=u^(2)+2gh ` we get : `(0)^(2)=(15)^(2)+2xx(-9.8) xx h` 0=225-19.6 h 19.6 h =225 `h=225/19.6` h=11.4 m Thus the ball will go the a maximum height of 11.4 meters before of begins to fall.

Discussion

1821.	Two identical spheres each of mass M and radius R are separated by a distance 3R. The force of attraction between them is proportional toA. `(1)/(R^(2))`B. `R^(4)`C. `R^(2)`D. `(1)/(R^(4))`
Answer» Correct Answer - B `F=(GM_(1)M_(2))/(d^(2))=(G[(4)/(3)pi R^(3)xx d][(4)/(3)pi R^(3)]d)/((3R)^(2))` i.e., `F prop (R^(6))/(R^(2))` or `F prop R^(4)`

Discussion

1822.	The force of attraction between two bodies of masses 100 kg and 1000 Kg separated by a distance of 10 m isA. `6.67xx10^(-7)N`B. `6.67xx10^(-8)N`C. `6.67xx10^(-9)N`D. `6.67xx10^(-10)`
Answer» Correct Answer - b `F=(GM_(1)M_(2))/(r^(2))` `=(6.67xx10^(-11)xx100xx1000)/(10^(2))` `=6.67xx10^(-8)N.`

Discussion

1823.	Identical packets are dropped from two areoplanes, one above the equator and the other above the north poole, both at height h. Assuming all condition are identical, will those packets take same time to reach the surface of earth. Justify your answer.
Answer» It is known that value of acceleration due to gravity (g) at poles is higer than its value at equator i.e.,`g_(p)gtg_(e)`. Therefore, from given height, the packet dropped above the north pole will reach the earth eariler than the packet dropped above the equator.

Discussion

1824.	A body of mass 50 kg and another body of mass 60 kg are separated by a distance of 2 m. What is the force of attraction between them?
Answer» m₁ = 50 kg, m₂ = 60 kg d = 2m, G = 6.67 × 10^-11 Nm² /kg² F = \(G\frac{m_1m_2}{d^2}\) = \(\frac{G\times50\times60}{2^2}=750\times G\) = 750 x 6.67 x 10^-11 = 5.0025 x 10^-8N

1824.

A body of mass 50 kg and another body of mass 60 kg are separated by a distance of 2 m. What is the force of attraction between them?

Answer»

m₁ = 50 kg,

m₂ = 60 kg

d = 2m, G = 6.67 × 10^-11 Nm² /kg²

F = \(G\frac{m_1m_2}{d^2}\)

= \(\frac{G\times50\times60}{2^2}=750\times G\)

= 750 x 6.67 x 10^-11 = 5.0025 x 10^-8N

Discussion

1825.	Why two children sitting close to each other do not come closer due to mutual force of attraction?
Answer» The attractive force between two children sitting close to each other are mutual. The force felt by each of them are equal and very short in magnitude. So they do not come closer to each other.

Discussion

1826.	Calculate the tension in a string that whirls a 2 kg toy in a horizontal circle of radius 2.5 iv when it moves at 3 m/s.
Answer» Mass of the toy m = 2 kg ; Radius of the circle = 2.5 m ; Speed of the toy = 3 m/s As the toy is moving in a horizontal circle, the necessary centripetal force is provided due to the tension in the string. ∴ The centripetal force , F_c = \(\frac{mv^2}{R}= \frac{2\,\times\,3\,\times3}{2.5}=\frac{18}{2.5}\) = 7.2 N

1826.

Calculate the tension in a string that whirls a 2 kg toy in a horizontal circle of radius 2.5 iv when it moves at 3 m/s.

Answer»

Mass of the toy m = 2 kg ;

Radius of the circle = 2.5 m ;

Speed of the toy = 3 m/s

As the toy is moving in a horizontal circle, the necessary centripetal force is provided due to the tension in the string.

∴ The centripetal force , F_c = \(\frac{mv^2}{R}= \frac{2\,\times\,3\,\times3}{2.5}=\frac{18}{2.5}\) = 7.2 N

Discussion

1827.	What makes the moon to move in a circular orbit around the earth?
Answer» The gravitational force between moon and earth act as centripetal force and makes the moon to revolve around the earth in uniform circular motion.

Discussion

1828.	What happens when the distance between the bodies is halved?
Answer» When the distance is halved, force of attraction becomes four times.

Discussion

1829.	In figure, we see that the moon falls around earth rather than straight into it. If the magnitude of velocity were zero, how would it move?
Answer» If the magnitude of velocity were zero, the moon would move towards earth due to acceleration due to gravity.

Discussion

1830.	If there is an attractive force between all objects, why we do not feel ourselves gravitating toward massive buildings in our vicinity?
Answer» Earth is massive than the building. Hence the gravitational force between ourselves and earth is more than that of between ourselves and building. Hence we do not feel gravitating towards massive building in our vicinity.

Discussion

1831.	The gravitational force between two bodies in 1 N if the distance between them is doubled, what will be the force between them ?
Answer» F = 1 F' = F/4 = 1/4 N.

1831.

The gravitational force between two bodies in 1 N if the distance between them is doubled, what will be the force between them ?

Answer»

F = 1

F' = F/4

= 1/4 N.

Discussion

1832.	What if the distance between the bodies is doubled?
Answer» When distance is doubled, attractive force reduced as to one fourth (1/4^th).

Discussion

1833.	According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies doubled?
Answer» Gravitational force F = \(\frac{GM_1M_2}{d^2}\) if mass of one of the bodies is doubled , M₁ =2M₁ Then F₁ = \(\frac{G2M_1M_2}{d^2}= \frac{2.GM_1M_2}{d^2}\) F₁ = 2F ∴ The gravitational force will be doubled.

1833.

According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies doubled?

Answer»

Gravitational force F = \(\frac{GM_1M_2}{d^2}\)

if mass of one of the bodies is doubled , M₁ =2M₁

Then F₁ = \(\frac{G2M_1M_2}{d^2}= \frac{2.GM_1M_2}{d^2}\)

F₁ = 2F

∴ The gravitational force will be doubled.

Discussion

1834.	What if the mass of both the bodies are doubled?
Answer» Force of attraction becomes 4 times.

Discussion

1835.	Two bodies are at a specific distance so as to attract each other. How many times will the mutual force of attraction be if the mass of one of them is doubled?
Answer» Mutual force of attraction doubled

Discussion

1836.	If the distance between two bodies that attract each other is trebled, how many times will their mutual force of attraction be? (9 time, 3 times, 1/3, 1/9)
Answer» Answer is 1/9 times

Discussion

1837.	Drill a hole at the bottom of an open bottle and fill it with water. Water goes out through the hole. Then allow the bottle to fall freely. What do you observe?
Answer» During free fall, water and the bottle possess some acceleration. So there is no reaction force experiences in water. As a result water does not go out.

Discussion

1838.	Write true or false for the following statements:The acceleration of a body thrown up is numerically the same as the acceleration of a downward falling body but opposite in sign.
Answer» True Explanation: When a body is thrown up and when a body comes down, in both the cases the only force acting on the body is the gravitational force attraction that acts towards the centre of the earth. F = m × a a = \(\frac{f}{m}\) Since, F and m both are same in both the cases, hence acceleration is also same. They are opposite in sign since, when a body comes down, then the acceleration is in direction of motion and hence positive, whereas when body is thrown up, then the acceleration is in opposite direction of motion and hence negative.

1838.

Write true or false for the following statements:The acceleration of a body thrown up is numerically the same as the acceleration of a downward falling body but opposite in sign.

Answer»

True

Explanation: When a body is thrown up and when a body comes down, in both the cases the only force acting on the body is the gravitational force attraction that acts towards the centre of the earth.

F = m × a

a = \(\frac{f}{m}\)

Since, F and m both are same in both the cases, hence acceleration is also same.

They are opposite in sign since, when a body comes down, then the acceleration is in direction of motion and hence positive, whereas when body is thrown up, then the acceleration is in opposite direction of motion and hence negative.

Discussion

1839.	The masses and radii of the Earth and the Moon are `M_1, R_1 and M_2,R_2` respectively. Their centres are at a distance d apart. The minimum speed with which a particel of mass m should be projected from a point midway between the two centres so as to escape to infinity is ........A. `sqrt((2G(M_(1)+M_(2)))/d)`B. `sqrt((4G(M_(1)+M_(2)))/d)`C. `sqrt((4GM_(1)M_(2))/d)`D. `sqrt((G(M_(1)+M_(2)))/d)`
Answer» Correct Answer - B Potential energy of mass `m` when it is midway between masses `M_(1)` and `M_(2) ` is `U=-(GM_(1)m)/(d//2)-(GM_(2)m)/(d//2)=-(2Gm)/d(M_(1)+M_(2))` According to law of conservation of energy `1/2mv_(e)^(2)=(2Gm)/d(M_(1)+M_(2))` Therefore, escape velocity `v_(e)=sqrt((4G(M_(1)+M_(2)))/d)`

Discussion

1840.	A planet of mass m is moving around the sun in an elliptical orbit of semi-major axis a :A. The total mechanical energy of the planet is varying periodically with timeB. The total energ of the planet is constant and equals `- (GmM_(s))/(2a), N_(s)` is mass of sunC. Total mechanical energy of the planet is constant and equals `- (GmM_(s))/(a), M_(2)` is mass of sunD. Date is insufficient of arrice at a conclusion
Answer» Correct Answer - B

Discussion

1841.	Statement I: Two satellites are following one another in the same circular orbit. If one satellite tries to catch another (leading one) satellite, then it can be done by increasing its speed without changing the orbit. Statement II: The energy of earth-satellite system in circular orbit is given by `E = (-Gms)//(2a)`, where `r` is the radius of the circular orbit.A. Statement I is True, Statement II is True: Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True: Statement II is Not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - D Here Statement 1 is incorrect because as speed of one satellite increases, its kinetic energy and hence total energy increases, i.e., total energy becomes less negative and hence r increases, i.e. orbit changes.

Discussion

1842.	Statement 1: The value of escape velocity from the surface of earth at `30^@` and `60^@` is `v_(1)=2v_(e), v_(2)=2//3v_(e)`. Statement II: The value of escape velocity is independent of angle of projection. A. Statement I is True, Statement II is True: Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True: Statement II is Not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - D The value of escape velocity is derived from the method of conservation of total mechanical energy and energy is independnet of direction.

Discussion

1843.	In our solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They (a) will not move around the sun since they have very small masses compared to sun. (b) will move in an irregular way because of their small masses and will drift away into outer space. (c) will move around the sun in closed orbits but not obey Kepler’s laws. (d) will move in orbits like planets and obey Kepler’s laws.
Answer» (d) will move in orbits like planets and obey Kepler’s laws.

Discussion

1844.	The escape velocity from earth is `v_(es)`. A body is projected with velocity `2v_(es)` with what constant velocity will it move in the inter planetary spaceA. `v_(es)`B. `3 v_(es)`C. `sqrt(3) v_(es)`D. `sqrt(5) v_(es)`
Answer» Correct Answer - C

Discussion

1845.	The escape velocity from earth is `v_(e)`. A body is projected with velocity `2v_(e)`. With what constant velocity will it move in the inter planetary space ?A. `v_(e)`B. `sqrt(2)v_(e)`C. `sqrt(3)v_(e)`D. `sqrt(5)v_(e)`
Answer» Correct Answer - C `U_(i)+K_(i)=U_(f)+K_(f)` `rArr-(GMm)/(R)+(1)/(2)m(2v_(e))^(2)=0+(1)/(2)mv^(2)` or `-(GM)/(R)+2v_(e)^(2)=(1)/(2)v^(2)` or `-(2GM)/(R)+(8GM)/(R)=v^(2)` or `v=sqrt((6GM)/(R))=sqrt(3((2GM)/(R)))=sqrt(3(2gR))=sqrt(3)v_(e)`.

Discussion

1846.	The escape velocity for a body of mass `1kg` from the earth surface is `11.2kms^(-1)`. The escape velocity for a body of mass `100kg` would beA. `11.2xx10^(2) kms^(-1)`B. `11.2 kms^(-1)`C. `11.2xx10^(-2) kms^(-1)`D. None of these
Answer» Correct Answer - B

Discussion

1847.	The escape velocity for a body of mass `1kg` from the earth surface is `11.2kms^(-1)`. The escape velocity for a body of mass `100kg` would beA. `11.2xx10^(-2)kms^(-1)`B. `111kms^(-1)`C. `11.2kms^(-1)`D. `11.2xx10^(-2)kms^(-1)`
Answer» Correct Answer - C Escape velocity is independent of mass of the object. It depends only on the mass of the planet as. `v=sqrt((2GM)/R)`

Discussion

1848.	A particle of mass m is projected from the surface of earth with a speed `V_(0)(V_(0) lt` escape velocity). Find the speed of particle at height `h = R` (radius of earth). (Take, `R = 6400 km` and `g = 9.8 m//s^(2))`A. `sqrt(gR)`B. `sqrt(v_(0)^(2)-2gR)`C. `sqrt(v_(0)^(2)-gR)`D. None of these
Answer» Correct Answer - C During motion of the particle, total mechanical energy remains constant. At the surface of earth, total mechanical energy is `E_(i) = (GmM)/(R) +(1)/(2)mv_(0)^(2)` `=- (GM)/(R^(2)) mR +(1)/(2)mv_(0)^(2)` `=- gmR +(1)/(2)mv_(0)^(2)` Total mechanical energy at height `h = R` is `E_(t) = (-GmM)/(2R) +(1)/(2)mv^(2)` `=- (gmR)/(2) +(1)/(2)mv^(2)` According to conservation principle of energy `E_(i) = E_(t)` `:. -gmR +(1)/(2)mv_(0)^(2) =- (gmR)/(2) +(1)/(2)mv^(2)` or `-2gR +v_(0)^(2) =- gR +v^(2)` `:. v = sqrt(v_(0)^(2)-gR)`

Discussion

1849.	If mass of a body is `M` on the earth surface, then the mass of the same body on the moon surface isA. 6 MB. `(M)/(6)`C. zeroD. M
Answer» Correct Answer - D Mass of a body is its inherent property. It is independent of location of the body. Thus, mass of the remains same (M) whether it is on the surface of the earth or on the surface of the moon.

Discussion

1850.	Kinetic energy of a particle on the surface of earth is `E_(0)` and the potential energy is `- 2E_(0)`. (a) Will the particle escape to infinity ? (ii) What is the value of potential energy where speed of the particle becomes zero?
Answer» Correct Answer - A::B (a) Total mechanical energy `= E_(0) - 2E_(0) = - E_(0)`. Since, it is nagative, it will not escape to infinity. (b) `E_(i) = E_(f) rArr E_(0) - 2E_(0) = 0 + U rArr U = - E_(0)`

Discussion

Explore topic-wise InterviewSolutions in .

If the radius of a planet is R and its density is `rho` , the escape velocity from its surface will beA. `v_(e) prop R sqrt(rho)`B. `v_(e) prop (1)/(sqrt(rhoR))`C. `V_(e) prop rhoR`D. `V_(e) prop (sqrt(rho))/( R)`

The escape velocity from the surface of the earth of radius `R` and density `rho`A. `sqrt(2pi g rhoR)`B. `2Rsqrt((2Gpirho)/(3))`C. `sqrt(4piG rho R)`D. `sqrt((4)/(3)piGrhoR)`

What is the radius of a planet of density `rho` if at its surface escape velocity of a body is `v`.

If the radius of a planet is R and its density is `rho` , the escape velocity from its surface will beA. `v_(e) prop rho R`B. `v_(e) prop sqrt(rho)R`C. `v_(e) prop sqrt(rho)/R`D. `v_(e) prop 1/(sqrt(rho) R)`

Escape velocity on the earthA. Is less than that on the moonB. Depends upon the mass of the bodyC. Depends upon the direction of projectionD. Depends upon the height from which it is projected

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would beA. `2 R`B. `4 R`C. `1/4 R`D. `1/2 R`

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is `R`, the radius of the planet would beA. `2R`B. `4R`C. `(1)/(4)R`D. `(1)/(2)R`

Say True or False.Aceleration is a scalar quantity.

Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.A. `(3Gm^(2))/(a)`B. `(3Gm^(2))/(2a)`C. `(Gm^(2))/(2a)`D. `(Gm^(2))/(a)`

Three particle of mass m each are placed at the three corners of an equilateral triangle of side a. Find the work which should be done on this system to increase the sides of the triangle to 2a.A. `(3Gm^(2))/(a)`B. `(3Gm^(2))/(2a)`C. `(4Gm)^(2)/(3a)`D. `(Gm^(2))/(a)`

A body starts from rest from a point distant `r_(0)` from the centre of the earth. It reaches the surface of the earth whose radius is `R`. The velocity acquired by the body isA. `2GMsqrt(1/R-1/(r_(0)))`B. `sqrt(2GM(1/R-1/(r_(0)))`C. `GMsqrt(1/R-1/(r_(0)))`D. `sqrt(GM(1/R-1/(r_(0)))`

An infinite number of particles each of mass `m` are placed on the positive X-axis of `1m, 2m, 4m, 8m,...` from the origin. Find the magnitude of the resultant gravitational force on mass `m` kept at the origin.A. `-8Gm`B. `-3Gm`C. `-4Gm`D. `-2Gm`

State whether the following statements are True or False : The mass of a body is the amount of matter present in it.

State any one characteristic of gravitational force.

State the SI and CGS units of G.

The Earth is acted upon by the gravitational force of attraction due to the sun. They why does the Earth not fall towards sun?

A ball is thrown up with a speed of 15 m/s . How high wiil it go before it begins to fall ? `(g=9.8 m//s^(2)`

Two identical spheres each of mass M and radius R are separated by a distance 3R. The force of attraction between them is proportional toA. `(1)/(R^(2))`B. `R^(4)`C. `R^(2)`D. `(1)/(R^(4))`

The force of attraction between two bodies of masses 100 kg and 1000 Kg separated by a distance of 10 m isA. `6.67xx10^(-7)N`B. `6.67xx10^(-8)N`C. `6.67xx10^(-9)N`D. `6.67xx10^(-10)`

Identical packets are dropped from two areoplanes, one above the equator and the other above the north poole, both at height h. Assuming all condition are identical, will those packets take same time to reach the surface of earth. Justify your answer.

A body of mass 50 kg and another body of mass 60 kg are separated by a distance of 2 m. What is the force of attraction between them?

Why two children sitting close to each other do not come closer due to mutual force of attraction?

Calculate the tension in a string that whirls a 2 kg toy in a horizontal circle of radius 2.5 iv when it moves at 3 m/s.

What makes the moon to move in a circular orbit around the earth?

What happens when the distance between the bodies is halved?

In figure, we see that the moon falls around earth rather than straight into it. If the magnitude of velocity were zero, how would it move?

If there is an attractive force between all objects, why we do not feel ourselves gravitating toward massive buildings in our vicinity?

The gravitational force between two bodies in 1 N if the distance between them is doubled, what will be the force between them ?

What if the distance between the bodies is doubled?

According to the equation for gravitational force, what happens to the force between two bodies if the mass of one of the bodies doubled?

What if the mass of both the bodies are doubled?

Two bodies are at a specific distance so as to attract each other. How many times will the mutual force of attraction be if the mass of one of them is doubled?

If the distance between two bodies that attract each other is trebled, how many times will their mutual force of attraction be? (9 time, 3 times, 1/3, 1/9)

Drill a hole at the bottom of an open bottle and fill it with water. Water goes out through the hole. Then allow the bottle to fall freely. What do you observe?

Write true or false for the following statements:The acceleration of a body thrown up is numerically the same as the acceleration of a downward falling body but opposite in sign.

The escape velocity from earth is `v_(es)`. A body is projected with velocity `2v_(es)` with what constant velocity will it move in the inter planetary spaceA. `v_(es)`B. `3 v_(es)`C. `sqrt(3) v_(es)`D. `sqrt(5) v_(es)`

The escape velocity from earth is `v_(e)`. A body is projected with velocity `2v_(e)`. With what constant velocity will it move in the inter planetary space ?A. `v_(e)`B. `sqrt(2)v_(e)`C. `sqrt(3)v_(e)`D. `sqrt(5)v_(e)`

The escape velocity for a body of mass `1kg` from the earth surface is `11.2kms^(-1)`. The escape velocity for a body of mass `100kg` would beA. `11.2xx10^(2) kms^(-1)`B. `11.2 kms^(-1)`C. `11.2xx10^(-2) kms^(-1)`D. None of these

The escape velocity for a body of mass `1kg` from the earth surface is `11.2kms^(-1)`. The escape velocity for a body of mass `100kg` would beA. `11.2xx10^(-2)kms^(-1)`B. `111kms^(-1)`C. `11.2kms^(-1)`D. `11.2xx10^(-2)kms^(-1)`

A particle of mass m is projected from the surface of earth with a speed `V_(0)(V_(0) lt` escape velocity). Find the speed of particle at height `h = R` (radius of earth). (Take, `R = 6400 km` and `g = 9.8 m//s^(2))`A. `sqrt(gR)`B. `sqrt(v_(0)^(2)-2gR)`C. `sqrt(v_(0)^(2)-gR)`D. None of these

If mass of a body is `M` on the earth surface, then the mass of the same body on the moon surface isA. 6 MB. `(M)/(6)`C. zeroD. M

Kinetic energy of a particle on the surface of earth is `E_(0)` and the potential energy is `- 2E_(0)`. (a) Will the particle escape to infinity ? (ii) What is the value of potential energy where speed of the particle becomes zero?