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2601.

A planet has twice the radius but the mean density is `1/4` th as compared to earth. What is the ratio of escape velocity from earth to that from the planetA. `1:1`B. `1:2`C. `2:1`D. `3:1`

Answer» Correct Answer - A
`v_(e )=sqrt((2GM)/(R ))=sqrt((2G4pi R^(3)d)/(3))=2sqrt(2)R sqrt((G pi d)/(3))`
`therefore (v_(e))/(v_(p))=(2sqrt(2)R sqrt((G pi d)/(3)))/(2sqrt(2)2R sqrt((G pi)/(3)(d)/(4)))=1`
2602.

Derive an expression for the acceleration due to gravity on the surface of the earth. State the factors on which g depends. How does it vary with depth? Where is it maximum on the surface of the earth?

Answer»

The acceleration produced in a freely falling body by the gravitational pull of the earth is called the acceleration due to gravitation.

We know that,

Force = Mass × Acceleration

F = m × a

a = \(\frac{F}{m}\)........................1

Where F is the force on the object of mass m dropped from a distance r from the centre of earth of mass M.

So, force exerted by the earth on the object is

F = G\(\frac{M\times m}{r^2}\)......................2

M = Mass of Earth

m = mass of object

r = distance of object from centre of earth

Now, from equation 1 and 2,

a = G\(\frac{M\times m}{r^2\times m}\)

a = G\(\frac{M}{r^2}\)

Now, from above,

a = g = Acceleration due to gravity

We also see that, although force is depending on the mass of the object,

F = G\(\frac{M\times m}{r^2}\)

But acceleration due to gravity is independent of the mass.

g = G\(\frac{M}{r^2}\)

Factors on which g depends are:

(i) Value of gravitational constant (G)

(ii) Mass of Earth (M)

(iii) Radius of Earth (r)

As gravitational constant G and mass of earth M are always constant, so the value f acceleration due to gravity g is constant as long as the radius of earth remains constant.

At the surface of earth also, the value of g is not constant.  \((g\,∝ \frac{1}{r^2})\)

At the poles, radius of earth is minimum, hence the g is maximum. Similarly, at the equator, the radius of earth is maximum, and hence the value of g is minimum. Also, as we go up from the surface of the earth, distance from the centre of the earth increases and hence value of g decreases.

The value of g also decreases as we go inside the surface of earth and g is zero at the centre of earth, as at the centre of the earth the object has mass around it, so net force cancels and thus net acceleration becomes zero.

2603.

Give at least two points of difference between G and g.

Answer»
Acceleration due to Gravity(g)Universal Gravitational constant (G)
An acceleration is produced on a freely falling body due to the gravitational force of earth is known as acceleration due to gravity.Gravitational constant G is the numerically equal to the force of gravitation that exists between two bodies of unit mass kept at a unit distance from each other.
value of g near earth's surface is 9.8 ms-2. It may vary from place to place.value of G is 6.67 x 10-11 N,m2kg-2 and it is an universal constant
Depends upon the distance between the massesIndependent of the distance between the masses
SI unit is ms-2SI unit is Nm2kg-2

2604.

If we go deeper inside the earth, what will change: weight or mass or both, why?

Answer»

As we go deeper inside the earth weight of object changes and mass remains invariant.

Since, mass of an object is the amount of matter contained in the object and as we know that going deep inside the earth will not change the amount of matter contained in the object, so mass remains constant. In fact, mass of an object always remains constant and does not change from one place to another.

On the other hand, weight of an object is the force with which earth pulls an object towards its centre.

W = m × g

From above, as we know that g decreases as we go deep inside the earth, so the value of the weight of the object also decreases.

2605.

Under what conditions a body becomes weightless?

Answer»

Weight of a body is nothing but the force with which it is attracted towards the centre of earth.

Force = Mass × Acceleration

W = m × g

Since, we have Force = Weight = W and acceleration is acceleration due to gravity or g.

A body becomes weightless when g becomes zero and this happens when acceleration due to gravity becomes zero and thus body becomes weightless.

Example:

At centre of earth, far away from earth’s surface, freely falling body

2606.

Give Reasons for the Following:As we go higher above the surface of the earth the weight of the body decreases

Answer»

We know that weight of a body is the force with which the body is attracted towards the center of earth.

W = m × g

Where, m = Mass of the body

And, g = Acceleration due to gravity

The mass of a body or object is constant and does not change from place to place.

But acceleration due to gravity is inversely proportional to the square of the distance from the center of earth. So, as we go up from surface of the earth, the distance from the center of earth increases and hence value of g decreases.

Since, W = m × g

So, as the g decreases, weight of the body also decreases with increase in height above the earth. Thus, as we go higher above the surface of the earth the weight of the body decreases

2607.

Differentiate between mass and weight.

Answer»
MassWeight
The mass of a body is the quantity of matter contained in it.The force with which the body is attracted towards the centre of earth is called  weight.
Mass is a scalar quantity, that is it has only magnitudeWeight is a vector quantity that is it has both magnitude as well as direction. It is always directed towards the centre of earth.
Mass of an object is constant and does not change from place to placeWeight of a body changes from place to place and is not constant
Mass can be determined via ordinary balanceWeight of a body is measured via a spring balance.
SI unit of mass is kilogram (kg)SI unit of weight is Newton(N)

2608.

Why does one feel giddy while moving on a merry go round ?

Answer»

When moving in a merry go round, our weight appears to decrease when we move down and increases when we move up, this change in weight makes us feel giddy.

2609.

Two planets `A` and `B` have the same material density. If the radius of `A` is twice that of `B`, then the ratio of the escape velocity `(v_(A))/(v_(B))` isA. `(1)/(2)`B. `2:1`C. `sqrt(2):1`D. `1:sqrt(2)`

Answer» Correct Answer - B
`(V_(e1))/(V_(e2))=sqrt((M_(1)R_(2))/(M_(2)R_(1)))=sqrt((R_(1)^(3))/(R_(2)^(3))xx(R_(2))/(R_(1)))=(R_(1))/(R_(2))=(2)/(1)`
2610.

Two planets `A` and `B` have the same material density. If the radius of `A` is twice that of `B`, then the ratio of the escape velocity `(v_(A))/(v_(B))` isA. 2B. `sqrt(2)`C. `1//sqrt(2)`D. `1//2`

Answer» Correct Answer - A
`V_(e )=sqrt((2GM)/(R )) rArr V_(e )= sqrt((2G.(4)/(3)piR^(3)rho)/(R )) rArr V_(e ) = sqrt(8G pi rho R)`
`(V_(A))/(V_(B))=(R_(A))/(R_(B)) rArr (V_(A))/(V_(B))=(2)/(1)`
2611.

a) Define the term ‘work’. Write the formula for the work done on a body when a force acts on the body in the direction of its displacement. Give the meaning of each symbol which occurs in the formula. b) A person of mass 50kg climbs a tower of height 72 meters. Calculate the work done.

Answer»

a) Work is done when an applied force produces motion in a body. 

Formula for work done is: 

W = Fs 

Where, 

W is the work done 

F is the force applied 

s is the distance travelled 

b) Mass of the person, m = 50kg 

Height of tower, h = 72m 

Acceleration due to gravity, g = 9.8 m/s

Work done, W = mgh = 35280 J

2612.

A weightlifter is lifting weights of mass 200kg up to a height of 2 meters. If g = 9.8 m/s2, calculate: a) potential energy acquired by the weight b) work done by the weightlifter

Answer»

Mass, m = 200 kg 

Height, h = 2m 

Acceleration due to gravity, g = 9.8 m/s2 

a) Potential energy = mgh = 3920 J 

b) Work done against gravity = potential energy gained by the weight 

Therefore, work done, W = mgh = 3920 J

2613.

If the acceleration due to gravity at the surface of theearth is g, the work done in slowly lifting a body of mass m from the earth's surface to a height R equal to the radius of the earth is.(a) 1/2 mgR(b) 2mgR(c) mgR (d) 1/4 mgR.

Answer»

The correct answer is (a) 1/2 mgR

EXPLANATION:   

The distance of the body from the center of the earth at a height R from the earth's surface = 2R. The acceleration due to gravity at distance x from the center of the earth ( for x>R) = GM/x² 

Force on the body by the earth =mGM/x² 

Since the body is lifted slowly, hence the moving force is ≈ mGM/x² 

Hence the work done by the force in taking the body from R to 2R  

= ∫m(GM/x²)dx 

(limit of integration from R to 2R) 

=mGM[-1/x] 

=mGM[-1/2R+1/R] 

=mGM/2R 

=½m(GM/R²)R 

= ½mgR

2614.

The time period of an earth-satellite in circular orbit is independent of(a) the mass of the satellite(b) radius of the orbit(c) none of them(d) both of them.

Answer»

(a) the mass of the satellite

EXPLANATION: 

If the radius of the orbit = a, then velocity of an earth satellite v = √(GM/a) and the time period T = 2πa/v.      

As we can see that both v and T are independent of the mass of the satellite but both depend on the radius of the orbit a.  

2615.

Take the effect of bulging of earth and its rotation in account. Consider the following statements: A. There are pints outside the earth where the earth of g is equal to its value at the equator. B.There are points outside the earth where the earth where the value of g is equla to its value at the poles.A. both A and B are trueB. A is true but B is falseC. B is true but A is falseD. both and B are false

Answer» Correct Answer - B
2616.

Take the effect of bulging of earth and its rotation in account. Consider the following statements :(A) There are points outside the earth where the value of g is equal to its value at the equator.(B) There are points outside the earth where the value of g is equal to its value at the. poles.(a) Both A and B are correct. 13.(b) A is correct but B is wrong.(c) B is correct but A is wrong.(d) Both A and B are wrong.

Answer»

(b) A is correct but B is wrong.

EXPLANATION:  

As we move towards poles the value of g  starts increasing. Another fact is that as we go up at a place the value of g starts decreasing. So at any point on the surface of the earth where the value of g is more than at the equator, there will be points outside the earth (at some altitude) where the value of g will be equal to the value at the equator. So (A) is correct.          

On the surface of the earth, the value of g is maximum at poles. So we will not get the same value at higher up (outside the earth).  

2617.

Take the effect of bulging of earth and its rotation in account. Consider the following statements:(A) There are points outside the earth where the value of g is equal to its value at the equator.(B) There are points outside the earth where the valus of g is equal to its value at the poles.(a) Both A and B are correct.(b) A is correct but B is wrong.(c) B is correct but A is wrong.(d) Both A and B are wrong.

Answer»

(b) A is correct but B is wrong.

EXPLANATION:  

As we move towards poles the value of g  starts increasing. Another fact is that as we go up at a place the value of g starts decreasing. So at any point on the surface of the earth where the value of g is more than at the equator, there will be points outside the earth (at some altitude) where the value of g will be equal to the value at the equator. So (A) is correct.          

On the surface of the earth, the value of g is maximum at poles. So we will not get the same value at higher up (outside the earth).  

2618.

Let V and E represent the gravitational potential and field at a distance r from the centre of a uniform solid sphere. Consider the two statements: (A) the plot of V against r is discontinuous. (B) The plot of E against r is discontinuous. (a) Both A and B are correct. (b) A is correct but.B is wrong. (c) B is correct but A is wrong. (d) Both A and B are wrong.

Answer»

(d) Both A and B are wrong.

EXPLANATION: 

(A) is wrong because the value of V at the center is -3GM/2a and it increases continuously to -GM/a at the surface and from -GM/a at the surface to zero at the infinity. So it is continuous.          

(B) is wrong because the value of E increases linearly from zero at the center to GM/a² at the surface, then decreases from GM/a² at the surface to zero at the infinity. So it is also continuous.

2619.

Let V and E be the gravitational potential and gravitational field at a distance r from the centre of a uniform spherical shell. Consider the following two statements :(A) The plot of V against r is discontinuous.(B) The plot of E against r is discontinuous.(a) Both A and B are correct.(b) A is correct but B is wrong.(c) B is correct but A is wrong.(d) Both A and B are wrong.

Answer»

(c) B is correct but A is wrong.

EXPLANATION: 

(A) is not correct because the gravitational potential inside the shell is constant =-GM/a, where a is the radius of the shell. As the distance increases from a, the potential gradually starts increasing till infinity to a value zero. So the curve is not discontinuous. (B) is correct because the gravitational field inside the shell is zero and just as it goes outside the value jumps to GM/a² and then gradually decreases to zero at infinity. So the curve at r =a is discontinuous.

2620.

Is it possibe for a body to have inertia but no weight?

Answer»

Yes, a body will always have mass but the gravitational force on it can be zero; for example, when it is kept at the centre of the earth.

2621.

How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?

Answer»

By Universal law of gravitation,

F = \(\frac{Gm_1m_2}{r^2}\)

The force acting between two point masses m1 and m2, is independent of the nature of medium between them. Therefore, Gravitational force acting between two masses will remain unaffected when they are dipped in water.

2622.

Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm find out gravitational force between themA. `5.3x10^(-10) N`B. `4.9xx10^(-9) N`C. `6.5xx10^(-8) N`D. `6.9xxx10^(-7) N`

Answer» Given `M_(1)=1 kg, M_(2)=2 kg` and r=50 cm `=1/2M`
`therefore` Gravitational force
`F=(GM_(1)M_(2))/(r^(2))=(6.67xx10^(11)xx1xx2)/(1/2)^(2)=5.3xx10^(-10)N`
2623.

If the radius of the earth shrinks by `0.2%` without any change in its mass, the escape velocity from the surface of the earthA. increases by `0.2 %`B. decreases by `0.2 %`C. increases by `0.1 %`D. increases by `0.4%`

Answer» Correct Answer - C
`v_(e)=sqrt((2GM)/R)`
`rArr v_(e)prop1/(sqrt(R))rArr (Deltav_(e))/(v_(e))xx100=-1/2((DeltaR)/Rxx100)`
2624.

The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K. (a) U < K. (b )U>K. (c) U = K

Answer»

(b ) ​U>K.

EXPLANATION: 

Let the mass of the moon = m, velocity v and its distance from the earth = a.  

v = √(GM/a), K.E. = K = ½mv² = ½m*GM/a 

→GMm/a = 2K 

The magnitude of gravitational potential energy of the moon  

U = GMm/a =2K 

Obviously U>K.

2625.

The magnitude of gravitational potential energy of the moon-earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K.(a) U < K .(b) U > K.(c) U = K.

Answer»

(b ) ​U>K.

EXPLANATION: 

Let the mass of the moon = m, velocity v and its distance from the earth = a.  

v = √(GM/a), K.E. = K = ½mv² = ½m*GM/a 

→GMm/a = 2K 

The magnitude of gravitational potential energy of the moon  

U = GMm/a =2K 

Obviously U>K.

2626.

The magnitude of gravitational potential energy of the moon earth system is U with zero potential energy at infinite separation. The kinetic energy of the moon with respect to the earth is K.A. `UltK`B. `UgtK`C. `U=KD.

Answer» Correct Answer - B
2627.

At what distance (in metre) from the centre of the Moon,the intensity of gravitational field will be zero? (Take, mass of Earth and Moon as `5.98xx10^(24)` kg and `7.35xx10^(23)` kg respectively and the distance between Moon and Earth is `3.85xx10^(8)`m)A. zeroB. `3.90xx10^(7)`C. `8xx10^(8)`D. `3.46xx10^(8)`

Answer» Given mass of earth `M_(e )=5.98xx10^(24)`kg
mass of moon `M_(m)=7.35xx10^(22)` kg
let x be the distance of the point from the centre of earth where gravitional intensity is zero therefore
`(GM_(e ))/(x^(2))=(GM_(m))/(3.85xx10^(8)-x)^(2)`
`(x)/(3.85xx10^(8)-x)=sqrt(M_(e))/(M_(m))=sqrt(5.98xx10^(24))/(7.35xx10^(2))=9`
`rarr x=9xx3.85xx10^(8)//10`
`=3.46xx10^(8)` m
2628.

The point at which the gravitational force acting on any mass is zero due to the earth and the moon system is (The mass of the earth is approximately `81` times the mass of the moon and the distance between the earth and the moon is `3,85,000km`).A. `36,000 km` from the moonB. `38,500 km` from the moonC. `34500 km` from the moonD. `30,000 km` from the moon

Answer» Correct Answer - B
distance of null point,`x=d/(sqrt((m_(2))/(m_(1)))+1)` from `m_(1)`, smaller mass
2629.

Consider a geosynchronous communications satellite of mass `m` placed in an equitorial circular orbit of radius `r_(0)`. These satellite have an "apogee engine" which provides the thrusts needed to reach the final orbit. Once an error by the ground controllers causes the apogee engine to be fired. the thrust happens to be directed the Earth and, despite the quick reaction of the ground crew to shut the engine off, an unwanted velocity variation `Deltav` is imparted on the satellite. we characterize this boost by the parameter `beta=Deltav//v_(0)`. the duration of the engine burn is always negligible with respect to any other orbital times, so that it can be considered as instantaneous.[Hing : under the action of central forces obeying the inverse square law, bodies follow trajectories descirbed by ellipses, parabolas or hyperbolas. in the approzimation `mlt ltM` the gravitating mass `M` is at one of the focuses. where `l` is a positive constant named the semilatus rectum and `epsilon` is the eccentricity of the curve. in terms of constants of motion: `l=(L^(2))/(GMm^(2))` and `epsilon=(1+(2EL^(2))/(G^(2)M^(2)m^(3)))^(1//2)` where `G` is the Newton constant, `L` is the modulus of the angular momentum of the orbiting mass, with respect to the origin, and `E` is its mechanical energy, with zero potential energy at infinity. suppose `betalt1` Determine the new time period of the satellite if `(beta=1/4)`A. `~~20 hrs`B. `~~26 hrs`C. `~~67.9 hrs`D. `~~48 hrs`

Answer» Correct Answer - b
2630.

Black Hole isA. super surface of atmosphereB. ozone layerC. super dense planetary material none of these .D.

Answer» Correct Answer - C
( c) Black hole is a super dense planetary material formed due to the continued compression of the core of a star during supernova explosion.
2631.

The direction of gravitational intensity at point P of a hemispherical shell of uniform mass desity is indicated by the arrow A. dB. eC. fD. g

Answer» Correct Answer - B
(b) The direction of gravitational intensity at point P will be along e. Hence, option (b) is correct.
2632.

A planet having mass equal to that of the earth `(M = 6 xx10^(24)kg)` has radius R such that a particle projected from its surface at the speed of light `(c=3xx10^(8)ms^(-1))` just fails to escape. Assuming Newton’s Law of gravitation to be valid calculate the radius and mass density of such a planet. Are the numbers realistic? Note: The radius that you calculated is known as Schwarzschild radius. Actually we need to use theory of general relativity for solving this problem.

Answer» Correct Answer - R = 8.9 mm: `d=2xx10^(30)kgm^(-3)`
2633.

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R form the surface of the earth is (g=acceleration due to gravity at the surface of the earth)A. `g/9`B. `g/3`C. `g/4`D. `g`

Answer» Correct Answer - A
2634.

Acceleration due to gravity on moon is `1//6` of the acceleration due to gravity on earth. If the ratio of densities of earth `(rho_(e))` and moon `(rho_(m))` is `((rho_(e))/(rho_(m)))=5/3` then radius of moon `(R_(m))` in terms of `R_(e)` will beA. `5/18 R_(e)`B. `1/6 R_(e)`C. `3/18 R_(e)`D. `1/(2sqrt(3)) R_(e)`

Answer» Correct Answer - A
2635.

The value of quantity G in the formula for gravitational force: a) depends on mass of the earth only b) depends on the radius of the earth only c) depends on both mass and radius of earth d) depends neither on mass nor on radius of earth

Answer»

The correct answer is d) depends neither on mass nor on radius of earth

2636.

The law of gravitation gives the gravitational force between: a) the earth and a point mass only b) the earth and the sun only c) any two bodies having some massd) any two charged bodies only

Answer»

The correct answer is c) any two bodies having some mass

2637.

A geostationary satellite is orbiting the earth at a height of `5R` above the surface of the earth, `2R` being the radius of the earth. The time period of another satellite in hours at a height of `2R` form the surface of the earth isA. 5B. 10C. `6sqrt(2)`D. `6//sqrt(2)`

Answer» Correct Answer - C
2638.

Discuss the variation of acceleration due to gravity at poles and equator due to latitude of the Earth.

Answer»

1. Effective acceleration due to gravity at P is given as,

g’ = g – Rω2cos2θ.

2. As the value of θ increases, cos θ decreases. Therefore g’ will increase as we move away from equator towards any pole due to the rotation of the Earth.

3. At equator θ = 0°

∴ cos θ = 1

∴ g’e = g – Rω2

The effective acceleration due to gravity (g’e) is minimum at equator, as here it is reduced by Rω2

4. At poles θ = 90° cos θ = 0

∴ g’P = g – Rω2 cos θ

= g – 0

= g

There is no reduction in acceleration due to gravity at poles, due to the rotation of the Earth as the poles are lying on the axis of rotation and do not revolve.

2639.

Assertion : The binding energy of a satellite does not depend upon the mass of the satellite. Reason : Binding energy is the negative value of total energy of satellite.A. If both Assertin and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true

Answer» Correct Answer - D
2640.

Why the value of acceleration due to gravity is more at the poles than at the equator?A. `g_(p) lt g_(e)`B. `g_(p) = g_(e) =g`C. `g_(p) = g_(e) lt g`D. `g_(p) gt g_(e)`

Answer» Correct Answer - D
2641.

The value of acceleration due to gravity is maximum at poles and minimum at equator.why ?

Answer» This is because earth is flattened at poles and it is bluged out at equator.
`:. R_(p)` is minimum and `R_(e)` is maximum.
As `g_(p)=(GM)/(R_(p)^(2))` and `g_(e)=(GM)/(R_(e)^(2))`
There, `g_(p)` is maximum and `g_(e)` is minimum.
2642.

The value of acceleration due to gravity of earth: a) is the same on equator and poles b) is the least on poles c) is the least on equator d) increases from pole to equator

Answer»

The correct answer is c) is the least on equator

2643.

A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal toA. `(GM_(0))/(D_(0)^(2))`B. `(4GmM_(0))/(D_(0)^(2))`C. `(4GM_(0))/(D_(0)^(2))`D. `(GmM)/(D_(0)^(2))`

Answer» Correct Answer - C
2644.

A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal toA. `4GM_(p)//D_(p)^(2)`B. `GM_(p)m//D_(p)^(2)`C. `GM_(p)//D_(p)^(2)`D. `4GM_(p)m//D_(p)^(2)`

Answer» Correct Answer - A
2645.

A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal to

Answer» `g_(p)=(GM_(p))/(R_(p)^(2))`
Here, `M_(p)=M_(0)`
`R_(p)=(1)/(2)D_(0)`
`therefore g_(p)=(GM_(0))/(((1)/(2)D_(0))^(2))`
`therefore g_(p)=(4GM_(0))/(D_(0)^(2))`
2646.

A spherical planet far out in space has a mass `M_(0)` and diameter `D_(0)`. A particle of mass m falling freely near the surface of this planet will experience an accelertion due to gravity which is equal toA. `GM_(0)//D_(0)^(2)`B. `4 mGM_(0)//D_(0)^(2)`C. `4GM_(0)//D_(0)^(2)`D. `GmM_(0)//D_(0)^(2)`

Answer» Correct Answer - C
2647.

Assertion : Gravitational force between two masses in air is F. If they are immersed in water, force will remain F Reason : Gravitational force does not depend on the medium between the masses.A. If both Assertin and Reason are correct and Reason is the correct explanation of AssertionB. If both Assertion and Reason are correct but Reason is not the correct explanation of AssertionC. If Assertion is true but Reason is falseD. If Assertion is false but Reason is true

Answer» Correct Answer - A
2648.

In vacuum all freely falling objects:A. have the same speed B. have the same velocityC. have the same accelerationD. have the same force

Answer»

In vacuum all falling objects will have same acceleration.

The constant acceleration is due to gravitational force of earth and is known as acceleration due to gravity given by: g = G\(\frac{M}{R^2}\)

where,

G = Gravitational Constant = 6.67 x 10-11 Nm2kg-2

M = Mass of Earth

R = Radius of Earth

From above, we see that all variables in the equation are constant. Thus g doesn’t depend on any other factors, and hence is constant.

Hence, option C is correct.

2649.

Two objects of different masses falling freely near the surface of moon would: a) have same velocities at any instant b) have different accelerations c) experience forces of same magnitude d) undergo a change in their inertia

Answer»

The correct answer is a) have same velocities at any instant

2650.

A spherical planet far out in space has mass 2M and radius a. A particle of mass m is falling freely near its surface. What will be the acceleration of that particle ?A. `(GM)/(a^(2))`B. `(3GM)/(a^(2))`C. `(2GM)/(a^(2))`D. `(4GM)/(a^(2))`

Answer» Correct Answer - C