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2651.

Two object of different masses falling freely near the surface of moon wouldA. have same velocities at any instantB. have different accelerationsC. experience forces forces of same magnitudeD. undergo a change in their inertia

Answer» Correct Answer - A
2652.

Two object of different masses falling freely near the surface of moon wouldA. have same velocity at any instantB. have different accelerationC. experience forces of same magnitudeD. undergo a change in their inertia

Answer» Correct Answer - A
Two object of different masses falling freely near the surface of moon will have same velocities at any instant. This is because for both,`u=0,a=g,v=u+at=0+g t=g t=`same for both.
2653.

Write the equations of motion for a freely falling body. Also, write the meaning of symbols used in the equations.

Answer»

Equations of motion of an object under free fall:

The initial velocity of the object remains zero when the object is under free fall. In this situation, the three equations of motion i.e. v = u + gt

s = ut + gt

v2 = u2 + \(\frac{1}{2}\) gt2

v2 = u2 + 2gh will become,

ν = gt

s = \(\frac{1}{2}\)gt2 and v2 = 2gh

Here, u = final velocity, t = time, g = acceleration due to gravity, h = height attained by the body.

2654.

What is the mass and weight of an object on the surface of the moon, if on the earth its mass is 12 kg?

Answer»

Mass of the object on earth is 12 kg. It will be 12 kg on the moon also. Value of ‘g’ on the moon is one-sixth of its value on the earth.

Weight of the object on earth = mg = 12 x 9.8 = 117.6

N Weight of the object on moon 1

= 12 x \(\frac{1}{2}\) x 9.8 = 19.6 N.

2655.

What is acceleration due to gravity? Write its formula.

Answer»

The acceleration produced in a body due to the gravitational pull of the earth, near its surface is called acceleration due to gravity. It is denoted by ‘g’. i.e., 

\(g=\frac{G.M_e}{R^2_e}\)

Here, Me = mass of earth, Re = radius of earth and G is universal gravitational constant.

2656.

If the moon attracts the earth, why does the earth not move towards the moon?

Answer»

The force exerted by the earth and the moon on each other is insufficient to move the earth, on account of its huge mass.

2657.

What is the difference between G and g?

Answer»

1. Any two objects in the universe attract each other by a force called gravitational force. 

2. It produces almost negligible acceleration. 

3. It is not the measure of weight, g (gravity) 

4. The force of gravitation exerted by the earth is called gravity. 

5. Gravity causes the object to accelerate. It measures the weight, i.e., w = mg.

2658.

What changes will arise in force of gravitation between the masses, if the distance between them is doubled?

Answer»

When d is doubled, the gravitational force will become one fourth.

2659.

Two bodies of different masses are falling freely near the surface of the earth. They: (A) have the same velocity at any instant (B) move with different accelerations. (C) experience forces of the same magnitude (D) undergo a change in their inertia.

Answer»

The answer is (A) have the same velocity at any instant

2660.

At the centre of the earth: (A) the mass of a body is zero but the weight is not zero (B) the mass of a body is not zero but the weight is zero (C) both mass and weight of a body are zero (D) both mass and weight of a body are not zero

Answer»

The answer is (B) the mass of a body is not zero but the weight is zero

2661.

What is the cause of ocean tides?

Answer»

The earth and the moon pull each, other due to the gravitational force of attraction. Water on the earth (ocean) is attracted by both the earth and the moon. The molecules in water are not rigidly fixed. Therefore, when the moon and the sun both lie on the same side of the earth (new moon) or on the opposite side of the earth (full moon) with the ocean, the combined gravitational pull causes a bulge in the ocean water. This bulging out of the ocean water in the direction of the moon is called tide.

2662.

At which place on the surface of the earth, the acceleration due to gravity is more and why?

Answer»

The value of g varies from point to point in the universe, according to the values of M and r. Since the earth is flattened at poles and V reduces. Now since \(g\alpha \frac{1}{r^2}\)

Therefore, the value of g is more at poles of the earth than the equator.

2663.

The law of gravitation is true for: (A) the earth and any object only (B) earth and the sun only (C) any two massive bodies (D) two charged bodies only

Answer»

The answer is (C) any two massive bodies

2664.

State Newton’s law of gravitation.

Answer»

“Every particle in the universe attracts every other particle with a force, which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them”.

2665.

Gravitation force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in Newtons of 10 kg object on the moon and on the earth?

Answer»

Mass of the object (m) = 10 kg

Acceleration due to gravity on earth = g 

acceleration due to gravity on moon = g

Now,

\(g_m=\frac{g_e}{6}=\frac{9.8}{6}=1.635m/sec^2\)

Weight of an object on earth We = mge = 10 × 9.8 = 98 Newton

Weight of an object on the moon

Wm = mgm = 10 × 1.635 = 16.35 Newton

2666.

What is the importance of the universal law of gravitation?

Answer»

The universal law of gravitation successfully explained the planetary motion in our solar system. The gravitational force is responsible for so many natural phenomenon like the formation of tides in seawater, the existence of the atmosphere near the earth and many more.

2667.

According to Newton’s law of gravitation, every object attracts every other object.Thus, if the earth attracts an apple towards itself, the apple also attracts the earth towards itself with the same force. Why then does the apple fall towards the earth, but the earth does not move towards the apple?

Answer»

The earth and the apple move towards each other, but the magnitude of the displacement of the earth is negligible relative to that of the apple. Also the observer is located on the earth.

[Note: The mass of the earth is far greater than that of an apple. Hence, the magnitude of the acceleration of the earth is negligible relative to that of the apple.]

2668.

According to Newton’s law of gravitation, the earth’s gravitational force is higher on an object of larger mass. Why doesn’t that object fall down with higher velocity as compared to an object with lower mass?

Answer»

F = ma and F = \(\frac{GMm}{r^2}\)

∴ Acceleration, a = \(\frac{GM}{r^2}\). This is independent of the mass (m) of the object. 

Hence, an object of larger mass and an object of lower mass fall down with the same velocity.

2669.

Inside a uniform sphere of density `rho` there is a spherical cavity whose centre is at a distance `l` from the centre of the sphere. Find the strength of the gravitational field inside the cavity.

Answer» Correct Answer - `[(4pi)/(3)Gpl]`
2670.

Inside a uniform sphere of density `rho` there is a spherical cavity whose centre is at a distance `l` from the centre of the sphere. Find the strength of the gravitational field inside the cavity.A. `E=(-2)/3 piGrhol`B. `E=(-4)/3piGrhol`C. `E=(-4)/3pi^(2)Grhol`D. `E=(-4)/3piGrho^(2)l^(2)`

Answer» Correct Answer - B
Let `E`= gravitational field at `x` due to the complete sphere `E_(1)`= field due to hole and `E_(2)`=field due to the remanning portion `E=E_(1)+E_(2) :. E_(2)=E-E_(1).....(i)`
`=(GM)/(x^(2))-(Gm)/((x-R/2)^(2))`
Here ,`M=4/3piR^(3)rho_(0)` and `m=4/3(R/2)^(3).rho_(0)`
2671.

A spherical hollow cavity is made in a lead sphere of radius `R` such that is surface touches the outside surface of the lead sphere and passes through its centre. The mass of the sphere before hollowing was `M` with what gravitational force will the hollowed out lead sphere attract a small sphere of mass m which lies at a distacne d from the centre of the lead sphere on the straight line connecting the centeres of the spheres and that of the hollow if `d =2R` .A. `(7GMm)/(18R^(2))`B. `(7GMm)/(36R^(2))`C. `(7GMm)/(9R^(2))`D. `(7GMm)/(72R^(2))`

Answer» Correct Answer - B
2672.

The figure represents a solid uniform sphere of mass `M` and radius `R`. A spherical cavity of radius `r` is at a distance a from the centre of the sphere. The gravitational field inside the cavity is A. non - uniformB. towards the center of the cavityC. directly proportional to `a`D. All of these

Answer» Correct Answer - C
Gravitational field at any point inside the cavity is uniform (both in magnitude as well as direction). So, let us find its value at centre of cavity .
`E_(R ) = E_(T) - E_(C)` (at centre of cavity)
`R rarr `Remaining, `T rarr` total, `C rarr` cavity
`E_(R ) = E_(T)` (as `E_(C) = 0)`
`= ((GM)/(R^(3))) a` or `E_(R) prop a`
2673.

If the gravitational field intensity at a point is given by `g = (GM)/(r^(2.5))`. Then, the potential at a distance `r` isA. `(2GM)/(3r^(1.5))`B. `(-GM)/(r^(3.5))`C. `-(2GM)/(3r^(1.5))`D. `(GM)/(r^(3.5))`

Answer» Correct Answer - A
`V=-intEdr`
2674.

Distinguish between the terms gravitation and gravity.

Answer»

Gravitation: It is the force of attraction between any two bodies in the universe. 

Gravity: It is the force of attraction between the earth and any object lying on or near its surface.

2675.

Define Gravitation and gravity.

Answer»

Gravitation is the force of attraction between any two bodies while gravity refers to the force of attraction between any body and the earth.

2676.

The magnitude of the gravitational field at distance `r_1 and `r_2` from the centre of a unifrom sphere of radius R and mass m are `F_1 and F_2` respectively. Then:A. `(F_1)/(F_2) = (r_1)/(r_2) if r_1 ltR and r_2 ltR`B. `(F_1)/(F_2) = (r_2^2)/(r_1^2) if r_1gt R and r_2 gtR`C. `(F_1)/(F_2) = (r_1)/(r_2) if r_1 gtR and r_2 gtR`D. `(F_1)/(F_2) = (r_1^2)/(r_2^2) if r_1 ltR and r_2 ltR`

Answer» Correct Answer - A::B
(a,b) `For rgt R, the gravitational field is F = (GM)/(r^2)`
`:.F_1 = (GM)/(r_1^2) and F_2 = (GM)/(r_2^2) rArr (F_1)/(r_2^2)/(r_1^2)`
For rltR, the gravitational field is `F = (GM)/(R^3)xxr`
`:.F_1 = (GM)/(R^3) xxr_1 and F_2 = (GM)/(R^3)xxr_2`
`rArr (F_1)/(F_2) - (r_1)/(r_2)`
2677.

A tuning is dug along the diameter of the earth. There is particle of mass `m` at the centre of the tunel. Find the minimum velocity given to the particle so that is just reaches to the surface of the earth. `(R =` radius of earth)A. `sqrt((GM)/(R))`B. `sqrt((GM)/(2R))`C. `sqrt((2GM)/(R))`D. it will reach with the help of negligible velocity

Answer» Correct Answer - A
`K_(i) + U_(i) = K_(f) + U_(f)`
`:. (1)/(2) m nu^(2) - (3)/(2) (GMm)/(R ) = 0 - (GMm)/(R )`
Solving, we get `nu = sqrt((GM)/(R ))`
2678.

A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `R/(n^(2))`B. `R/((1-n^(2))`C. `(Rn^(2))/((1-n^(2)))`D. `Rn^(2)`

Answer» Correct Answer - C
`1/2mv^(2)=(GMm)/R=(GMm)/(R+h)=(GMmh)/(R(R+h))=(mghR)/(R+h)`
Since `v=nv_(3)` and `v_(3)=sqrt(2gR)` hence `=(Rn^(2))/(1+n^(2))`
2679.

A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `(1-k^(2))/R`B. `R/(1-k^(2))`C. `R(1-k^(2))`D. `R/(1+k^(2))`

Answer» Correct Answer - B
Using conservation of energy
`1/2m(Kv_(e))^(2)-(GMm)/R=-(GMm)/(R+h)`
Use `v_(e)=sqrt(2gR)` and `Gm=gR^(2)`
and to solve we get `h+R=R//(1=K^(2))`.
2680.

A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `(I-k^(2))/R`B. `R/(1-k^(2))`C. `R(1-k)^(2)`D. `R/(1+k^(2))`

Answer» Correct Answer - B
If a body is projected from the surface of the earth with a velocity v and reaches a height h, then according to law of conservation of energy
`1/2mv^(2)=(mgh)/(1+h//R)`
Here `v=kv_(e)=ksqrt(2gR)`
Therefore, `1/2mk^(23)gR=(mg(r-R))/(1+(r-R)/R)`
`k^(2)R[1+(r-R)/R]=r-R` or `k^(2)r=r-R`
or `r=(R/(1-k^(2)))`
2681.

A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `(R)/(k^(2))`B. `(2R)/(1-k^(2))`C. `(2R)/(k^(2))`D. `(R)/(1-k^(2))`

Answer» Correct Answer - D
2682.

A projectile is fired from the surface of earth of radius `R` with a velocity `k upsilon_(e)` (where `upsilon_(e)` is the escape velocity from surface of earth and `k lt 1)`. Neglecting air resistance, the maximum height of rise from centre of earth isA. `(R )/(k^(2) - 1)`B. `k^(2) R`C. `(R )/(1 - k^(2))`D. `kR`

Answer» Correct Answer - C
`K_(i) + U_(f) = K_(f) + U_(f)`
`(1)/(2) m (k nu_(e))^(2) - (GMm)/(R ) = 0 - (GMm)/(r )`
`:. (1)/(2)k^(2) (2gR)- (GM)/(R ) = - (GM)/(r)`
or `(GM)/(r ) (1 - k^(2)) = (GM)/(r )` (as `gR = (GM)/(R ))`
So, `r = (R )/(1 - k^(2))`
2683.

An object weights W newton on earth. It is suspended from the lower end of a spring balance whose upper end is fixed to the ceiling of a space capsule in a stable orbit around the earth. The reading of the spring balance will beA. WB. less than WC. more than WD. zero

Answer» Correct Answer - d
Since the object in the space capsule is in a state of weightlessness (or zero gravity), the reading of the spring balance will be zero.
2684.

The height a which the weight of a body becomes `1//16th` its weight on the surface of earth (radius `R`) isA. 3RB. 4RC. 5RD. 15R

Answer» Correct Answer - A
2685.

The height a which the weight of a body becomes `1//16th` its weight on the surface of earth (radius `R`) isA. `5R`B. `15R`C. `3R`D. `4R`

Answer» Correct Answer - C
2686.

The height a which the weight of a body becomes `1//16th` its weight on the surface of earth (radius `R`) isA. `4R`B. `5R`C. `15R`D. `3R`

Answer» Correct Answer - D
`g/16=g/((1+h/R)^(2))`
`implies (1+h/R)=4implies h=3R`
2687.

Four particles each of mass m are placed at the vertices of a square of side l. the potential at the centre of square isA. `-sqrt(32)(GM)/L`B. `-sqrt(64)(GM)/(L^(2))`C. zeroD. `-sqrt(16)(GM)/L`

Answer» Correct Answer - A
`U=-4(GM)/(L//sqrt(2))=-sqrt(32)(GM)/L`
2688.

The earth `E` moves in an elliptical orbit with the sun `S` at one of the foci as shown in figure. Its speed of motion will be maximum at the point A. `C`B. `A`C. `B`D. `D`

Answer» Correct Answer - B
Speed of the earth will be maximum when its distance form the sun is minimum because `mvr`=constant
2689.

Different points in the earth are at slightly different distance from the sun and hence experience different force due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the `CM` (centre of mass) causing translation and a net torque at the `CM` causing rotation around an axis through the `CM`. for the earth-sun system (approximating the earth as a uniform density sphere).A. the torque is zeroB. the torque causes the earth to spinC. the rigid body result is not applicable since the earth is not even approximately a rigid bodyD. the torque causes the earth to move around the sun

Answer» Correct Answer - A
As the earth is revolving around the sun in a circular motion due to gravitational attraction. The force of attraction will be of radial nature i.e. angle between position vector `r` and force `F` is zero. So, torque=`|tau|=|rxxF|=rFsin0^(@)=0`
2690.

To an astronaut is a spcaceship, the sky appearsA. blackB. whiteC. greenD. blue

Answer» Correct Answer - A
2691.

Define gravitational field strength. Which plane in the solar system has the greatest gravitational field strength? What is the gravitational field strength of a planet where a 60 kg astronaut weighs 300 N?

Answer»

The gravitational field strength at a point in a field is defined as the force experienced by a body of unit mass placed at that point.

I = \(\frac {GM}{R^2}\) ⇒ I = g

The value of I is maximum for Saturn.

I = \(\frac {300N}{60kg}\)

I = 5 ms-2.

2692.

What is a geostationary satellite? Write any four applications of artificial satellites.

Answer»

1 Part: A satellite which revolves round the earth with the same angular speed and the same direction as that of earth around its axis is a geostationary satellite. 

Uses of artificial satellites:

1. for mapping of forests (or to study deforestation) 

2. for mapping of rivers, reservoirs 

3. for pollution monitoring. 

4. for weather forecasting, prediction, and warning of cyclones, mapping of flood-affected areas 

5. geostationary satellites are used for telephones, mobiles, broadcasting of radio and television.

2693.

If an artificial satellite is moving in a circular orbit around earth with speed equal to one fourth of `V_(e)` from earth, then height of the satellite above the surface of the earth isA. `7R`B. `4R`C. `3R`D. `R`

Answer» Correct Answer - A
`v_(0)=(v_(e))/4, V_(0)=(sqrt((GM)/(R+h)))`
2694.

A body of mass `m` is lifted up from the surface of earth to a height three times the radius of the earth `R`. The change in potential energy of the body isA. `mgR`B. `(3)/(4)mgR`C. `(1)/(3)mgR`D. `(2)/(3)mgR`

Answer» Correct Answer - B
2695.

What is the orbital period of revolution of an artificial satellite revolving in a geostationary orbit?

Answer»

It is 24 hours.

2696.

A satellite of mass 2000 kg is revolving around the earth in a circular orbit of radius `4xx10^(4)` km. The gravitational force exerted by the satellite on the earth is ________. [Mass of the earth `= 6xx10^(24)kg, G=6.67xx10^(-11)Nm^(2)//kg^(2)`]A. `5xx10^(8)N`B. `5xx10^(3)N`C. `5xx10^(5)N`D. `5xx10^(4)N`

Answer» Correct Answer - A
`F=(GMm)/(d^(2))=5xx10^(8)`
2697.

A ring of mass `m_(1)` and radius R is fixed in space at some location. An external agent brings a point mass `m_(2)` from infinity to centre of the ring. Work done by the external agent will beA. `-(Gm_(1)m_(2))/(R)`B. `(Gm_(1)m_(2))/(R)`C. `(Gsqrt(m_(1)^(2)+m_(2)^(2)))/(R)`D. `(Gm_(1)m_(2))/(R(m_(1)^(2)+m_(2)^(2)))`

Answer» Correct Answer - A
Work done, `W=DeltaU=U_(f)-U_(i)=U_(f)("as " U_(i)=0)`
`rArr W = -(Gm_(1)m_(2)//R)`.
2698.

A satellite can be in a geostationary orbit around earth in an orbit of radius `r`. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit aroun earth radiusA. `(r)/(2)`B. `(r)/(2sqrt2)`C. `(r)/((4)^(1//3)`D. `(r)/((2)^(1//3))`

Answer» Correct Answer - C
2699.

A satellite can be in a geostationary orbit around earth in an orbit of radius `r`. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit aroun earth radiusA. `r/2`B. `r/(2sqrt(2))`C. `r/((4)^(1//3))`D. `r/((2)^(1//3))`

Answer» Correct Answer - C
Angular speed of earth=angular speed of geostationary satellite
`T prop1/(omega)`
`implies (T_(2))/(T_(1))=(omega_(1))/(omega_(2))=1/2 implies T_(2)=(T_(1))/2`
Also, `T prop r^(3//2)`
`:. ((r_(2))/(r_(1)))^(3//2)=(T_(2))/(T_(1))=1/2`
`(r_(2))/r_(1)=(1/2)^(2//3)=1/(4^(1//3))`
`r_(2)=(r_(1))/(4^(1//3))`
2700.

A person brings a mass of `1 kg` from infinty to a point `A`. Initially, the mass was at rest but it moves at a speed of `3 m//s` as it reaches `A`. The work done by the person on the mass is `- 5.5 J`. The gravitational potential at `A` isA. `- 1 J//kg`B. `- 4.5 J//kg`C. `- 5.5 J//kg`D. `- 10 J//kg`

Answer» Correct Answer - D
`W = DeltaE = E_(A) - E_(a)`
`= (K_(A) + U_(A)) - (K_(a) + U_(a))`
`= (K_(A) + mV_(A)) - (0 + 0)`
`:. - 5.5 = (1)/(2) xx 1 xx (3)^(2) + (1) V_(A)`