2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 11 InterviewSolution

501.	Find radius of such planet on which the man escapes through Jumping. The capacity of Jumping of person on earth is `1.5m` Density of planet is same as that of earth .
Answer» For a planet: `(1)/(2) mv^(2) - (GM_(P)m)/(R_(P)) =0 rArr (1)/(2) mv^(2) = (GM_(P)m)/(R_(P))` On earth `rarr (1)/(2) mv^(2) = m ((GM_(E))/(R_(E)^(2))) h` `therefore (GM_(P)m)/(R_(P)) = (GM_(E)m)/(R_(E)^(2))h rArr (M_(P))/(R_(P)) = (M_(E)h)/(R_(E)^(2))` `:.` Density (`rho`) is same `rArr (4//3piR_(P)^(3)rho)/(R_(P)) =(4//3piR_E^2hrho)/R_E^2 rArr sqrt(R_(E)h)` .

Discussion

502.	The potential energy os a body of mass 100 kg is circulating in orbit at a height of 600 km above the surface of earth is (Radius of earth 6400 km, mass of earth `=6xx10^(24)" kg and G"=2//3xx10^(-10)Nm^(2)//kg^(2))`A. `-5.16xx10^(9)J`B. `-5.50xx10^(9)J`C. `-5.70xx10^(9)J`D. `-2.95xx10^(9)J`
Answer» Correct Answer - c `"P.E. "=-("GMm")/((R+h))=-(2xx10^(-10)xx6xx10^(24)xx100)/(3xx7xx10^(6))` `=-0.57xx10^(10)=-57xx10^(9)J`

Discussion

503.	How much work is done in circulating a small object of mass m around a sphere of mass m in a circle of radius R.
Answer» Correct Answer - [0]

Discussion

504.	The gravitational potential due to a mass distribution is `V = 3X^(2) Y+ Y^(3) Z`. Find the gravitational filed.
Answer» Correct Answer - A::B::C `E = - [(delV)/(delX) hati + (delV)/(delY) hatj +(del V)/(delz)hatk]`

Discussion

505.	Distance between the centres of two stars is `10alpha.` The masses of these stars are M and 16M and their radii a and 2a, respectively. A body of mass m is fired straight form the surface of the larger star towards the smaller star. What should be its minimum inital speed to reach the surface of the smaller star? Obtain the expression in terms of G,M and a.
Answer» Correct Answer - `[(3)/(2) sqrt((5Gm)/(a))]`

Discussion

506.	Gravitational potential at `X= 2 m` is decreasing at a rate of `10 J//kg - m` along the positive x- direction. If implies that the magnitude of gravitational filed at `X = 2m` is also `10 n//kg`. Is this statement true or false?
Answer» `\|E\| = sqrt[(-(delV)/(del_(X)))^(2) + (-(delV)/(del_(Y)))^(2)+((-delV)/(del_(Z)))^(2))` `-(delV)/(delx) = 10 J//kg -m` is given No information is given about `- (delV)/(del_(Y))` and `-(delV)/(del_(Z))` So, `\|E\| ge 10 N//kg`

Discussion

507.	An object is dropped from height `h=2 R` on the surface of earth. Find the speed with which it will collide with ground by neglecting effect of air. (Where R is radius of earth, take mass of earth M)
Answer» The initial potential energy of object is `U_(i)=-(GMm)/(3R)` Final potential energy `U_(f)=-(GMm)/(R)` By law of conservation of energy, `Delta KE = - Delta PE` `rArr (1)/(2)mv^(2)=-(U_(f)-U_(i))=U_(i)-U_(f)` `rArr (1)/(2)mv^(2)=-(GMm)/(3R)+(GMm)/(R)` `rArr (1)/(2)v^(2)=(2GM)/(3R)` `rArr v=sqrt((4GM)/(3R))=2sqrt((GM)/(3R))`.

Discussion

508.	Consider the two identical particles shown in the given figure. They are released from rest and may move towards each other influence of mutual gravitational force. Gravitational potential energy of the two particle system A. is zeroB. is contant `(ne 0)`C. decreases as the separation decreasesD. increases as the separation decreases
Answer» Correct Answer - C When particles are released from rest their separation decreases. Therefore graivitational potential energy of the system decreases.

Discussion

509.	For a uniform ring of mass M and radius R at its centreA. field and potential both are zeroB. field is zero but potential is `(GM)/(R)`C. field is zero but potential is `-GM//R`D. magnitude of field is `(GM)/(R^(2))` and potential `-(GM)/(R)`
Answer» Correct Answer - C For a uniform ring of mass M and radius R at it centre, the field is zero but potential is `-(GM)/(R)`.

Discussion

510.	A point mass m is placed inside a spherical shell of radius R and mass M at a distance `R/2` form the centre of the shell. The gravitational force exerted by the shell on the point mass isA. `(GMm)/( R^(2))`B. `(2GMm)/(R^(2))`C. zeroD. `(4Mm)/(R^(2))`
Answer» Correct Answer - C ( c) If a point mass is placed inside a uniform spherical shell, the gravitational force on the point mass is zero. Hence, the gravitational force exerted by the shell on the point mass is zero.

Discussion

511.	By which curve will the variation of gravitational potential of a hollow sphere of radius R with distance be depictedA. B. C. D.
Answer» Correct Answer - C

Discussion

512.	A point mass m is placed inside a spherical shell of radius R and mass M at a distance `R/2` form the centre of the shell. The gravitational force exerted by the shell on the point mass isA. `(2Gm^(2))/(R^(2))`B. `(Gm^(2))/(R^(2))`C. `(Gm^(2))/(2R)`D. zero
Answer» Correct Answer - D Inside spherical shell gravitational field due to shell is zero.

Discussion

513.	By which curve will be variation of gravitational potential of a hollow sphere of radius R with distance be depicted ?A. B. C. D.
Answer» Correct Answer - C For hollow sphere `V_("in")=(-GM)/(R),V_("surface")=(-GM)/(R),V_("out")=(-GM)/(r)` i.e., potential remains constant inside the sphere and it is equal to potential at the surface and increase when he point moves away from the surface of sphere.

Discussion

514.	A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. `gx`B. `(gR^(2))/(R+x)`C. `(gR^(2))/(R+x)`D. `(gR)/(R-x)`
Answer» Correct Answer - B `V_(0)=sqrt((GM)/(R+h))=sqrt((gR^(2))/(R+x))`

Discussion

515.	A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. gxB. `(gR)/(R-x)`C. `(gR^(2))/(R+x)`D. `((gR^(2))/(R+x))^(1//2)`
Answer» Correct Answer - D `(mv^(2))/(R+X)=(GM)/((R+X)^(2))=(gR^(2))/((R+X)^(2))` or `V=[(gR^(2))/(R+X)]^(1//2)`

Discussion

516.	Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. GMm/12RB. GMm/3RC. GMm/8RD. GMm/6R
Answer» Correct Answer - A `TE=-(GMm)/(2r)` `=(-GMm)/(2(3R))-((-GMm)/(2(2R)))=(GMm)/(12R)`

Discussion

517.	Which of the following statements are true? For a particle on the surface of the earth:A. the linear speed is minimum at the equatorB. the angular speed is maximum at the equatorC. the linear speed is minimum at the polesD. the angular speed is `7.3xx10^(-5) rads^(-1)` at the equator
Answer» Correct Answer - C::D Linear speed is `v=romega`. At the equator, the radius of the earth `r` is maximum. Therefore `omega=v/r` is minimum. Also, `omega=(2pi)/T=(2xx3.14)/(24xx60xx60)=7.3xx10^(-5)rads^(-1)` Hence the correct choices are c and d.

Discussion

518.	A small mass `m` is moved slowly from the surface of the earth to a height `h` above the surface. The work done (by an external agent) in doing this isA. `mgh`, for all values of hB. `mgh`, for `hlt ltR`C. `1//2, mgR`, for `h=R`D. `-1//2mgR`, for `h=R`
Answer» Correct Answer - B::C `W=U_(2)-U_(1)=(GMm)/(R+h)(-(GMm)/R)=-GMm[1/R-1/(R+h)]` `=gR^(2)m[h/(R(R+h))]=(mgRH)/(R+h)` For `hlt ltR, W=mgh` for `h=R, W=(mgR)/2`

Discussion

519.	Which of the following are correct?A. An astronaut going from the earth to the Moon will experience weightlessness once.B. When a thin uniform spherical shell gradually shrinks maintaining its shape, the gravitational potential at its centre decreases.C. In the case of a spherical shell, the plot of `V` versus `r` is continuous.D. In the case of a spherical shell, the plot of gravitational field intensity `I` versus `r` is continuous
Answer» Correct Answer - A::B::C a. For a proper explanation , the earth Moon system will be regared as an isolated system. At a particular point in space, the gravitational force of attraction of the earth on astronaut will be balanced by the gravitational force of attraction of the Moon. b. `V=-GM//r` . As `R` decreases, `GM//R` increases or `-GMM//R` decreases. c. In case of the sphereical shell, plot of `I` versus `r` is dicontinous.

Discussion

520.	Treating the earth as a symmetrical sphere of radius `R = 6400 km` with field `9.8 N//kg` at its surface, calcualate the vertical speed with which a rocket should be fired so as to reach a height `4R` from the surface.
Answer» Correct Answer - `10 km//s`

Discussion

521.	What should be the radius of a planet with mass equal to that of earth and escape velocity on its surface is equal to the velocity of light. Given that mass of earth is `M=6xx10^(24)kg`.
Answer» Correct Answer - `9 mm`

Discussion

522.	What is the weight of a body in a geostationary satellite?
Answer» The weight of a body is zero in a geostationary satellite.

Discussion

523.	You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 Kg?
Answer» Even gases (like air) exert an upward force (or buoyant force) on the objects placed in them. Now, when we stand on a weighing machine, then the air exerts an upward force (buoyant force or upthrust) and makes us slightly lighter than we actually are. So, if a weighing machine shows our mass to be 42 kg, then our actual mass will be slightly more than 42 kg.

Discussion

524.	You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Answer» Our mass is more than 42 kg.

Discussion

525.	You find your mass to be 42 kg on a weighing machine. In your mass more or less than 42kg?
Answer» In fact, a weighing machine is a sort of spring balance which measures the weight (and not the mass) of a body. When we stand on the weighing machine, our weight (which is due to gravitational attraction of the Earth) act vertically downwards. But the buoyancy due to air on our body acts vertically upwards. As a result of this, our apparent weight (true weight-buoyancy force) is less than the true weight. since the weighing machine measure the apparent weight, our true weight is more, i.e., more than 42 kg.

Discussion

526.	What is the SI unit of mass ?
Answer» kilogram (kg).

Discussion

527.	A balloon is rising up with an acceleration of 10`m//s^(2)`. A body is dropped from the balloon when its velocity is 200 `m//s`. The body strickes the ground in half a minute. With that velocity did the body hit the ground ? Take `g=9.8m//s^(2)`.
Answer» Initial velocity of body `u=200 m//s` `a=g=9.8 m//s^(2)` `t=(1)/(2)min=30s` `v=?` From `v-u+at` `v=-200+9.8xx30=94 m//s`.

Discussion

528.	What do you mean by acceleration due to gravity?
Answer» In free fall, three is no change in the direction of objects. But the magnitude of velocity of falling objects goes on increasing. It means acceleration is produced in a body in free fall. This acceleration is called acceleration due to gravity.We may define acceleration due to gravity as the acceleration produced in free fall due to gravitational forced of Earth on the body.

Discussion

529.	What is the SI unit of weight ?
Answer» Newton`(N)`.

Discussion

530.	What is inverse square rule ?
Answer» The gravitational fotce of attraction (F) between any two object is inversely proportinal to the square of the distance (r ) between them, i.e.,`F prop 1//r^(2)`.

Discussion

531.	What is the weight of a body of mass 1 kg on the surface pf earth?
Answer» `W=mg=1xx9.8=9.8` newton.

Discussion

532.	Distinguish between gravitational and gravity.
Answer» Gravitation is the phenomenon of attraction between any two object in the universe. Gravity is the phenomenon of attraction between earth and any other body in the universe.

Discussion

533.	What is the ratio pf weight of an object on moon to its weight on earth?
Answer» `("Weight of object on moon")/("Weight of object on earth")=(1)/(6)`

Discussion

534.	What is centripetal force ? What is its function ? Illustrate with one example.
Answer» Centripetal force is the force required to move a body unifornly in a circle. Its function is to change the direction of motion of the body. For example, gravitational force of attraction of earth on the moon provides the neccesary centripetal force for orbiting of moon around earth.

Discussion

535.	The weight of a person on the surface of the Earth is 490 N. Find his weight on the surface of Jupiter and the moon. Also, compare it with his weight on the Earth. Mass of moon `= 7.3 xx 10^(22)` kg, Mass of Jupiter `= 1.96 xx 10^(27)` kg, Radius of moon `= 1.74 xx 10^(6)` m and radius of Jupiter `= 7 xx 10^(7)` m `G = 6.67 xx 10^(-11) N m^(2) kg^(-2)` ltBrgt Given, `(1)/(1.742) = 0.33 "and ge" = 9.8 m s^(-2)`
Answer» The mass of the person is `(W)/(g) = (490)/(9.8) = 50 kg` Acceleration due to gravity on the moon, `g_(m)` is given by, `g_(m) = (GM)/(R^(2))` substituting the values, `g_(m)=(6.67 xx 10^(-11) xx 7.3 xx 10^(22))/((1.74 xx 10^(6))^(2))=161 ms^(-2)` weight on the moon `= mg = 50 xx 1.61 = 80.5 N` the ratio of weight on the moon to earth is `= (80.5)/(490) = 0.16` Acceleration due to gravity on Jupiter is, `(g_(p))` `g_(p) = (GM)/(R^(2))` Substitution the values, `g_(p) = (6.67 xx 10^(-11) xx 1.96 xx 10^(27))/((7 xx 10^(7))^(2))` `= (6.67 xx 1.96 xx 10^(16))/(49 xx 10^(14)) = 26.68 ms^(-2)` weight of person on Jupiter is ` = m xx g_(p) = 50 xx 26.68 = 1334 N` ratio of weight on Jupiter to weight on earth is

Discussion

536.	Calculate mass of earth taking it to ne a sphere of radius 6400 km. given `g=9.8 m//s^(2)` and `G=6.67xx10^(-11)N m^(2)kg^(-2)`
Answer» Here, `R=6400 km=6.4xx10^(6)m, g=9.8m//s^(2)` and `G=6.67xx10^(-11)Nm^(2)kg^(-2)` From `g=(GM)/(R^(2))` `M=(g=R^(2))/(G)=(9.8(6.4xx10^(6))^(2))/(6.67xx10^(-11)) kg=6xx10^(24)kg`

Discussion

537.	If earth is taken as a sphere of radius 6400 km and mass `6xx10^(24)` kg, what would be the value of g on the surface of earth ?
Answer» Here, `R=6400 km=6.4xx10^(6)m` `M=6xx10^(24) kg` As `g=(GM)/R^(2)=(6.67xx10^(-11)xx6xx10^(24))/(6.4xx10^(6))^(2)=9.8m//s^(2)`

Discussion

538.	Why doesn’t the leaning tower of Pisa topple over?
Answer» The centre of gravity of the leaning tower of Pisa is very close to earth. Even though it is leaning, the line of action of total weight passes through the base. The base area of leaning tower of Pisa is very large. So it doesn’t topple over.

Discussion

539.	Height of a weather satellite: `8.7m//s^(2)` :: Height of Communication satellite:_______________.
Answer» `0.225m//s^(2)` The value of g changs with the change in the height of the satellites.

Discussion

540.	Mass: Scalar quantity :: Weight:_______________.
Answer» Vector quantity-Weight being a force is a vector quantity and its direction is towards the centre of the earth.

Discussion

541.	When an objects is thrown upward, the force of gravity_____________.A. is opposite to the direction of motionB. is in the same direction as that of motionC. becomes zero at higher pointD. increase as it rise up
Answer» Correct Answer - A Is opposite to the direction of motion

Discussion

542.	The mass of objects_______________ at any place on the surface on the Earth.A. remains constantB. is non-unifromC. changesD. increases
Answer» Correct Answer - A remains constant

Discussion

543.	An artificial satellite is at a height of 36,500 km above earth’s surface. What is the work by earth’s gravitational force in keeping it in its orbit?
Answer» No energy is required by a satellite to keep it orbiting because the work done by the centripetal force is zero.

Discussion

544.	Give on example each central force and non-central force.
Answer» Examples of Central force: gravitational force of a point mass, electrostatic force due to a point change. Example of Non-central force: spin-dependent nuclear forces, magnitude force between two current carrying loops.

544.

Give on example each central force and non-central force.

Answer»

Examples of

Central force: gravitational force of a point mass, electrostatic force due to a point change.

Example of

Non-central force: spin-dependent nuclear forces, magnitude force between two current carrying loops.

Discussion

545.	Figure shows the variation of energy with the orbit radius r of a satellite in a circular motion. Mark the correct statement. A. C shows the total energy, B the kinetic energy and A the potential energy of the satelliteB. A shows the kinetic energy, B the total energy and C the potential energy of the satelliteC. A and B are the kinetic and potential energies and C the total energy of the satelliteD. C and A are the kinetic and potential energies respectively and B the total energy of the satellite
Answer» Correct Answer - C

Discussion

546.	If the radius of the earth were increased by a factor of 2 keeping the mass constant, by what factor would its density have to be changed to keep g the same?A. `(1)/(8)`B. 4C. `(1)/(2)`D. `(1)/(4)`
Answer» Correct Answer - C

Discussion

547.	A planet is moving in an elliptical path around the sun as shown in figure. Speed of planet in positions P and Q are `V_(1)` and `V_(2)` respectively with `SP=r_(1)` and `SQ=r_(2)` , then `v_(1)//v_(2)` is equal to A. `(r_(1))/(r_(2))`B. `(r_(2))/(r_(1))`C. `((r_(2))/(r_(1)))^(2)`D. `((r_(1))/(r_(2)))^(2)`
Answer» Correct Answer - B

Discussion

548.	What is the direction of areal velocity of the earth around the sun?
Answer» Area velocity of earth around the sun is – \(\frac{dA}{dt}\)= \(\frac{L}{2m}\) [L = angular momentum, m = mass of earth] But angular momentum, L = \(\vec{r}\times\vec{p}\) = \(\vec{r}\times m{\vec{v}}\) Areal velocity, \(\big(\frac{dA}{dt}\big)=\frac{1}{2m}\)\((\vec{r}\times m\vec{v})\) = \(\frac{1}{2}(\vec{r}\times \vec{v})\) Therefore, the direction of \((\vec{r}\times \vec{v})\) areal velocity is in direction of i.e. perpendicular to the plane of \(\vec{r}\) and \(\vec{v}\) (as by Maxwell’s right hand grip rule).

548.

What is the direction of areal velocity of the earth around the sun?

Answer»

Area velocity of earth around the sun is – \(\frac{dA}{dt}\)= \(\frac{L}{2m}\)

[L = angular momentum, m = mass of earth]

But angular momentum, L = \(\vec{r}\times\vec{p}\)

= \(\vec{r}\times m{\vec{v}}\)

Areal velocity, \(\big(\frac{dA}{dt}\big)=\frac{1}{2m}\)\((\vec{r}\times m\vec{v})\)

= \(\frac{1}{2}(\vec{r}\times \vec{v})\)

Therefore, the direction of \((\vec{r}\times \vec{v})\) areal velocity is in direction of i.e. perpendicular to the plane of \(\vec{r}\) and \(\vec{v}\) (as by Maxwell’s right hand grip rule).

Discussion

549.	The potential energy of a body mass m is `U=ax+by` the magnitude of acceleration of the body will be-A. `(ab)/(m)`B. `((a+b)/(m))`C. `(sqrt(a^(2)+b^(2)))/(m)`D. `(a^(2)+b^(2))/(m)`
Answer» Correct Answer - C

Discussion

550.	Two small balls of mass m each are suspended side by side by two equal threds to length L. If the distance between the upper ends of the threads be a, the angle `theta` that the threads will make with the vertical due to attraction between the balls is : A. `"tan"^(-1)((a-x)g)/(mG)`B. `"tan"^(-1)(mG)/((a-x)^(2)g)`C. `"tan"^(-1)((a-x)^(2)g)/(mG)`D. `"tan"^(-1)((a^(2)-g^(2))g)/(mG)`
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

Find radius of such planet on which the man escapes through Jumping. The capacity of Jumping of person on earth is `1.5m` Density of planet is same as that of earth .

The potential energy os a body of mass 100 kg is circulating in orbit at a height of 600 km above the surface of earth is (Radius of earth 6400 km, mass of earth `=6xx10^(24)" kg and G"=2//3xx10^(-10)Nm^(2)//kg^(2))`A. `-5.16xx10^(9)J`B. `-5.50xx10^(9)J`C. `-5.70xx10^(9)J`D. `-2.95xx10^(9)J`

How much work is done in circulating a small object of mass m around a sphere of mass m in a circle of radius R.

The gravitational potential due to a mass distribution is `V = 3X^(2) Y+ Y^(3) Z`. Find the gravitational filed.

Gravitational potential at `X= 2 m` is decreasing at a rate of `10 J//kg - m` along the positive x- direction. If implies that the magnitude of gravitational filed at `X = 2m` is also `10 n//kg`. Is this statement true or false?

An object is dropped from height `h=2 R` on the surface of earth. Find the speed with which it will collide with ground by neglecting effect of air. (Where R is radius of earth, take mass of earth M)

For a uniform ring of mass M and radius R at its centreA. field and potential both are zeroB. field is zero but potential is `(GM)/(R)`C. field is zero but potential is `-GM//R`D. magnitude of field is `(GM)/(R^(2))` and potential `-(GM)/(R)`

A point mass m is placed inside a spherical shell of radius R and mass M at a distance `R/2` form the centre of the shell. The gravitational force exerted by the shell on the point mass isA. `(GMm)/( R^(2))`B. `(2GMm)/(R^(2))`C. zeroD. `(4Mm)/(R^(2))`

By which curve will the variation of gravitational potential of a hollow sphere of radius R with distance be depictedA. B. C. D.

A point mass m is placed inside a spherical shell of radius R and mass M at a distance `R/2` form the centre of the shell. The gravitational force exerted by the shell on the point mass isA. `(2Gm^(2))/(R^(2))`B. `(Gm^(2))/(R^(2))`C. `(Gm^(2))/(2R)`D. zero

By which curve will be variation of gravitational potential of a hollow sphere of radius R with distance be depicted ?A. B. C. D.

A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. `gx`B. `(gR^(2))/(R+x)`C. `(gR^(2))/(R+x)`D. `(gR)/(R-x)`

A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite isA. gxB. `(gR)/(R-x)`C. `(gR^(2))/(R+x)`D. `((gR^(2))/(R+x))^(1//2)`

Energy required to move a body of mass m from an orbit of radius 2R to 3R isA. GMm/12RB. GMm/3RC. GMm/8RD. GMm/6R

Which of the following statements are true? For a particle on the surface of the earth:A. the linear speed is minimum at the equatorB. the angular speed is maximum at the equatorC. the linear speed is minimum at the polesD. the angular speed is `7.3xx10^(-5) rads^(-1)` at the equator

A small mass `m` is moved slowly from the surface of the earth to a height `h` above the surface. The work done (by an external agent) in doing this isA. `mgh`, for all values of hB. `mgh`, for `hlt ltR`C. `1//2, mgR`, for `h=R`D. `-1//2mgR`, for `h=R`

Treating the earth as a symmetrical sphere of radius `R = 6400 km` with field `9.8 N//kg` at its surface, calcualate the vertical speed with which a rocket should be fired so as to reach a height `4R` from the surface.

What should be the radius of a planet with mass equal to that of earth and escape velocity on its surface is equal to the velocity of light. Given that mass of earth is `M=6xx10^(24)kg`.

What is the weight of a body in a geostationary satellite?

You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 Kg?

You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

You find your mass to be 42 kg on a weighing machine. In your mass more or less than 42kg?

What is the SI unit of mass ?

A balloon is rising up with an acceleration of 10`m//s^(2)`. A body is dropped from the balloon when its velocity is 200 `m//s`. The body strickes the ground in half a minute. With that velocity did the body hit the ground ? Take `g=9.8m//s^(2)`.

What do you mean by acceleration due to gravity?

What is the SI unit of weight ?

What is inverse square rule ?

What is the weight of a body of mass 1 kg on the surface pf earth?

Distinguish between gravitational and gravity.

What is the ratio pf weight of an object on moon to its weight on earth?

What is centripetal force ? What is its function ? Illustrate with one example.

Calculate mass of earth taking it to ne a sphere of radius 6400 km. given `g=9.8 m//s^(2)` and `G=6.67xx10^(-11)N m^(2)kg^(-2)`

If earth is taken as a sphere of radius 6400 km and mass `6xx10^(24)` kg, what would be the value of g on the surface of earth ?

Why doesn’t the leaning tower of Pisa topple over?

Height of a weather satellite: `8.7m//s^(2)` :: Height of Communication satellite:_______________.

Mass: Scalar quantity :: Weight:_______________.

When an objects is thrown upward, the force of gravity_____________.A. is opposite to the direction of motionB. is in the same direction as that of motionC. becomes zero at higher pointD. increase as it rise up

The mass of objects_______________ at any place on the surface on the Earth.A. remains constantB. is non-unifromC. changesD. increases

An artificial satellite is at a height of 36,500 km above earth’s surface. What is the work by earth’s gravitational force in keeping it in its orbit?

Give on example each central force and non-central force.

If the radius of the earth were increased by a factor of 2 keeping the mass constant, by what factor would its density have to be changed to keep g the same?A. `(1)/(8)`B. 4C. `(1)/(2)`D. `(1)/(4)`

What is the direction of areal velocity of the earth around the sun?

The potential energy of a body mass m is `U=ax+by` the magnitude of acceleration of the body will be-A. `(ab)/(m)`B. `((a+b)/(m))`C. `(sqrt(a^(2)+b^(2)))/(m)`D. `(a^(2)+b^(2))/(m)`