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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
Which of the following does not possess the ability to do work not because of motion? a) a sparrow flying in the sky b) a sparrow moving slowly on the ground c) a sparrow in the nest on a tree d) a squirrel going up a tree |
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Answer» The correct answer is c) a sparrow in the nest on a tree |
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| 302. |
Two sphere of masses `m` and `M` are situated in air and the gravitational force between them is `F`. The space around the masses in now filled with a liquid of specific gravity `3`. The gravitational force will now beA. `F/9`B. `3F`C. `F`D. `F/3` |
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Answer» Correct Answer - C Gravitational force is independent of the medium. Hence, this will remain the same. |
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| 303. |
A satellite of the earth is revolving in a circular orbit with a uniform speed `v`. If the gravitational force suddenly disappears, the satellite willA. Continue to move with velocity `v` along the original orbitB. Move with a velocity `v` tangentially to the original orbitC. Fall down with increases velocityD. Ultimately come to rest somewhere on the original orbit |
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Answer» Correct Answer - B Due to inertia of direction. |
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| 304. |
Two sphere of masses `m` and `M` are situated in air and the gravitational force between them is `F`. The space around the masses in now filled with a liquid of specific gravity `3`. The gravitational force will now beA. `F`B. `F//3`C. `F//9`D. `3F` |
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Answer» Correct Answer - A Gravitational force is indpendent of medium between the particles. |
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| 305. |
Three equal masses of 1 kg each are placed at the vertices of an equilateral `Delta PQR` and a mass of 2kg is placed at the centroid O of the triangle which is at a distance of `sqrt(2)`m from each the verticles of the triangle. Find the force (in newton) acting on the mass of 2 kg.A. 2B. `sqrt(2)`C. 1D. zero |
| Answer» Correct Answer - D | |
| 306. |
Two masses of mass m each are fixed at a separation distance of `2d`. A small mass `m_(s)` placed midway, when displaced slightly, starts oscillating. Then : A. Frequency of simple harmonic motion is given by `1/(2pi)sqrt((4Gm)/(d^(3)))`B. Frequency of simple harmonic motion is given by `1/(2pi)sqrt((2Gm)/(d^(3)))`C. Accleration of the mass `m_(s)`, is given by `(Gm)/(d^(2))`D. Time period of vibration is `2pisqrt((GM)/((2d)^(3))` |
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Answer» Correct Answer - B |
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| 307. |
Mass M is uniformly distributed only on curved surface of a thin hemispherical shell. A, B and C are three points on the circular base of hemisphere, such that A is the centre. Let the gravitational potential at poins A, B and C be `V_(A), V_(B), V_(C)` respectively. Then : A. `V_(C) gt V_(B) gt V_(C)`B. `V_(C) gt V_(B) gt V_(A)`C. `V_(B) gt V_(A)` and `V_(B) gt V_(A)`D. `V_(A) = V_(B) = V_(C)` |
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Answer» Correct Answer - D |
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| 308. |
A body weighs 10 kg on earth, where `g=9.8m//s^(2)`. What would be its mass and weight on moon, where `g=1.6m//s^(2)` ? |
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Answer» Mass on moon=mass on earth=10kg weight on moon =`mg=10xx1.6=16N` |
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| 309. |
An object weighs 60 N at the surface of earth. How much will it weigh at the surface of the moon ? `["Given" g_(m)=(1)/(6)g_(e)]`A. 1NB. 10NC. 6ND. 360N |
| Answer» Correct Answer - B | |
| 310. |
A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? |
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Answer» Given, Force = 20 N, Weight W = 9.8 N We know, W = mg So, 9.8 = m × 9.8 Or m = 1 kg And, F = ma So, 20 = 1 × a Or a = 20 m/s2 |
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| 311. |
A force of 20N acts upon a body whose weight is 9.8N. What is the mass of the body and how much is its acceleration? |
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Answer» Weight = 9.8N W = mg m = 1kg Force, F = ma a = 20 m/s2 |
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| 312. |
Can a body have mass but no weight ? Give reasons for your answer. |
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Answer» Yes, weight of a body is not constant, it varies with the value of acceleration due to gravity, g. Weight of a body is zero, when it is taken to the centre of the earth or in the interplanetary space, where g=0. |
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| 313. |
A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration ? (g = 9.8 m s-2 ). |
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Answer» Weight= 9.8N W= m x g 9.8 =m x 9.8 m= 1kg Force, F= mass x acceleration 20 N = 1kg x a Acceleration, a=20m/s2 |
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| 314. |
A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone ? What is its mass ? (g = 10 m/s2 ). |
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Answer» Weight of the stone = Gravitational force acting on it = 20 N Weight, W= m x g 20=m x 10 m=2 kg |
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| 315. |
What does low values of escape velocity indicate? |
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Answer» Little probability of atmosphere on the planet. |
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| 316. |
Mass of an object is 10 kg. What is its weight on the earth? |
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Answer» m = 10 kg, g = 9.8 m/s2. W = mg = 10 × 9.8 = 98 N. m = 10 kg, g = 9.8 m/s2.W = mg = 10 × 9.8 = 98 N |
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| 317. |
How does the force of gravitation between two objects change when the distance between them is reduced to half ? |
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Answer» The force of gravitation between two objects is inversely proportional to the square of distance between them. That is, `F prop (1)/(r^(2))`.Now when the distance between two objects is reduced to half ,that is made `1/2` then the force between them will becomes 4 times `["because"((1)/(1/2))^(2)=4]`. |
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| 318. |
What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 1023 kg and radius of the earth 6.4 x 106m) |
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Answer» Gravitational force between the earth and an F = \(\frac{GMm}{R^2}\) |
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| 319. |
Gravitational force acts on all objects in proportion, to their masses. Why then a heavy object does not fall faster than a light object? |
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Answer» Each object falls towards the earth with the acceleration equal to acceleration due to gravity, which is constant (9.8ms-2) and does not depend on the mass of the object. So heavy object does not fall faster than the light object. |
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| 320. |
Two particle of mass m and M are initially at rest at inifinte distance . Find their relative velocity of approch due to grvitational atraction when d is their separation at any instant . [Hint : From the principle of conservation of energy `(GmM)/(d) =(1)/(2)mv_(1)^(2) +(1)/(2)v_(2)^(2).` . From the principle of conservationl of momentum , `mv_(1)-Mv_(2)=0` . Relative velocity of approch =` V_(1)+V_(2)`] |
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Answer» Correct Answer - `sqrt(2G(m+M)//d)` |
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| 321. |
Gravitational force that acts on all objects is proportional to their masses. It is when a heavy object does not fall faster than a light object. |
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Answer» The acceleration due to gravity attained by an object due to gravitational force does not depend on mass of the object. So heavy object and lighter object falls at a time on earth. |
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| 322. |
Gravitational force acts on all objects in properties to their masses. Why then, a heavy object does not fall faster than a light object? |
| Answer» Acceleration `=("force")/("mass")` and force prope mass, therefo re,acceleration is constant/same for heavy and light objects. | |
| 323. |
Two heavy sphere each of mass `100 kg` and radius `0.10 m` are placed `1.0 m` apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the sphere? Is an object placed at that point in equilibrium? If so, is the equilbrium stable or unstable. |
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Answer» Here `G=6.67xx10^(-11)Nm^(2)kg^(-2)M=100kg,R=0.1M`. Distance between the two spheres, d=1.0M Suppose that the distance of either sphere from the mid point of the line joining their centre is r. then r=d/2=0.5m. The gravitational field at the mid-point due to two spheres will be eual and opposite. |
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| 324. |
Mass of 20 kg is distributed uniformly over a ring of radius 2m. Find the the grvitational field at a point lies an th axis of the ring at a distance of `2 sqrt(3)` m from the centre. |
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Answer» Given, Mass of the ring, `M=2kg` Radius of the ring, `R = 2m` Distance of the point, `r = 2sqrt(3)m` `:.` Required gravitational field `E(r=2sqrt(3))=(GM)/((R^(2)+r^(2))^(3//2))=(Gxx2)/((sqrt(4+12))^(3))` `=(2G)/(64)=(G)/(32)=(6.67 xx10^(-11))/(32)"Nkg"^(-1)` `=0.21 xx10^(-11)"Nkg"^(-1)=2.1xx10^(-12)"Nkg"^(-1)`. |
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| 325. |
A heavy point of mass 15 kg is placed at the origin of coordinate axis. Find the gravitational potential at a point located at `x=5 cm` on X-axis. |
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Answer» Given, `M=15` kg Gravitation potential at `x=5 m` `V=(-GM)/(x)=(-6.67xx10^(-11)xx15)/(5)=-20.01 xx10^(-11)"J kg"^(-1)`. |
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| 326. |
The curves for potential energy ( U ) and kinetic energy `(E_(k))` of a two particle system are shown in figure. At what points the system will be bound? A. Only at point DB. Only at point AC. At point D and AD. At points A , B and C |
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Answer» Correct Answer - D |
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| 327. |
Figure shows the kinetic energy `(E_(k))` and potential energy (`E_(p)`) curves for a two-particle system. Name the point at which the system is a bound system. A. `A`B. `B`C. `C`D. `D` |
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Answer» Correct Answer - A::B::C::D For all the points `E_(P)gtE_(K)` (numerically), So, the total energy is negative. Thus, The system is a bound system corresponding to all the points. |
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| 328. |
The magnitude of the Sun’s gravitational field as experienced by Earth is ….. (a) same over the year (b) decreases in the month of January and increases in the month of July (c) decreases in the month of July and increases in the month of January (d) increases during day time and decreases during night time. |
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Answer» (c) decreases in the month of July and increases in the month of January |
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| 329. |
The radius of the earth is 6000 km . What will be the weight of a 120 kg body if it is taken to a height of 2000 km above the surface of the earth? |
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Answer» Correct Answer - 67.5 kg f |
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| 330. |
Where will it be profitable to purchase 1 kilogram sugarA. At polesB. At equatorC. At `45^(@)` latitudeD. At `45^(@)` latitude |
| Answer» Correct Answer - b | |
| 331. |
Find the self gravitationl potential of (a) a thin unifrom spherical shell of radius R snd mass M, (b) a solid sphere of the same radius and mass . |
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Answer» Correct Answer - `-(GM^(2))/(2R), -(3)/(5) (GM^(2))/(R)` |
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| 332. |
Radius of earth is around 6000 km . The weight of body at height of 6000 km from earth surface becomesA. HalfB. One-fourthC. One thirdD. No change |
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Answer» Correct Answer - B |
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| 333. |
Where will it be profitable to purchase 1 kilogram sugarA. At polesB. At equatorC. At `45^(@)` latitudeD. At `40^(@)` latitude |
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Answer» Correct Answer - B |
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| 334. |
Derive an expression fro the gravitational field due to a unifrom rod of length L and M at a point on its perpendicular bisector at a distance d from the centre. |
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Answer» Correct Answer - B::D A small section of rod is condered at x distance mass of the element `=(M/Lxxdx=dm)` `=dE_1=(G(dm)xx1)/((d^2+x^2))=dE_2` Resultant `eE=2dE_1 sintheta` `2xx(G(dm))/(d^2+x^2)xxd/(sqrt(d^2+x^2))` `=(2xxGMxxd dx)/(L(d^2+x^2({(sqrt(d^2+x^2))})` Total gravitatioN/Al field `E=int_0^(L/2) (2Gmd dx)/(L(d^2+x^2)^(3/2))` integrating the above the equation it can be found that `E=(2Gm)/(dsqrt(L^2+4d^2)0` |
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| 335. |
A planet is revolving around the sun in an elliptical orbit. Which out of the following remains constant. (a) Linear speed (b) angular momentum ( c) kinetic energy (d) potential energy (e) total energy throughout its orbit.A. linear speedB. angular momentum about the sunC. kinetic energyD. potential energy |
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Answer» Correct Answer - B In elliptical orbit sun is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet of do not remain constant |
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| 336. |
A particle of mass m is placed at the centre of a unifrom spherical shell of mass 3 m and radius R The gravitational potential on the surface of the shell is .A. `(GM)/( R)`B. `-(GM)/ ( R)`C. `-(4GM)/(R )`D. `-(2GM)/(R )` |
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Answer» Correct Answer - C ( c) Gravitational potential on the surface of the shell is V=Gravitational potential due to particle `(V_(1))`+ Gravitational potential due to shell itself `(V_(2))` `=-(GM)/( R)+(-(G3m)/(R ))= - (4Gm)/(R )` |
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| 337. |
Assertion: Generally the path of projectile form the earth is parabolic but it is elliptical for projection going to a very large height. Reason: The path of projectile is independent of the gravitational force of earth.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false. |
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Answer» Correct Answer - C Up to ordinary heights, the change in the distance of a projectile form the centre of earth is negligible compared to the raduis of earth. Hence the projectile moves under a nearly uniform gravitational force and the path is parabolic. But for the projectiles moving to a large height, the gravitational force, the path of the projectile becomes elliptical. |
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| 338. |
If a satellite is revolving around a planet of mass Min an elliptical orbit of semi-major axis a. Show that the orbital speed of the satellite when it is at a distance are from the planet will be given by `v^(2)=GM[2/r-1/a]` |
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Answer» In elliptical orbit, total energy `E=-(GMm)/(2a)=` constant At position `r`,if `v` is orbital speed of satellite the `KE=1/2mv^(2)` and `PE =(GMm)/r` `:. 1/2mv^(2)-(GMm)/r=-(GMm)/(2a)impliesv^(2)=GM(2/r-1/a)` |
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| 339. |
A satellite with kinetic energy `E_(k)` is revolving round the earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer spaceA. `E_(k)`B. `2E_(k)`C. `1/2E_(k)`D. `3E_(k)` |
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Answer» Correct Answer - A Binding energy =-kinetic energy and if this amount of energy `(E_(k))` given to satellite then it will escape into outer space. |
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| 340. |
A satellite with kinetic energy `E_(k)` is revolving round the earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer spaceA. `2E_(K)`B. `3E_(k)`C. `E_(K)`D. `E_(K//2)` |
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Answer» Correct Answer - C `K.E.=(1)/(2)mV_(0)^(2), K.E.` for escape `=(1)/(2)mV_(e)^(2)` `=(1)/(2)m(sqrt(2)V_(0))^(2)=2((1)/(2)mV_(0)^(2))` |
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| 341. |
The energy of a satellite revolving round the earth in a circular orbit is increased but keeping its energy less than zero, then the average radius of the new orbit of the satellite willA. increaseB. decreaseC. remain sameD. increase decrease depending on the direction of rotation of satellite |
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Answer» Correct Answer - a It may jump into higher orbit (or) it may attain elliptical orbit. |
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| 342. |
A satellite with kinetic energy `E_(k)` is revolving round the earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer spaceA. `E_(k)`B. `2 E_(k)`C. `1/2 E_(k)`D. `3 E_(k)` |
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Answer» Correct Answer - A |
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| 343. |
Two earth-satellite are revolving in the same circular orbit round the centre of the earth. They must have the sameA. massB. angular momentumC. kinetic enrgyD. velocity |
| Answer» Correct Answer - D | |
| 344. |
A hole is drilled through the earth along a diameter and stone is dropped into it. When the stone is at the centre of the earth, it has finite (a) weight (b) acceleration(C) Potential Energy (D) MassA. `a&b`B. `b&c`C. `a,b&c`D. `c&d` |
| Answer» Correct Answer - D | |
| 345. |
A hole is drilled from the surface of earth to its centre. A particle is dropped from rest in the surface of earth in terms of its escape velocity on the surface of earth `v_(e)` is :A. `(v_(e))/(2)`B. `v_(e)`C. `sqrt(2)v_(e)`D. `(v_(e))/(sqrt(2))` |
| Answer» Correct Answer - D | |
| 346. |
The figure shows a spherical shell of mass `M`. The point `A` is not at the centre but away from the centre of the shell. If a particle of mass `m` is placed at `A`, then A. The particle will move towards the centre.B. The particle will move away from the centre, towards the nearest wall.C. The particle will move towards the centre if `m lt M` and away from the centre if `m lt M`D. The particle will remain stationary |
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Answer» Correct Answer - D |
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| 347. |
The time perio of revolution of geo-stationary satellite with respect to earth isA. `24 hrs`B. `1yr`C. infinityD. zero |
| Answer» Correct Answer - C | |
| 348. |
Water stored in a dam possesses: a) no energy b) electrical energyc) kinetic energy d) potential energy |
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Answer» The correct answer is d) potential energy |
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| 349. |
Why do we feel light on our feet when standing in a swimming pool with water up to our armpits ? |
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Answer» As more and more volume of our body is immersed in water, the apparent weight of the body goes on decreasing and the body seems to become lighter. This is due to the increase in upward buoyant force acting on the body. |
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| 350. |
Two satellites `A` and `B` revolve around a plant in two coplanar circular orbits in the same sense with radii `10^(4) km` and `2 xx 10^(4) km` respectively. Time period of `A` is `28` hours. What is time period of another satellite? |
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Answer» Correct Answer - B `T prop r^(3//2)` `:. ((T_(2))/(pi)) = ((r_(2))/(r_(1)))^(3//2)` or `T_(2) = ((r_(2))/(r_(1)))^(3//2) T_(1)` `=((2 xx 10^(4))/(10^(4)))^(3//2) (28)= 56sqrt(2) h` |
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