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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.
| 701. |
If R is the radius of a planet and g is the acceleration due to gravity, then the mean density of the planet is given byA. `(3g)/(4pi GR)`B. `(4piGR)/(3g)`C. `(4piGR)/(3G)`D. `(3G)/(4piGR)` |
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Answer» Correct Answer - a `g=(GM)/(R^(2))=(Grho)/(R)(4pi)/(3)R^(3)` `g=Grho(4pi)/(3)R` `rho=(3g)/(G4pi R)` |
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| 702. |
There is no atomosphere on moon becauseA. it is close to the earthB. its surface temperature is `-10^(@)C`C. the escape velocity of gas molecules is less than their rms velocity on the moonD. the escape velocity of the gas molecules is more than their rms velocity on the moon |
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Answer» Correct Answer - C There is no atmosphere on the moon because the escape velocity of the gas molecules is less than their rms velocity on the moon. [if `v_("rms")gt v_(e)`, the gas molecules will escape.] |
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| 703. |
The escape velocity of a body from the surface of the earth is expressed in terms of density and diameter of the earth, it is found that it isA. directly proportional to the mass of the bodyB. directly proportional to the diameter of the earthC. inversely proportional to the diameter of the earthD. inversely proportional to the density of the earth |
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Answer» Correct Answer - B For the escapvelocity of a body from the surface of the earth is `v_(e)=sqrt((2GM)/(R ))=sqrt(2G.(4)/(3)(pi R^(3)rho)/(R ))` `= sqrt((4R^(2).(2piG rho))/(3))=2R sqrt((2pi G rho)/(3))` `therefore v_(e)prop 2R` or the diameter of the earth. |
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| 704. |
The acceleration due to gravity on the surface of the earth is g. If a body of mass m is raised from the surface of the earth to a height equal to the radius R of the earth, then the gain in its potential energy is given byA. mgRB. 2 mgRC. `(1)/(2) mgR`D. `(1)/(4)mgR` |
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Answer» Correct Answer - C P.E. on the surface `=-(GMm)/(R )` P.E. at a height `h =R =-(GMm)/(R+R)` `therefore` Gain in P.E. `=-(GMm)/(2R)+(GMm)/(R )=(GMm)/(2R)` But `GM=gR^(2)therefore` Gain `=(gR^(2)m)/(2R)=(1)/(2)gmR` |
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| 705. |
Explain the term gravitational force. What is gravitation? |
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Answer» There exists a force of attraction between any two particles of matter in the universe such that the force depends only on the masses of the particles and the separation between them. It is called the gravitational force and the mutual attraction is called gravitation. |
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| 706. |
Identify the law shown in and state the three respective laws.(Schematic diagram) |
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Answer» (a) From the given description we understand Kepler’s three laws. (b) Kepler’s law of areas: The line joining the planet and the Sun sweeps equal areas in equal intervals of time. (c) Kepler’s law of periods: The square of the period of revolution or a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun. |
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| 707. |
A body, the mass and the weight of which were already determined at the Equator, is now placed at the Pole. In this context, choose the correct statement from thefollowing:(a) Mass does not change, weight is maximum (b) Mass does not change, weight is minimum (c) Both mass and weight are maximum (d) Both mass and weight are minimum |
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Answer» Answer is (a) Mass does not change, weight is maximum |
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| 708. |
If this body is allowed to fall freely, will there be any change in the force experienced by the body? |
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Answer» Weight lessens occurs |
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| 709. |
What is the weight of a body of mass 10 kg? |
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Answer» Weight F = mg = 10 × 9.8 = 98 N |
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| 710. |
A cricket ball is dropped from a height of 20 metres. (a) Calculate the speed of the ball when it hits the ground. (b) Calculate the time it takes to fall though this height `(g=10 m//s^(2))` |
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Answer» Here ,Initial speed u=0 Final speed v=? (to be calculated ) Acceleration due to gravity `g=10m//s^(2)` (Ball comes down) height h=20 m Now we know that for a freely falling body `v^(2)=u^(2)+2gh` `v^(2)=(0)^(2)+xx10 xx20` `v^(2)=400` `v=sqrt(400)` v=m//s Thus the speed of cricket ball when it hits the ground will be 20 meters per second (b ) Now , Intial speed u=0 n Final speed v=20 m/s (calculated above ) Acceleration due to gravity `g=10 m//s^(2)` And Time,t =? (to be calculated ) Putting these values in the formula : we v=u+gt `20 =0+10 xx t` 10 t= 20 ltnrgt `t=20/10` t= 2 s Thus the ball takes 2 second to fall through a height of 20 metres . |
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| 711. |
Suppose gravity of earth suddenly become zero, then in which direction will the moon begin to move if no other celestial body affects it? |
| Answer» The moment the gravity of earth become zero, centripetal force required by the moon to move in circular path around earth is no longer provided. The moon will, therefore, move along the tangent to the circular orbit at that instant. | |
| 712. |
Why does the moon not fall to the ground? |
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Answer» The gravitational force in the moon due to earth is equal to the gravitational force on the earth due to moon. Hence the moon does not fall to the ground. |
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| 713. |
How do you feel during the free-fall of your body from the height? |
Answer»
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| 714. |
As a car speeds up when rounding a curve, does its centripetal acceleration increase? Use an equation to defend your answer. |
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Answer» Centripetal acceleration ac = \(\frac{v^2}{r}\) As v increases, its centripetal acceleration also increases. |
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| 715. |
What is the gravitational force on a body inside a spherical shell? Why? |
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Answer» Inside the shell, the net gravitational force is zero. This is because there is no mass inside, the gravitational field is zero, thereby force is zero. |
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| 716. |
A planet’s radius is reduced by 5%, with its mass unchanged. What is the percentage change in ‘g’? |
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Answer» We know that g ∝ \(\frac {1}{R^2}\) So, \(\frac {Δg}{g}\) = – 2 \(\frac {Δg}{R}\) ⇒ \(\frac {Δg}{g}\) = – 2(5%) = – 10% |
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| 717. |
Define gravitational field strength. Find the field at a point distance ‘x’ form a mass ‘m1‘. |
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Answer» The gravitational field is defined as the gravitation force experienced by an object of unit mass at any point. The Gravitational force on an object of mass m due to mass ‘m1‘ at a distance x. F = \(\frac {Gm_1m}{x^2}\) Field = \(\frac {F}{m}\) = \(\frac {Gm_1}{x^2}\). |
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| 718. |
Is the earth really spherical in shape? |
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Answer» No, the earth is not spherical in shape. |
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| 719. |
Average density of the earthA. is inversely proportional to gB. does not depend upon gC. is a complex function of gD. is directly proportional to g |
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Answer» Correct Answer - D `because g=(GM)/(R^(2))=(G)/(R^(2))[(4)/(3)pi R^(3)rho]` where `rho` is the average density of the earth `=(4)/(3)pi rho RG` `therefore rho=(3g)/(4pi RG)`. Thus `rho prop g`. |
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| 720. |
Find the expression for the weight of the body at centre of the Earth. |
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Answer» We know that, the value of acceleration due to gravity at depth d is given by gd = g [1- d/R] where R is radius of Earth. At depth = R, gd = g[1 – R/R] = 0 So, weight of the body at centre of Earth is 0. |
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| 721. |
Is it possible to keep a satellite, so that it is always over Kashmir? Why? |
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Answer» No, It is possible to place a geostationary satellite on an equatorial plane. Since Kashmir is not on the equator it is not possible to place a satellite above it. |
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| 722. |
A cylinder and a hollow sphere are placed on a surface. If the height of the cylinder is equal to the diameter of the sphere, what is the ratio of the heights of centre of gravity of the cylinder and the sphere from the surfaced? |
| Answer» Ratio of the height of the centre of gravity `= (h)/(2) : (d)/(2) = (h)/(2) : (h)/(2) = 1 : 1` | |
| 723. |
In motion of an object under the gravitational influence of another object. Which of the following quantities is not conserved ?A. Angular momentumB. Mass of an objecttC. Total mechanical energyD. Linear momentum |
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Answer» Correct Answer - D (d) Linear momentum is not conserved. |
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| 724. |
Does the value of g depend on the mass of the falling body? Why? |
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Answer» The value of g does not depend on the mass of the falling body. The reason is the gravitational force on a body due to the earth is directly proportional to the mass of the body and for a given force, the acceleration of a body is inversely proportional to the mass of the body. |
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| 725. |
Define mass. State its SI and CGS units. |
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Answer» The mass of a body is the amount of matter present in it. Its SI unit is the kilogram (kg) and CGS unit is the gram (g). [Note: Mass has only magnitude, not direction. Thus, it is a scalar quantity.] |
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| 726. |
State the three equations of motion of a body under the influence of gravitational force of earth. When is g positive and when is it negative? What is maximum height? |
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Answer» The three equation of motion of body under the infulece of gravitational force of earth are (i) `v=u+g t` (ii) `s=ut+(1)/(2)g t^(2)` (iii) `v^(2)-u^(2)=2gs` the value of g is positive when body is falling freely towards earth, and the value of g is negative when a body is thrown up vertically. maximum height is the height at which up ward velocity (of projection) of the body reduce to zero. |
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| 727. |
As per the request of one of his friends from the equator, Rahul buys 100 grams of silver at the north pole. He hands it over to his friend at the equator. Will the friend agree with the weight of the silver bought? If not, why? |
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Answer» The weight of a body is given by W = mg, where m is the mass of the body and g is the acceleration due to gravity, g varies from place to place. The value of g at the equator is less than that at the north pole (as well as the south pole). Hence, the weight of the silver bought at the north pole would be less when the silver is weighed at the equator. Therefore, Rahul’s friend will disagree about the weight of the silver. [Note: The mass being independent of the value of g, Rahul’s friends will agree about the mass of the silver.] |
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| 728. |
Define weight. State its SI and CGS units. |
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Answer» The weight of a body is defined as the force with which the earth attracts it. Its SI unit is the newton and CGS unit is the dyne. [Note : In the usual notation, the magnitude of the weight of a body on the earth s surface is W = \(\frac{GmM}{R^2}\) = \(m\frac{GM}{R^2}\) = mg. Thus, W ∝ g. Hence, weight varies just like the acceleration due to gravity. It is maximum at the poles and minimum at the equator. It decreases with altitude (ft) and depth (d) below the earth’s surface. It becomes zero at the earth’s centre. At a height above the earth’s surface, W = \(\frac{GmM}{(R+h)^2}\) at a depth d below the earth’s surface, W = \(\frac{GmM(R-d)}{R^2}\). Weight has magnitude and direction (towards the earth’s centre). It is a vector quantity.] |
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| 729. |
What is the relation between the SI unit of weight and the CGS unit of weight? |
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Answer» The relation between the SI unit of weight (the newton) and the CGS unit of weight (the dyne) is 1 newton = 105 dynes. |
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| 730. |
A water tank has a capacity of 1000 litres. It is in the shape of a cylinder. The length of the cylinder is 1 metre. An electric motor pump set is used to fill the water in the tank. This pumpset lifts 120 litres of water per minute. What is the velocity in the shift of centre of gravity of the water tank? |
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Answer» The values of the cylinder having 1000 litres capacity is 1 `m^(3)`. Find the area of the cross section and the height of the cylinder. If 120 litres of water is pumped into the cylinder for every one minute, then the level of water rises up by the rate of `(120)/(1000)`m. Will the C.G. of the water tank rise by `(120)/(1000)` m in one minute? Now, find the change in C.G. of the water tank in one second by dividing (2) by 60 seconds. |
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| 731. |
A Person stood in an accelerating elevartor. Explain how the apparent weight of this person varies. |
| Answer» When the elevator accelerates upwards, the apparent weight of the person who stood inside it increases. When the elevator acceleartes downwards, the apparent weight of the person decreases. | |
| 732. |
(i) what is the mass of body of weight 1 kg ? (ii) what is the weight of a body of mass 1 kg ? |
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Answer» (i) mass of a body of weight 1 kg is one kilogram. (ii) weight of a body of mass 1 kg is 1 kg `wt =9.8N`. |
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| 733. |
Prove that time taken by a body to rise to highest point is always equal to the time taken by it to fall through he same height. |
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Answer» (i) Time taken by body to rise to highes point from `v=v+at` `0=u-g t, t=(u)/(g)` From `v^(2)-u(2)=2as` `0-u^(2)=(-g)h,h=u^(2)/2g` this is the maximum height. (ii) time taken by the bnody to fall to ground From `s=ut+(1)/(2)at^(2)` `u^(2)/2g=0+(1)/(2)g t^(2)` `r^(2)=u^(2)/g^(2),t=(u)/(g)` which is same as in (i) above. |
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| 734. |
Two objects of masses `m_(1)` and `m_(2)` having the same size are dropped simultaneously from heights `h_(1) and h_(2)` respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid and (ii) both of them are hollow, size remaining the same in each case. give reason. |
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Answer» As the two masses are dropped,`u=0,a=g` (which does not depend upon mass) From `s =ut+ (1)/(2) at^(2)` `h_(1) = 0 +(1)/(2) g t_(1)^(2)` and `h_(2) = 0 +(1)/(2) g t_(2)^(2)` `:. (h_(1))/(h_(2))=(t_(1)/t_(2))^(2) or (t_(1))/(t_(2))=sqrt(h_(1)/(h_(2)))` This ratio of time shall remain the same if (i)one of the objects is hollow and the other one is solid-as the ratio is independent of masses. (ii) both the objects are hollow, as acc. due is gravity (g) does not depent upon mass and also whether the body is hollow/solid. As size remains the same in each case, there is no effect of resistance due to air in the two cases. |
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| 735. |
Mass of a body is 5 kg .What is its weight ? |
| Answer» Thus the weight of the body is 49 newtons . | |
| 736. |
What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 kmA. `7.8xx10^(-4)"rad/s"`B. `7.8" rad/s"`C. `0.8xx10^(-4)"rad/s"`D. 1 rad/s |
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Answer» Correct Answer - a `g_(e)=g-R omega^(2)` `(3)/(5)g-g=-Romega^(2)` `(5g-3g)/(5)=R omega^(2)` `(2)/(5)g=Romega^(2)` `omega^(2)=(2g)/(5R)=(2xx9.8)/(5xx6.4xx10^(6))` `omega^(2)=(196xx10^(-6))/(5xx64)` `=(14)/(8sqrt5)xx10^(-3)=0.78xx10^(-3)` `omega=7.8xx10^(-4)"rad/s."` |
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| 737. |
State whether the following statements are True or False : Outside the earth, g varies as 1/(R + h)2. |
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Answer» Answer is True. |
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| 738. |
State whether the following statements are True or False : The value of G changes from place to place. |
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Answer» False. (The value of G is the same throughout the universe.) |
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| 739. |
A planet revolves around sum whose mean distance is `1.588` times the mean distance between earth and sun. The revolution time of planet will beA. 1.25 yearsB. 1.59 yearsC. 0.89 yearsD. 2 years |
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Answer» Correct Answer - D |
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| 740. |
State whether the following statements are True or False : The value of g increases with altitude. |
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Answer» False. (The value of g decreases with altitude.) |
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| 741. |
Name the force due to which the earth revolves around the Sun. |
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Answer» The earth revolves around the Sun due to the gravitational force of attraction exerted on it by the Sun. |
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| 742. |
Name the force that keeps a satellite in the orbit around the earth. |
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Answer» The gravitational force due to the earth keeps a satellite in the orbit around the earth. |
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| 743. |
State whether the following statements are True or False : The escape velocity of a body does not depend on the mass of the body. |
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Answer» Answer is True |
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| 744. |
If `M` is the mass of the earth and `R` its radius, the ratio of the gravitational acceleration and the gravitational constant isA. `(R^(2))/(M)`B. `(M)/(R^(2))`C. `MR^(2)`D. `(M)/(R)` |
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Answer» Correct Answer - B Gravitational acceleration is given by `g = (GM)/(R^(2))` where, `G =` gravitational constant `(g)/(G) = (M)/(R^(2))` |
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| 745. |
For a satellite moving in an orbit around the earth, ratio of kinetic energy to potential energy isA. `2`B. `1/2`C. `1/sqrt(2)`D. `sqrt(2)` |
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Answer» Correct Answer - B |
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| 746. |
A particle of mass m is projected upword with velocity `v=(v_(e))/(2)` (`v_(e))` escape of the particle isA. `-(GMm)/(2R)`B. `-(GMm)/(4R)`C. `-(3GMm)/(4R)`D. `-(2GMm)/(3R)` |
| Answer» Correct Answer - C | |
| 747. |
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them, to take the particle far away from the sphere (you may take `G=6.67xx10^(-11)Nm^(2)//kg^(2))`A. `6.67xx10^(-9) J`B. `6.67xx10^(-10) J`C. `13.34xx10^(-10) J`D. `3.33xx10^(-10) J` |
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Answer» Correct Answer - B |
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| 748. |
Which of the following options are correct? (a) Acceleration due to gravity decreases with increasing altitude. (b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density). (c) Acceleration due to gravity increases with increasing latitude. (d) Acceleration due to gravity is independent of the mass of the earth. |
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Answer» (a) Acceleration due to gravity decreases with increasing altitude. (c) Acceleration due to gravity increases with increasing latitude. |
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| 749. |
Particles of masses 2M, m and M are respectively at points A, B and C with AB = ½ (BC). m is much-much smaller than M and at time t = 0, they are all at rest (Fig. 8.1). At subsequent times before any collision takes place:(a) m will remain at rest.(b) m will move towards M.(c) m will move towards 2M.(d) m will have oscillatory motion. |
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Answer» (c) m will move towards 2M. |
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| 750. |
Choose the wrong option. (a) Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational mass is relevant in determining the gravitational force on it by an external mass. (b) That the gravitational mass and inertial mass are equal is an experimental result. (c) That the acceleration due to gravity on earth is the same for all bodies is due to the equality of gravitational mass and inertial mass. (d) Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot. |
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Answer» (d) Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot. |
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