

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.
1001. |
The forces of action and reaction appear only when the bodies are at rest.the statement isA. trueB. falseC. sometimes true and sometimes falseD. cannot predict. |
Answer» Correct Answer - B No,the statement is false. |
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1002. |
Explain why, big boulders can be moved easily by flood. |
Answer» Big boulders weig much less while in water and as such are easily moved by the flood. | |
1003. |
Why does a sharp knife cut objects more effectively than a blunt knife ? |
Answer» A sharp knife cuts objects easily because due to its very thin edge, the force of our hand falls on a very small area of the object producing large pressure. |
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1004. |
The value of g at the centre of the Earth is zero. Explain? |
Answer» The acceleration due to gravity is given as `g=(GM)/(R^(2))` At the centre of the Earth, the mass under consideration is zero. i.e. `M=O` `:.g=0` Hence acceleration due to gravity, at the centre of the earth is zero. |
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1005. |
Statement-1: A satellite is moving in a circular orbit of the earth. If the gravitational pull suddenly disappears. Then it moves with the same speed tangential to the orginal orbit. Statement-2: the orbital speed of a satellite increases with the increase in radius of the orbit.A. Statement-1 is true, statement-2 is true: Statement-2 is a correct explanation for statement-1B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
Answer» Correct Answer - C | |
1006. |
If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why? |
Answer» To pull an object along the floor, it is necessary to do work against the force of friction between the object and the surface of the floor. This force of friction is proportional to the weight, mg, of the object. If the value of g becomes twice its value, the weight of the object and hence the force of friction will become double. Therefore, it will become two times more difficult to pull a heavy object along the floor. |
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1007. |
At what height from the surface of the earth, the total energy of satellite is equal to its potential energy at a height `2R` from the surface of the earth (`R`=radius of earth)A. `2R`B. `R//2`C. `R//4`D. `4R` |
Answer» Correct Answer - B `(-GMm)/(2r)=((-GMm)/(3R)), r=R+h` |
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1008. |
A satellite of mass `m` is orbiting around the earth at a height `h` above the surface of the earth. Mass of the earth is `M` and its radius is `R`. The angular momentum of the satellite is independent of :A. `m`B. `M`C. `h`D. none of these |
Answer» Correct Answer - D `L=mv_(0)(R+h)` `msqrt((GM)/(R+h)) (R+h) =sqrt(GM(R+h))` i.e, `L` depends on `m,M` as well as `h`. |
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1009. |
A satellite of mass `m` is orbiting around the earth at a height `h` above the surface of the earth. Mass of the earth is `M` and its radius is `R`. The angular momentum of the satellite is independent ofA. `m`B. `M`C. `h`D. none of these |
Answer» Correct Answer - D `L=mv_(0)(R+h)` `=msqrt((GM)/((R+h)))(R+h)=msqrt(GM(R+h))` i.e., `L` depends on `m,M` as well as `h` |
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1010. |
Two concentric shells have masses `M` and `m` and their radii are `R` and `r`, respectively, where `R gt r`. What is the gravitational potential at their common centre?A. `-(GM)/R`B. `-(GM)/r`C. `-G[M/R-m/r]`D. `-G[M/R+m/r]` |
Answer» Correct Answer - D `V=(GM)/R+(-(Gm)/r)=-G[M/R+m/r]` |
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1011. |
The time of revolution of planet `A` round the sun is `8` times that of another planet `B`. The distance of planet `A` from the sun is how many `B` from the sunA. `4 : 1`B. `1 : 2`C. `3 : 1`D. `1 : 4` |
Answer» Correct Answer - D `T^(2)prop r^(3) " " n_(1)^(2)d_(1)^(3)=n_(2)^(2)d_(2)^(3)" " (because T=(1)/(n))` `therefore ((n_(1))/(n_(2)))^(2)=((d_(2))/(d_(1)))^(3)=((8)/(1))^(3)` `therefore (d_(2))/(d_(1))=(2^(3))^(2//3)=4 " " therefore (d_(1))/(d_(2))=(1)/(4)` |
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1012. |
The time of revolution of planet `A` round the sun is `8` times that of another planet `B`. The distance of planet `A` from the sun is how many `B` from the sunA. 2B. 3C. 4D. 5 |
Answer» Correct Answer - C |
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1013. |
The period of revolution of planet A round from the sun is 8 times that of B. The distance of A from the sun is how many times greater then tht of B from the sun ?A. 2B. 4C. 3D. 5 |
Answer» Correct Answer - C As `T_(2) prop r^(3)` so `(T_(A)^(2))/(T_(B)^(2))=(r_(A)^(3))/(r_(B)^(3))` or `(r_(a))/(r_(B))=(T_(A))/(T_(B))^(2//3)=(8)^(2//3)=4 rarr r_(A)=4r_(B)` so `r_(A)-r_(B)=4r_(B)-r_(B)=3r_(B)` |
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1014. |
Identify the position of sun in the following diagram if the linear speed of the planet is greater at C than at D. |
Answer» Sun should be at B as speed of planet is greater when it is closer to sun. |
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1015. |
Is it possible to place an artificial satellite in an orbit so that it is always visible over New Delhi ? |
Answer» No, A satellite will be always visible only if it revolves in the equatorial plane, but New Delhi does not lie in the region of equatorial plane. |
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1016. |
A satellite does not require any fuel to orbit the earth. Why ? |
Answer» The gravitational force between satellite and earth provides the necessary centripetal force for the satellite to orbit the earth. |
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1017. |
A satellite of small mass burns during its descent and not during ascent. Why ? |
Answer» The speed of satellite during descent is much larger than during ascent, and so heat produced is large. |
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1018. |
Prove that if a body is thrown vertically upwards, then the time of ascent is equal to the time of descent. |
Answer» Let, u = Initial velocity v = Final velocity h = Height attained a = acceleration t = time Ascent: Ball thrown upward with velocity u. So, v = 0 at maximum height h. Also, a = -g Now, using v = u + at 0 = u - gta ta = \(\frac{u}{g}\).......................1 Where ta = time of ascent Also, by using V2 = U2 + 2as 0 = u2 - 2gh u = \(\sqrt {2gh}\).............2 From equations 1 and 2, we get, ta = \(\sqrt \frac{2h}{g}\)..........A Descent: Ball comes down from height h. So, u = 0 and a = g. So, using v = u + at v = gtd ….......3 Where td = Time of descent Now, using v2 = u2 + 2as v2 = 0 + 2gh v = \(\sqrt {2gh}\)................4 From equations 3 and 4, we see that, td = \(\sqrt \frac{2h}{g}\)..........B From equations A and B, we see that, ta = td |
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1019. |
Weight of 3 kg body isA) 3 NB) 3.2 NC) 29.4 ND) 19.6 N |
Answer» Correct option is C) 29.4 N |
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1020. |
Gravitational force between two bodies exists A) when they are in contact only B) when they are not in contact only C) any of the above two casesD) none of these |
Answer» C) any of the above two cases |
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1021. |
When a stone moves up, its speed …………….. A) increases B) decreases C) does not change D) none of these |
Answer» Correct option is B) decreases |
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1022. |
Then why is the Earth not seen moving towards the stone ? |
Answer» Because mass of earth is too lage. Its acceleration `(a=F//m)` under the same force is negligble. | |
1023. |
A body is projected upwards. What is its initial velocity at maximum height? |
Answer» At maximum height, velocity of body=zero. | |
1024. |
A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entrie time of motion? What is the velocity at 5 seconds after the projection? Take g = 10 m/s2. |
Answer» Initial speed u = 40 m/s ; g = 10 m/s2 Maximum height reached (h) = \(\frac{u^2}{2\,g}=\frac{40\,\times\,40}{2\,\times\,10}\) = 80 m Entire time of motion (T) = \(\frac{2u}{g}=\frac{2\,\times\,40}{10}\) = 8 s Entire time of motion is 8 seconds. ∴ It starts to fall down after 4 seconds. At 5 seconds the body is in downward direction. u = 0 m/s, a = g = 10 m/s2, t = 5 – 4 = 1 sec. v = u + at = 0 + 10 × 1 = 10 m/s ∴ The velocity at 5 seconds is 10 m/s downward. |
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1025. |
If `W_(1) W_(2)` and `W_(3)` represent the work done in moving a particle from `A` to `B` along three different paths `1.2` and`3` respectively (asshown ) in the gravitational fieled of a point mass m, find the correct relation between ` `W_(1) W_(2)` and `W_(3)` A. `W_(1)=W_(2)=W_(3)`B. `W_(1)gt W_(2)gt W_(3)`C. `W_(1)=W_(2)gt W_(3)`D. `W_(1)lt W_(2)lt W_(3)` |
Answer» Correct Answer - A Since gravitational force is conservative so, work done will independent of path Therefore `w_(1)=w_(2)=w_(3)` |
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1026. |
Why an astronaut in an orbiting space craft is not zero gravity although he is in weightlessness ? |
Answer» The astronaut is in the gravitational field of the earth and experiences gravity. However, the gravity is used in providing necessary centripetal force, so is in a state of free fall towards the earth. |
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1027. |
Why is the weight of a body at the poles more than the weight at the equator ? Explain. |
Answer» As g = GM/R2 and the value of R at the poles is less than that the equator, so g at poles is greater than that g at the equator. Now, gp > ge , hence mgp > ge i.e., the weight of a body at the poles is more than the weight at the equator. |
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1028. |
If the density of a planet is doubled without any change in its radius, how does ‘g’ change on the planet. |
Answer» ‘g’ gets doubled as g ∝ ρ (density). |
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1029. |
Statement I: In free space a uniform spherical planet of mass `M` has a smooth narrow tunnel along its diameter. This planet and another superdense small particle of mass `M` start approaching towards each other from rest under action of their gravitational forces. When the particle passes through the centre of the planet, sum of kinetic energies of both the bodies is maximum. Statement II: When the resultant of all forces acting on a particle or a particle like object (initially at rest) is constant in direction, the kinetic energy of the particle keeps on increasing.A. Statement I is True, Statement II is True: Statement II is a correct explanation for Statement I.B. Statement I is True, Statement II is True: Statement II is Not a correct explanation for Statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True. |
Answer» Correct Answer - A Till particle reaches the centre of the planet, force on both the bodies are in direction on increasing. After the particle crosses the centre of the planet, forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kinetic energies of oth the bodies is maximum. Therefore Statement 1 is true, Statement 2 is true: Statement 2 is a correct explanation for Statement 1. |
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1030. |
Give any two applications of remote sensing. |
Answer» 1. remote sensing is used for mapping of forests. 2. remote sensing is used for groundwater exploration. |
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1031. |
How is a satellite launched? |
Answer» In order to launch a satellite, it is taken vertically upwards to the required orbit and then appropriate horizontal velocity is imparted to the satellite so that it revolves round the earth. |
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1032. |
What is the actual shape of the orbit of a planet around the sun ? What assumption was made by Newton regarding the shape of an orbit of a planet around the sun for deriving his inverse square rule from Kepler’s third law of planetary motion ? |
Answer» The actual shape of the orbit of a planet around the sun is elliptical. The assumption made by the Newton regarding the shape of an orbit of a planet around the sun was that the orbit of a planet around the sun is ‘circular’. |
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1033. |
The values of g at six distances A, B, C, D, E and F from the surface of the earth are found to be 3.08 m/s2, 9.2.3 m/s2, 0.57 m/s2, 7.34 m/s2 , 0.30 m/s2 and 1.49 m/s2, respectively. (a) Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first) (b) If the value of distance F be 10000 km from the surface of the earth, state whether this distance is deep . inside the earth or high up in the sky. Give reason for your answer. |
Answer» (a) 9.23 m/s2 , 7.34 m/s2 , 3.08 m/s2 , 1.49 m/s2 , 0.57 m/s2 , 0.30 m/s2 (b) This distance F of 10000 km is high up in the sky. The distance of 10000 km cannot be deep inside the earth because the radius of earth is only about 6400km and the value of g at the centre of earth becomes zero. |
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1034. |
Write the common unit of density. |
Answer» Grams per cubic centimtre (g/cm3). |
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1035. |
What is the density of water in SI units ? |
Answer» Density of water =1000kg/m3 . |
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1036. |
What is the value of relative density of water ? |
Answer» Relative density of water is 1. | |
1037. |
Name the quantity whose one of the units is pascal (Pa) |
Answer» Correct Answer - Pressure |
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1038. |
Name the quantity whose one of the units is pascal (Pa). |
Answer» Pressure has unit of Pascal (Pa). | |
1039. |
When we hold a suitcase steady at some heightA. weight of suitcase stops actingB. suitcase alone applies forceC. suitcase is under the action of balanaced forcedD. none of the above |
Answer» Correct Answer - C The suitcase is under the action of balanced forced-weight of suitcash and force applied by our hand. |
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1040. |
What happens to the force between two object, if (i) the mass of one object is doubled ? (ii) the distance between the object is doubled and tripled? (iii) the masses of both object are doubled? |
Answer» As gravitational force between two object, `F prop (m_(1)m_(2))/r^(2)` thereforce, (i) when mass of one object is doubled, the force become twice. (ii) when distance between the objects is doubled, force become `(1//4)` of its previous value. When distance between the objects is tripled, the force become `(1//9)` of its previous value . (iii) when masses of both object are doubled, force becomes 4 times. |
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1041. |
State the units in which pressure is measured. |
Answer» Pressure is measured in newtons per square metre (N/m2 ) i.e., pascal (Pa). |
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1042. |
State whether the following statements are true or false : (a) The buoyant force depends on the nature of object immersed in the liquid (b) Archimedes’ principle can also be applied to gases. |
Answer» (a) False (b) True |
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1043. |
What is the other name of buoyant force |
Answer» Correct Answer - Upthrust |
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1044. |
In which direction does the buoyant force on an object due to a liquid act ? |
Answer» Buoyant force on an object due to a liquid act s in the vertically upward direction. | |
1045. |
A moving body cannot stop on its own. This is due toA. inertia of restB. inertia of motionC. inertia of directionD. all of theese. |
Answer» Correct Answer - B A moving body cannot stop on its own due to inertia of motion. |
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1046. |
What is the other name of buoyant force ? |
Answer» Upthrust is the other name of buoyant force. | |
1047. |
What is the imprtance of universal law of gravitation? |
Answer» universal law of gravitation is important as it accounts for motion of planets around the sun , motion of moon and other artificial satellites around the earth, snowfall and rainfall on earth, flow of water in rivers and so many othe phenomena. | |
1048. |
A block is placed on a horizontal table. Identify the force of actionA. weight of blockB. support of table on the blockC. either (a) or (b)D. neither (a) nor (b). |
Answer» Correct Answer - A Action is the forced due to weight of the block. |
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1049. |
Name the force which makes heavy objects appear light when immersed in a liquid. |
Answer» Buoyant force. | |
1050. |
What is upthrust ? |
Answer» The upward force acting on an object immersed in a liquid is called upthrust. | |