

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.
2501. |
Statement-1: Pendulum clock stops working on the spaceship Statement-2: Pendulum of the pendulum clock falls down on the spaceship.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 | |
2502. |
Statement-1: Escape velocity is independent of the angle of projection. Statement-2: Escape velocity from the surface of earth is `sqrt(2gR)` where `R` is radius of earth.A. Statement-1 is true, statement-2 is true, Statement-2 is a correct explanation for statement-1.B. 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 - a Using only energy conservation `1/2mv_(e)^(2)=(GMm)/R=0+0`. |
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2503. |
Three particles are projected vertically upward from a point on the surface of the earth with velocities `sqrt(2gR//3)` surface of the earth. The maximum heights attained are respectively `h_(1),h_(2),h_(3)`.A. `h_(1):h_(2)=2:3`B. `h_(2):h_(3)=3:4`C. `h_(1):h_(3)=1:4`D. `h_(2)=R` |
Answer» Correct Answer - C::D Case: I `U_(i)+K_(f)=U_(f)+K_(g)` `(-Gm_(e)m)/(R)+(1)/(2)mv^(2)=-(GM_(e)m)/(R+h_(1))+0` `(-GM_(e)m)/(R)+(1)/(2)m(2GM_(e))/(R^(2))(R)/(3)=-(GM_(e)m)/(R+h_(1))` `-(1)/(R)+(1)/(3R)=-(1)/(R+h_(1))impliesh_(1)=(R)/(2)` Case II: `U_(i)+K_(i)=U_(f)+K_(g)` `-(GM_(e)m)/(R)+(1)/(2)mv^(2)=-(GM_(e)m)/(R+h_(2))+0` `(-GM_(e)m)/(R)+(1)/(2)m(2GM_(e))/(R^(2))R=-(GM_(e)m)/(R+h_(2))` `-(1)/(R)+(1)/(2R)=-(1)/(R+h_(2))impliesh_(2)=R` Case III: `U_(i)+K_(i)=U_(f)+K_(g)` `-(GM_(e)m)/(R)+(1)/(2)m(4GM_(e))/(R^(2))(R)/(3)=-(GM_(e)m)/(R+h_(3))` `-(1)/(R)+(1)/(3R)=-(1)/(R+h_(3))impliesh_(3)=2R` |
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2504. |
Statement-1: Escape velocity is independent of the angle of projection. Statement-2: Escape velocity from the surface of earth is `sqrt(2gR)` where `R` is radius of earth.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 - B | |
2505. |
If a feather and a stone are released from the top of a building simultaneously, the stone reaches the ground earlier than the feather. |
Answer» (a) The motion of a body falling in air accelerated due to the earth’s gravitational force on the body. The force due to buoyancy of air acts on the body in the upward direction. As the body falls, the friction with air opposes its motion. (b) This opposition due to air depends on the size, shape, density and velocity of the body. It Is greater for a feather than for a stone. Hence, the stone has greater downward acceleration than the feather. Therefore, the stone reaches the ground earlier than the feather though ‘they are released simultaneously from the same height. |
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2506. |
A particle of mass 10g is kept on the surface of a uniform sphere of masss 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particel 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` |
Answer» Correct Answer - B Potential energy of system of two masses `U=(-GMm)/R=(-6.67xx10^(-11)xx100xx10xx10^(-3))/(10xx10^(-2))` `U=-6.67xx10^(-10)J` So, the amount of work done to take the particle up to infinite will be `6.67xx10^(-10)J` |
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2507. |
A particle of mass 10g is kept on the surface of a uniform sphere of masss 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particel 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` |
Answer» Correct Answer - 4 `W=(GMm)/r=6.67xx10^(-11)xx100xx10//0.1=6.67xx10^(-7)J` |
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2508. |
Two spherical bodies of mass M and 5M & radii R & 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smallar body just before collision isA. `2.5R`B. `4.5R`C. `7.5R`D. `1.5R` |
Answer» Correct Answer - C If `x_(1)` and `x_(2)` are the distance covered by the two bodies, then `x_(1)+x_(2)=9R`. Also `Mx_(1)=5mx_(2)impliesx_(2)=(x_(1))/5` `x_(1)+(x_(1))/5=9Rimpliesx_(1)=7.5R` |
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2509. |
A particle of mass 10g is kept on the surface of a uniform sphere of masss 100kg and radius 10cm. Find the work to be done against the gravitational force between them to take the particel far away from the sphere (you may take `G = 6.67xx10^(-11) Nm^2 /kg^2)`A. `13.34xx10^(-10)J`B. `3.33xx10^(-10)J`C. `6.67xx10^(-9)J`D. `6.67xx10^(-10)J` |
Answer» Correct Answer - D Work done `=` change in `GPE =U_(oo)-U_(R)` `W=0-((GMm)/R)=(GMm)/R` `=(6.67xx10^(-11)xx100xx10^(-2))/10^(-1)=6.67xx10^(-10)J` |
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2510. |
What are Newton’s laws of motion? |
Answer» (1) Newton’s first law of motion: An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it. (2) Newton’s second law of motion: The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force. (3) Newton’s third law of motion: Every action force has an equal and opposite reaction force that acts simultaneously. [Note: Equal in magnitude and opposite in direction.] |
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2511. |
A cosmic body `A` moves to the sun with velocity `v_(0)`(when far from the sun) and aiming parameter `l`, the arm of the vector `v_(0)`, relative to the centre of the sun. find the minimum distance by which this body will get to the sun. mass of the sum is `M`. |
Answer» Correct Answer - `[(GM)/(v_(0)^(2))(sqrt((1+(d^(2)v_(0)^(4))/(G^(2)M^(2))))-1)]` |
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2512. |
State the SI unit of density. Relate it with its cgs unit. |
Answer» SI unit of density is kilogram per cubic meter (kg/m3 ). Its cgs unit is gram per cubic centimeter (g/cm3 ). One g/cm3 is equal to one thousand kg/m3 . 1 kg/m3 = 1000 g/cm3 |
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2513. |
A body lying on a mattress stands up on it. He observes that the mattress is now depressed deeper down. Why does this happen? |
Answer» When the person was lying on the mattress, his weight was distributed throughout his body that is his surface area in contact with the mattress increases and as we know the pressure is inversely proportional to area, therefore, less pressure will be exerted on the mattress. As soon as he stands on his feet, the area decreases which is in contact with the mattress and therefore pressure increases causing the mattress to be depressed more. |
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2514. |
An empty plastics bottle float, while a plastic bottle filled with sand sinks in water. Why? |
Answer» The gravitational pull on the empty plastic bottle is less than the upthrust exerted on it by the water. In other words, its density is less than the density of water therefore it floats. On the other hand, a plastic bottle filled with sand sinks as the gravitational pull on it (weight) is more than the upthrust of water. |
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2515. |
A solid ball and a hollow ball of the same size, are dropped from a height. The solid ball falls faster to the ground. Why? |
Answer» The mass of the solid ball is more as compared to a hollow ball. Therefore gravitational force exerts more force on the solid ball. Also, air resistance acts on both the objects but it will act more on lighter objects and therefore decrease its velocity to fall down. |
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2516. |
No one can sink in the Dead Sea. Do you agree? Give reason |
Answer» No one can sink in the Dead Sea is a misconception. The Dead Sea is highly saline that is salt content is very high. Due to the high salt content, the water of Dead Sea is denser than the freshwater. So objects which are denser than the salty water will sink while those which are not will float. |
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2517. |
A hollow plastic ball is taken to the bottom of a through of water and released there (see adjoining figure).(a) What happens to the ball? (b) Give a reason for this phenomenon. |
Answer» (a) The ball will come on the surface of the water and will float there. (b) The density of the ball is less than the density of water. The another reason could be the gravitational pull on the ball is less than the upthrust exerted by water. |
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2518. |
A projectile of mass m is thrown vertically up with an initial velocity v from the surface of earth (mass of earth = M). If it comes to rest at a height h, the change in its potential energy isA. `GMmh//R(R+h)`B. `GMmh^(2)//R(R+h)^(2)`C. `GMmhR//R(R+h)`D. `GMm//hR(R+h)` |
Answer» Correct Answer - A | |
2519. |
Take a piece of thread. Tie a small piece of stone to one end of the thread. Hold the other end of the thread in hand and whirl it round. Note the motion of the stone. If the thread breakes suddenly, the stone files off along tangent to the circle at the instant. Read the above passage and answer the following question: (i) which force is whirling the stone ? who is providing this force ? (ii) why does the stone fly off along the tangnet to the circle at that instant? (iii) what value of life do you learn from this? |
Answer» (i) The force whirling the stone is the centripetal force acting on the stone. This force is being provided by our hand whirling the stone. (ii) when the thread breakes, centripetal force is no longer provided. Therefore, inertia of direction takes the stone along the straight line path , which is along tangnet to the circle at the instant. (iii) Form these observation,we learn that to keep our children on track ,motivational has to be continous.the applies equally to our lives too. the moment our efforts and motivation cease, our life/children are likely to go off the track. |
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2520. |
A satellite orbits around the earth in a circular orbit with a speed `upsilon` and orbital radius `r`. If it loses some energy, then `upsilon` and `r` changes asA. `d` will increases, `v` will increasesB. `d` will increases, `v` will decreasesC. `d` will decreases, `v` will decreasesD. `d` will decreases, `v` will increases |
Answer» Correct Answer - D `(mv^(2))/d=(GMm)/(d^(2)) implies v=sqrt((GM)/d) implies v prop1/(sqrt(d))` |
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2521. |
A satellite of mass m is going around the earth in a circular orbital at a height `(R)/(2)` from the surface of the earth. The satellite has lived its life and a rocket, on board, is fired to ake it leave the gravity of the earth. The rocket remains active for a very small interval of time and imparts an impulse in the direction of motion of the satellite. Neglect any hange in mass due to firing of the rocket. Find the minimum impulse imparted by the rocket to the satellite. Find the minimum work done by the rocket engine. Mass of the earth = M, Radius of the earth = R |
Answer» Correct Answer - (a) `(sqrt(2)-1) m sqrt((2GM)/(3R))` (b) `(GMm)/(3R)` |
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2522. |
To launch a satellite at a height h above the surface of the earth (radius R) a two stage rocket is used . The first stage is used to lift the satellite to the desired height and the econd stage is used to impart it a tangential velocity so as to put it in a circular orbital. Assume (incorrectly) that the mass of rocket is negligible and that there is no tmospheric resistance. If `E_(1)` and `E_(2)` are the energies delivered by the first and the second stage of the rocket. Calculate the ratio `(E_(2))/(E_(2))` Calculate the time period of the satellite if it is given that `(E_(2))/(E_(2))=1.` Take mass of the earth to be M. |
Answer» Correct Answer - (a) `(2h)/R` `(b) sqrt((27pi^(2)R^(3))/(2GM))` |
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2523. |
Why a multi-stage rocket is required to launch a satellite? |
Answer» This is done to save fuel. |
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2524. |
A `3kg` mass and `4kg` mass are placed on `X` and `Y` axes at a distance of `1` meter from the origin and a `1kg` mass is placed at the origin. Then the resultant gravitational force on `1kg` mass isA. `7G`B. `G`C. `5G`D. `3G` |
Answer» Correct Answer - C `theta=90^(@), F_(1)=(3G)/(1^(2)), F_(2)=(4G)/(1^(2))` `F=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)cos theta)` |
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2525. |
Gravitational force is a conservation and medium independent force. Its nature is attractive. Gravitational field intensity and gravitational potential gives information about gravitational fied in vector and scaler form respectively. Actually gravitational fied intensity is equal to the negative of the negative of teh potential gradient. potential energy is defined for only conservation force. it is also equal to the total energy in escaping condition. gravitational potential is either negative or zero but can never be positive due to attractive nature of gravitational force. A person brings a mass of `1 kg` from infinity to a point `A`. Initially the mass was at rest but it moves with a speed of `2 m//s` as it reaches `A`. The work done by the person on a mass is `-3 J`. The potential of `A` is:A. `-3 J//kg`B. `-2 J//kg`C. `-5 J//kg`D. `-7 J//kg` |
Answer» Correct Answer - c | |
2526. |
A small planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force between the planet and the stae were proprtional to `R^(-5//2)`, then T would be proportional toA. `R^(3//2)`B. `R^(3//5)`C. `R^(7//2)`D. `R^(7//4)` |
Answer» Correct Answer - c `"AS "Fprop (1)/(R^(2))" "therefore R^(2)prop (1)/(F)` `T^(2) prop R^(3) therefore T^(2) prop R^(2)xxR therefore T^(2) prop (R)/(F)` `therefore T^(2) prop (R)/(R^(-5//2)) therefore R^(7//2)` |
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2527. |
`F=(Gm_1 m_2)/R^2` is validA. Between bodies with any shapeB. Between particlesC. Between any bodies with uniform densityD. Between any bodies with same shape |
Answer» Correct Answer - B | |
2528. |
Find the false statementA. Gravitational force acts along the line joining two interacting particlesB. Gravitaional force is independent of mediumC. Gravitational force forms an action-reactionD. Gravitational force does not obey the principle of superposition. |
Answer» Correct Answer - D | |
2529. |
If `F_g` and `F_e` are gravitational and electrostatic forces between two electrons at a distance `0.1m` then `F_g//F_e` is in the order ofA. `10^(43)`B. `10^(-43)`C. `10^(35)`D. `10^(-35)` |
Answer» Correct Answer - B | |
2530. |
`F_g, F_e` and `F_n` represent the gravitational electro-magnetic and nuclear forces respectively, then arrange the increasing order of their strengthsA. `F_(n),F_(c),F_(g)`B. `F_(g),F_(e),F_(n)`C. `F_(e),F_(g),F_(n)`D. `F_(g),F_(n),F_(e)` |
Answer» Correct Answer - B | |
2531. |
Read the following statements : `S_(1)` : An object shall weigh more at pole than at equator when weighed by using a physical balance. `S_(2)` : It shall weigh the same at pole and equator when weighed by using a physical balance. `S_(3)` : It shall weigh the same at pole and equator when weighed by using a spring balance. `S_(4)` : It shall weigh more at the pole than at equator when weighed using a spring balance. Which of the above statements is/are correct ?A. `S_(1)` and `S_(2)`B. `S_(1)` and `S_(4)`C. `S_(2)` and `S_(3)`D. `S_(2)` and `S_(4)`. |
Answer» Correct Answer - D | |
2532. |
Explain why a tennis ball bounces higher on hills than in plane ? |
Answer» The value of ‘g’ on hills is less than at the plane, so the weight of tennis ball on the hills is lesser force than at planes that is why the earth attract the ball on hills with lesser force than at planes. Hence the ball bounces higher. |
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2533. |
Why the space rockets are generally launched west to east ? |
Answer» Since the earth revolves from west to east, so when the rocket is launched from west to east the relative velocity of the rocket increases which helps it to rise without much consumption of fuel. |
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2534. |
The motion of a planet around sun in an elliptical orbit is shown in the following figure. Sun is situated at one focus. The shaded areas are equal. If the planet takes time `t_1` and `l_2` in moving from `A` to `B` and from `C` to `D` respectively, then A. `t_(1)gtt_(2)`B. `t_(1)ltt_(2)`C. `t_(1)=t_(2)`D. incomplete information |
Answer» Correct Answer - C | |
2535. |
Figure shows the motion of a planet around the Sun `S` in an elliptical orbit with the Sun at the focus. The shaded areas `A` and `B` are also shown in the figure which can be assumed to be equal. If `t_(1)` and `t_(2)` represent the time taken for the planet to move from `a` to `b` and `c` to `d`, respectively then A. `t_(1)ltt_(2)`B. `t_(1)gtt_(2)`C. `t_(1)=t_(2)`D. from the given information the relation between `t_(1)` and `t_(2)` cannot be determined |
Answer» Correct Answer - C Equal time is taken to cover equal area. |
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2536. |
Choose the wrong optionA. Inertial mass is a measure of difficulty of accelerating a body by an external force whereas the gravitational force on it by an external massB. That the gravitational mass and inertial mass are equal is an experimental resultC. That the acceleration due to gravity on the earth is the same for all bodies is due to the equality of gravitational mass and inertial massD. Gravitational mass of a particle like proton can depend on the presence of neighbouring heavy objects but the inertial mass cannot |
Answer» Correct Answer - D Gravitational mass of proton is equivalent to its inertial mass and is independent of presence neighbouring heavy objects. |
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2537. |
Both the earth and the moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moonA. will be ellipticalB. will not be strictly elliptical because the total gravitational force on it is not centralC. is not elliptical but will necessarily be a closet curveD. deviates considerably from being elliptical due to influence of planets other than the earth |
Answer» Correct Answer - B As observed from the sun, two types of forces are acting on the moon one is due to gravitational attraction between the attraction between the earth and the moon. Hence, total force on the moon is not central. |
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2538. |
The density of water is `1000 kg//m^3` and the density of copper is `8900 kg//m^3`. Which of the following statements is incorrect ?A. `(a)=("The density of a certain volume of copper")/("The density of the same volume of water" )=8.9`B. `(a)=("The density of a certain volume of copper")/("The density of the same volume of water" )=8.9`C. `(a)=("The density of a certain volume of copper")/("The density of the same volume of water" )=8.9`D. `(a)=("The density of a certain volume of copper")/("The density of the same volume of water" )=8.9` |
Answer» Correct Answer - B |
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2539. |
The diagrams represent four measuring cylinders containing liquids. The mass and volume of the liquid in each cylinder are stated. Which two measuring cylinders could contain an identical liquid? A. W and XB. W and YC. X and YD. X and Z |
Answer» Correct Answer - D |
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2540. |
A satellite is revolving round the earth at a height of 600 km. find a. the speed f the satelite and b. the time period of thesatellit. Radius of the earth =6400 km and mass of the earth `=6xx10^24kg`. |
Answer» The distance of the satellite form the centre of the earth is 6400 km+600 km=7000 km. The speed of the satellite is `sqrt((GM)/a)` `sqrt(6.67xx10^-11N-m^2/kg^2x6xx10^24kg)/(7000xx10^3m)` `=7.6xx10^3ms^-=7.6s^-1` The time period is `T=(2pia)/v` `=(2pixx7000xx10^3m)/(7.6xx1063ms^-1)=5.8xx10^3s`. |
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2541. |
A particle is throws vertically upwards from the surface of earth and it reaches to a maximum height equal to the radius of earth. The radio of the velocity of projection to the escape velocity on the surface of earth isA. `(1)/(sqrt2)`B. `(1)/(2)`C. `(1)/(4)`D. `(1)/(2sqrt2)` |
Answer» Correct Answer - A Change in kinetic energy = change in potential energy `:. (1)/(2) m nu^(2) = - (GMm)/(R + R) - (-(GMm)/(R))` `t((GM)/(R)) = sqrt(2GM//R)/(sqrt(2))` `= (nu_(e))/(sqrt(2))` where, `nu_(e) =` escape velocity |
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2542. |
A satellite moves estwards very near the surface of the Earth in equitorial plane with speed `(v_(0))`. Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If `R` =radius of the Earth and `omega` be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the earth.A. `(4piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))`B. `(2piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`C. `(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))`D. `(2piomegaR^(2))/(v_(0)^(2)+R^(2)omega^(2))` |
Answer» Correct Answer - C `T_(west)=(2piR)/(v_(0)+Romega)` and `T_("east")=(2piR)/(v_(0)-Romega) implies DeltaT=T_("east")` `implies T_("east")-T_(west)=2piR[(2piR)/(v_(0)^(2)-R^(2)omega^(2))]=(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))` |
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2543. |
A particle is prjected vertically upwards the surface of the earth (radius `R_(e))` with a speed equal to one fourth of escape velocity what is the maximum height attained by it from the surface of the earth?A. `(16)/(15)R_(e)`B. `(R_(e))/(15)`C. `(4)/(15)R_(e)`D. None of these |
Answer» Correct Answer - B From conservation of mechanical energy `(1)/(2)mv^(2)=(GMm)/(R_(e))-(GMm)/(R)` Where `R=` maximum distance from centre of the earth also `v=(1)/(4)v_(e)=(1)/(4)sqrt((2GM)/(R_(e)))` `implies(1)/(2)mxx(1)/(16)xx(2GM)/(R_(e))=(GMm)/(R_(e))-(GMm)/(R)impliesR=(16)/(15)R_(e)impliesh=R-R_(e)=(R_(e))/(15)`. |
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2544. |
A satellite moves estwards very near the surface of the Earth in equitorial plane with speed `(v_(0))`. Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If `R` =radius of the Earth and `omega` be its angular speed of the Earth about its own axis. Then find the approximate difference in the two time period as observed on the earth. |
Answer» `T_("west")=(2piR)/(v_(0)+Romega)` and `T_("east")=(2piR)/(v_(0)-Romega)impliesDeltaT=T_("east")-T_("west")=2piR[(2Romega)/(v_(0)^(2)-R^(2)omega^(2))]=(4piomegaR^(2))/(v_(0)^(2)-R^(2)omega^(2))` | |
2545. |
In the solar system, which is conservedA. Total EnergyB. K.E.C. Angular VelocityD. Linear Momentum |
Answer» Correct Answer - A |
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2546. |
A body revolved around the sun 27 times faster then the earth what is the ratio of their radiiA. `1//3`B. `1//9`C. `1//27`D. `1//4` |
Answer» Correct Answer - B |
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2547. |
A satellite of mass m goes round the earth along a circular path of radius r. Let `m_(E)` be the mass of the earth and `R_(E)` its radius then the linear speed of the satellite depends on.A. `m, m_(E)` and `r`B. `m, R_(E)` and `r`C. `m_(E)` onlyD. `m_(E)` and `r` |
Answer» Correct Answer - D | |
2548. |
Figure shows the orbit of a planet `P` round the sun `S`. `AB` and `CD` are the minor and major axed of the ellipse. If `t_(1)` is the time taken by the planet to travel along `ACB` and `t_(2)` the time along `BDA`, thenA. `t_(1)=t_(2)`B. `t_(1)gtgt t_(2)`C. `t_(1)lt t_(2)`D. nothing can be concluded |
Answer» Correct Answer - b Since serial velocity is constant so `t_(1)gtt_(2)`. |
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2549. |
Figure shows the orbit of a planet `P` round the sun `S`. `AB` and `CD` are the minor and major axed of the ellipse. If `U` is the potential energy and `K` kinetic energy then `|U|gt|K|` atA. only `D`B. only `C`C. both `D & C`D. neither `D`nor `C` |
Answer» Correct Answer - c Since total energy is always negative and `K` is always positive so `|U|gt|K|` |
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2550. |
The maximum and minimum distance of a comet form the sun are `8xx10^(12)m and 1.6xx10^(12)m`. If its velocity when nearest to the sun is `60m//s`, what will be its velocity in m/s when it is farthestA. `12`B. `60`C. `112`D. `6` |
Answer» Correct Answer - A |
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