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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.
| 1. |
A 3-kg steel ball strikes a wall with a speed of `10.0ms^(-1)` at an angle of `60.0^(@)` with the surfaces of the wall. The ball bounces off with the same speed and same angle. If the ball was in contact with the wall for 0.2s, find the average force exerted by the wall on the ball. A. `300 N`B. zeroC. `150 sqrt(3) N`D. `150N` |
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Answer» Correct Answer - D As is clear from figure , components of momentum perpendicular to wall cancel out, and components along the wall add. As `Fxxt=` change in momentum `=2 m upsilon cos theta` `:. F=(2 m upsilon cos theta)/( t)=(2xx3xx10cos 60^(@))/( 0.2)` `(60)/(2)xx(10)/(2)=150N` |
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| 2. |
The distance x moved by a body of mass 0.5 kg under the action of a force varies with time t as`x(m)=3t^(2)+4t+5` Here, t is expressed in second. What is the work done by the force in first 2 seconds? |
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Answer» From the displacement equation, one can determine the expression for force acting on the object It has been calculated below `x=3t^(2)+4t+5` Differentiating once, w.r.t. time, we get `(dx)/(dt)=6t+4 or` `v=6t+4` Differentiating again, w.r.t. time we get `(dv)/(dt)=6` `impliesa=6 m//s^(2)` So, the force is `F=ma=0.5xx6=3N` The body m oves in a straight line and positive sign in force and velocity for the interval `0-2` shows that the angle between force and displacement is `0^(@).` The displacement `S=x_(2)-x_(g)` `=[3(2)^(2)+4(2)+5]-[3(0)^(2)+4(0)+5]` `=12+8+5-5=20m` So, work done is `W=3xx20=60J` |
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| 3. |
The power (P) of an engine lifting a mass of 100 kg upto a height of 10 m in 1 min isA. P=163.3 WB. P=9800 WC. P=10000 WD. P=5000 W |
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Answer» Correct Answer - A (a) `"Power"=("Work done")/("Time")` Work done `=mgh=100xx9.8xx10` Time =1 min=60 s `:. " " P=(100xx9.8xx10)/(60)=163.3 W` |
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| 4. |
A weight lifter lifts a weight off the ground and holds is up, thenA. work is done in lifting as well as holding the weight.B. no work is done in both lifting and holding the weight.C. work is done in lifting the weight but no work is required to done in holding it up.D. no work is done in lifting the weight but work is required to be done in holding it up. |
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Answer» Correct Answer - C When a weightlifter lifts a weight work done by the lifting force `W_1=Fdcos0^@=Fd` Work done in holding it up `W_2 =0` (because displacement is zero) |
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| 5. |
Calculate the worked done in lifting a 300N weight to a height of 10m with an accelertaion `0.5ms^(2)`. Take g=`10ms^(-2)`. |
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Answer» Given, Weight, w=300 N, Height, h=10 m Acceleration, a=0.5 `ms^(-2)` Here, `m=(w)/(g)=(300)/(10)`=30 kg, `F_(net)`=mg+ma `:.` Work done, W=mgh+mah=m(a+g)h =30(0.5+10)10=3150 J |
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| 6. |
A perosn is holding a bucket by applying a force of 20N. He moves a horizontal distance of 15m first and then climbs up a distance of 12m. What is the total work done by him? |
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Answer» Here, `F=20N, s_(1)=15m, theta_(1)=90^(@)` `s_(2)=12m, theta_(2)=0^(@), W=?` `W=W_(1)+W_(2)=Fs_(1)costheta_(1)+Fs_(2)costheta_(2)` `=20xx15cos90^(@)+20xx12cos0^(@)` `W=0+240=240J` |
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| 7. |
What is tidal energy ? How is it harnessed ? |
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Answer» As is known, hydropower is the least expensive way to generate electricity, and that too with least pollution problems caused by burning coal and oil to produce electricity. Twice each day, the Atlantic Ocean pours into the Bay of Fundy (in eastern Canada) producint the highest tides in the world ranging rom 12m to an extreme of 17m. The huge amount of kinetic energy available in these tides can be converted into electrical energy by building a dam and hydroelectric power plant at the entrance to the bay. Such a dam would obstruct the natural flow of tides. Every twelve hours, the incoming tides will rush through the open flood gates in the dam and be impounded behind the dam. When the tides recede, water would exit through turbines generating hydroelectric power. |
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| 8. |
A weight lifter does……………in lifting the weight………………..the ground but ……………in………………..the weight up. |
| Answer» some work `,` above `,` does no work `,` holding | |
| 9. |
While catching a cricket ball of mass `200g` moving with a velocity of `20ms^(-1)`, the player draws his hands backwards through `20cm`. Find the work done in catching the ball and the average force exerted by the ball on the hand. |
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Answer» Here, `m=200g=(1)/(5)kg, u=20m//s, upsilon=0,` `s=20cm=(1)/(5)m, W=?F=?` Work done `=` Loss of KE `=0-(1)/(2)m u^(2)=-(1)/(2)xx(1)/(5)(20)^(2)=-40J` As `W=Fxxs, F=(W)/(s)=(40)/(1//5)=200N` |
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| 10. |
Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m. |
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Answer» In this case, work is done only against friction between the road and tyres. Against gravity. `W=Fs cos 90^(@)=Zero .` |
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| 11. |
A car of mass 500 kg is driven with acceleration 1 `m//s^(2)` along straight level road against constant external resistance of 1000 N. When the velocity is `5 m//s` the rate at which the engine is working is :A. 5 kWB. 7.5 kWC. 2.5 kWD. 10 kW |
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Answer» Correct Answer - B |
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| 12. |
A block of mass m is attached to the two springs in vertical plane as shown in the figure - 3.95. If initially both the springs are at their natural lengths then velocity of the block is maximum at displacement x given as A. `x=(mg)/(2k)`B. `x= (mg)/k`C. `x=(mg)/(4k)`D. `x=(3mg)/(2k)` |
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Answer» Correct Answer - A |
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| 13. |
In the diagram shown, the block A and B are of the same mass M and the mass of the block C is `(M_(1))`. Friction is present only under the block A. the whole system is suddenly released from the state of rest. The minimum coefficient of friction on keep the block A in the state of rest is equal to . A. (a) `(M_(1))/(M)`B. (b) `(2M_(1))/(M)`C. (c) `(M_(1))/(2M)`D. (d) None of these` |
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Answer» Correct Answer - B `Let X_(m)` is maximum elongation of spring. Then, incerase in potential energy of spring`=`decrease in potential energy of `C`. `:. 1/2KX_(m)^(2) =M_(1)gX_(m)` or `KX_(m) =`maximum spring force `=2M_(1)g -mu_("min") Mg` `:. mu_("min") =(2M_(1))/(M)` |
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| 14. |
Two masses `m_(1) =10 Kg` and `m_(2)=5kg` are connected by an ideal string as shown in the figure. The coefficient of friction between `m_(1)` and the surface is `mu=0.2` Assuming that the system is released from rest calculate the velocity of blocks when `m_(2) has descended by `4m`. `(g=10 m//s^(2))` . |
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Answer» Correct Answer - D `v_(m_1) =v_(m_(2))` `d_(m_1) =h_(m_2) =4m` Using the equation, `E_(i)-E_(f) =` Work done against friction `0-[(1)/(2) xx (10 + 5) (v^(2))-5 xx 10 xx 4]` `=0.2 xx 10 xx 10 xx 4` Solving we get, `v=4m//s` |
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| 15. |
For example 32 consider target body B initially at rest, get the expressin for velocities of bodies after the collision `(a)m_(1)=m_(2)(b)m_(2)gt gtm_(1)(c)m_(2)ltltm_(1)` |
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Answer» Put `m_(1)=m_(2)and u_(2)=0` in equations (x) & (xi) (b) Put `m_(1)=0` in equations (x) &n (xi) (c ) Put `m_(2)=0` in equations (x) & (xi) `u_(2)=0` (given) From equation (10) `v_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))" "...(12)` From equation (11) `v_(2)=(2m_(1)u_(1))/(m_(1)+m_(2))" "...(13)` (a) Put `m_(1)=m_(2)=m` (say) in equations (12) & (13) `v_(1)=((m-m)u_(1))/(m+m)=0` `v_(2)=(2m u_(1))/(m+m)=u_(1)` i.e., body A comes to rest and body B starts mobing with the initial velocity of A `100%` KE of A is transferred to the body B. (b) `m_(2)gt gtm_(1)` `m_(1)` can be ignored Put `m=0` in equations (12) & (13) `v_(1)=(-m_(2))/(0+m_(2))u_(1)=-u_(1)` `v_(2)=(2xx0xxu_(1))/(0+m_(1))=0` When a light body a collides against a heavy body B at rest. A rebounds with its own velocity and B continous to be at rest e.g., a ball rabounds with same speed (direction of velocity is opposite) on striking a floor. (c ) (When target body B at rest has negligible mass) `m_(2) lt lt m_(1),m_(2)` can be ignored. Put `m_(2)=0` in equations (12) and (13) `v_(1)=-((m_(1)-0))/(m_(1)+0)u_(1)=u_(1)` `v_(2)=(2m_(1)u_(1))/(m_(1)+0)=2u_(1)` When a heavy body AS undergoes an elasticv collision witha light body at rest, A keeps on moving with the same velocity of its own and B starts moving with double the initial velosity of A. |
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| 16. |
If two objects collide and one is initially at rest (a) is it possible for both to be at rest after collision ? (b) is it possible for any one to be at rest after collision.? |
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Answer» (a) No, because momentum will not be conserved in that cae. (b) Yes, when masses of two objects are equal and collision is perfectly elastic. |
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| 17. |
A ball of mass m moving with vel. V strikes head on elastically with a number of balls of same mass at rest in a line. Only one ball from other side moves with same velocity. Explain wy not two balls move simultaneously, each with ve. `v//2` ? |
| Answer» When one ball of mass m moves with ve. V, both linear momentum and K.E. are conserved. If two balls were to move simultaneously, each with vel. `v//2` , linear momentum will be conserved, but K.E. will not be conserved. | |
| 18. |
Is it possible to have a collision in which the whole of KE is lost? |
| Answer» Yes. For example, in perfectly inelastic collision of two bodies moving towards each other with equal linear momenta. | |
| 19. |
1 kilowatt hour (kWh) is equal toA. `2.25xx10^22` eVB. `2.25xx10^23` eVC. `2.25xx10^25` eVD. `2.25xx10^27` eV |
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Answer» Correct Answer - C 1 kilowatt hour (kW h) (`10^3` W) x (3600 s) `=3.6 xx 10^(6) J=(3.6xx10^6)/( 1.6xx10^(-19))` eV ( `because` 1 eV= `1.6 xx 10^(-19 )`) `=2.25 xx 10^25 eV` |
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| 20. |
if work done by a conservative force is positive the select the correct option(s). (a) potential energy will decrease. (b) potential energy may increase or decrease. (c )kinetic energy will increase. (d) kinetic energy may increase or decrease. |
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Answer» Correct Answer - A::D `DeltaU =-W` so, if work done by conservative force is positive then `DeltaU` is negative or potential energy will decrease. But there is no straight forward rule regarding the kinetic energy. |
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| 21. |
A body of mass m is moving in a circle of radius r with a constant speed v, The force on the body is `(mv^(2))/(r )` and is directed towards the centre what is the work done by the from in moving the body over half the circumference of the circle?A. `(mv^(2))/(pi r^(2))`B. zeroC. `(mv^(2))/(r^(2))`D. `(pi r^(2))/(mv^(2))` |
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Answer» Correct Answer - B Work done by centripetal force is alwase zero because force and instantaneous displacement are always perpendicular `W = vecF - vec S = F s cos theta s cos (90^(@)) = 0` |
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| 22. |
A body of mass m is moving in a circle of radius r with a constant speed v, The force on the body is `(mv^(2))/(r )` and is directed towards the centre what is the work done by the from in moving the body over half the circumference of the circle?A. `(mv^(2))/(pir^(2))`B. zeroC. `(mv^(2))/(r^(2))`D. `(pir^(2))/(mv^(2))` |
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Answer» Correct Answer - B |
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| 23. |
If the unit of length and force be increased four times, then the unit of energy isA. 16 timesB. 8 timesC. 2 timesD. 4 times |
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Answer» Correct Answer - A |
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| 24. |
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare ? |
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Answer» Work done in stopping the body `=` force `xx` distance `=` K.E. of body , which is same for two bodies. As retarding force applied is the same, therefore, distance moved by both the bodies before coming to rest must be the same. |
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| 25. |
What happened to the ball when the moving car hit it? |
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Answer» Ball moves forward |
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| 26. |
An aeroplane’s velocity is doubled.(a) What happened to its momentum? Is the law of conservation of momentum obeyed?(b) What happens to its kinetic energy? Is the law of conservation of energy obeyed? |
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Answer» (a) The momentum of aeroplane will be doubled. Yes, the law of conservation of momentum will also be obeyed because increase in momentum of aeroplane is simultaneously accompanied by increase in momentum of exhaust gases. (b) K.E. becomes four times. Yes, the law of conservation of energy is obeyed with the increase in K.E. coming from the chemical energy of fuel, i.e., from the burning of its fuel. |
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| 27. |
What type of energy is stored in the spring of watch? |
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Answer» Potential energy. |
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| 28. |
A bomb of 12 kg divides in two parts whose ratio of masses is 1 : 3. If kinetic energy of smaller part is 216 J , then momentum of bigger part in kg m / sec will beA. 36B. 72C. 108D. Data is incomplete |
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Answer» Correct Answer - A |
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| 29. |
Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta isA. `4:1`B. `sqrt(2):1`C. `1:2`D. `1:16` |
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Answer» Correct Answer - C |
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| 30. |
Two bodies of masses 2 m and m have their K.E. in the ratio 8 : 1, then their ratio of momenta isA. `1:1`B. `2:1`C. `4:1`D. `8:1` |
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Answer» Correct Answer - C |
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| 31. |
Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of their linear momenta is thenA. `3:1`B. `9:1`C. `1:1`D. `sqrt(3):1` |
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Answer» Correct Answer - D |
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| 32. |
If a body of mass 200 g falls a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface. Then what is the decrease in P.E. of the body at the contact. `(g=10m//s^(2))`A. 200 JB. 400 JC. 600 JD. 900 J |
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Answer» Correct Answer - B |
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| 33. |
Consider the following two statements: A. Linear momentum of a system of partcles is zero. B. Kinetic energ of a system of particles is zero.A. 1 implies 2 and 2 implies 1B. 1 does not imply 2 and 2 does not imply 1C. 1 implies 2 but 2 does not imply 1D. 1 does not imply 2 but 2 implies 1 |
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Answer» Correct Answer - D |
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| 34. |
A light and a heavy body have equal momenta. Which one has greater K.EA. The light bodyB. The heavy bodyC. The K.E. are equalD. Data is incomplete |
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Answer» Correct Answer - A |
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| 35. |
A force of 5 N, making an angle `theta` with the horizontal, actig on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1J. The horizontal component of the force isA. 1.5NB. 2.5 NC. 3.5 ND. 4.5 N |
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Answer» Correct Answer - B |
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| 36. |
Tripling the speed of the motor car multiplies the distance needed for stopping it byA. 3B. 6C. 9D. Some other number |
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Answer» Correct Answer - C |
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| 37. |
A light body and a heavy body have same kinetic energy. Which one has greater linear momentum?A. The light bodyB. The heavy bodyC. Both have equal momentumD. It is not possible to say anything without additional information |
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Answer» Correct Answer - B |
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| 38. |
If momentum is increased by `20%` then K.E. increases byA. 0.44B. 0.55C. 0.66D. 0.77 |
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Answer» Correct Answer - A |
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| 39. |
A force of 5 N acts on a 15 kg body initially at rest. The work done by the force during the first second of motion of the body isA. 5JB. `(5)/(6)J`C. `6J`D. `75J` |
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Answer» Correct Answer - B |
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| 40. |
If the kinetic energy of a body increases by `0.1 %` the percent increase of its momentum will beA. 0.0005B. 0.001C. 0.01D. 0.1 |
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Answer» Correct Answer - A |
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| 41. |
If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy isA. 1.5B. 2C. 2.25D. 3 |
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Answer» Correct Answer - D |
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| 42. |
If the increase in the kinetic energy of a body is 22%, then the increase in the momentum will beA. 0.22B. 0.44C. 0.1D. 3 |
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Answer» Correct Answer - C |
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| 43. |
A free body of mass `8 kg` is travelling at `2` mater per second in a straight line. At a certain instant , the body splits into two equal parts due to internal wxplosion which releases `16 joules of energy . Neither part leves the original line of motion. FinallyA. Both parts continue to move in the same direction as that of the original bodyB. One part comes to rest and the other moves in the same direction as that of the original bodyC. One part comes to rest and the other moves in the direction opposite to that of the original bodyD. One part moves in the same direction and the other in the direction opposite to that of the original body |
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Answer» Correct Answer - B |
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| 44. |
A body of mass` 5 kg` is moving with a momentum of `10 kg m//s`. A force of `0.2 N` acts on it in the direction of motion of the body for `10 sec.` The increase in its kinetic energy.A. 2.8 joulesB. 3.2 joulesC. 3.8 joulesD. 4.4 joules |
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Answer» Correct Answer - D |
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| 45. |
If the momentum of a body increases by 0.01%, its kinetic energy will increase byA. 0.0001B. 0.0002C. 0.0004D. 0.0008 |
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Answer» Correct Answer - B |
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| 46. |
A block of mass `m` attached with a massless spring of force constant k The block is placed over a rough inclined surface for which the coefficient of friction is `0.5` the system is released from rest when the spring was unstretched The minimum mass of the hanging block to move the small block "m" up the plane is (neglect mass of spring and pulley and friction in pulley) |
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Answer» Correct Answer - `[[mg(sintheta+mu_(s)costheta)]^(2)/(2k)]` |
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| 47. |
A body of mass 10kg at rest is acted upon simultaneously by two forces 4 N and 3 N at right angles to each other. The kinetic energy of the body at the end of 10 sec isA. 100 JB. 300 JC. 50 JD. 125 J |
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Answer» Correct Answer - D |
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| 48. |
A block of mass `m` at rest is acted upon by a force `F` for a time t. The kinetic energy of block after time t isA. `(F^(2)t^(2))/(m)`B. `(F^(2)t^(2))/(2m)`C. `(F^(2)t^(2))/(3m)`D. `(Ft)/(m)` |
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Answer» Correct Answer - B |
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| 49. |
Point out the correct alternative: 1. When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains same. 2. Work done by a body against friction always results in a loss of its kinetic/potential energy. 3. The rate of change of total momentum of a many-particle system is proportional to the external force/ sum of the Internal forces on the system. 4. In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies. |
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Answer» 1. Potential energy decreases 2. Kinetic energy 3. External force 4. Total linear momentum/total energy of two bodies. |
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| 50. |
State if each of the following statements is true or false. Give reasons for your answer. 1. In an elastic collision of two bodies, the momentum and energy of each body is conserved. 2. Total energy of a system is always conserved, no matter what internal and external forces on the body are present. 3. Work done in the motion of a body over a closed loop is zero for every force in nature. 4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. |
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Answer» 1. False, the momentum and energy of each body is conserved. 2. False, the external force on the system may increase or decrease the total energy of the system. 3. False, for the nonconservative forces (friction) the work done in closed loop is not zero. 4. True, usually in an inelastic collision the final kinetic energy is always less than initial kinetic energy of the system. |
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