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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 weight lifter jerks `220 kg` vertically through `1.5` metre and holds still at that height for two minutes. The work done by him in lifting and in holding it still are respectivelyA. `220 J, 330 J`B. `3234J, 0 J`C. `2334 J, 10 J`D. `0 J, 3234 J` |
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Answer» Correct Answer - B `W = vec(F).vec(S) = FS cos theta` In lifting the weight `F = mg, theta = 0^(@)`, in holding the weight, `S = 0` |
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| 2. |
A board is moving with velocity `v` on a smoother horizontal plane. The upper surface of the board is rough on which a ball falls with velocity `v` and rebounds with velocity `(v)/(2)`. The mass of the board is same as that of ball. After the collision, the board comes to state of rest. The co-efficient of friction between the board and the ball isA. `(1)/(2)`B. `(2)/(3)`C. `(1)/(4)`D. `(3)/(5)` |
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Answer» Correct Answer - B Impluse due to normal force on the ball `N(dt) = m(v+(v)/(2)) , N dt = (3mv)/(2)` impluse due to frictional force on plank `mu N (dt) = m(v-0), mu((3mv)/(2)) = mv , mu = (2)/(3)` |
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| 3. |
An object of mass 5 kg falls from rest through a vertical distance of `20 m` and attains a velocity of 10 m/s. How much work is done by the resistance of the air on the objrct? `(g=10m//s^(2))`.A. `750J`B. `-750 J`C. `850 J`D. `-650 J` |
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Answer» Correct Answer - B `W_("mg")+W_("air") = Delta K.E`. |
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| 4. |
A massless spring with a force constant `K = 40N//m` hangs vertically from the ceiling.A `0.2 kg` block is attached to the end of the spring and held in such a position that the spring has its natural length and suddenyl released. The maximum elastic strain energy stored in the spring is `(g=10 m//s^(2))`A. `0.1J`B. `0.2J`C. `0.05J`D. `0.4J` |
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Answer» Correct Answer - B `PE = (1)/(2)kx^(2),x = 2(mg)/(k)` |
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| 5. |
A ball of mass m is thrown in air with speed `v_(1)` from a height `h_(1)` and it is at a height `h_(2)(gth_(1))` when its speed becomes `v_(2)` . Find the work done on the ball the air resistance.A. `m g(h_(2)-h_(1))+(1)/(2)m(v_(2)^(2)-v_(1)^(2))`B. `m g(h_(2)-h_(1))`C. `(1)/(2)m(v_(2)^(2)-v_(1)^(2))`D. `m g (h_(2)-h_(1))-(1)/(2)(v_(2)^(2)-v_(1)^(2))` |
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Answer» Correct Answer - A `W_("mg")+W_("air") = Delta K.E`. |
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| 6. |
A stone of mass `6 kg` is revolved in a vertical circle of diameter `6m`., such that its speed is minimum at a point. If the `K.E` at the same point is `250 J`, then minimum `PE` at that point isA. `200 J`B. `150 J`C. `100 J`D. `450 J` |
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Answer» Correct Answer - A `T.E_("bottom") = T.E_("given point")(1)/(2)m(sqrt(5 gr))^(2) = P.E+K.E` |
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| 7. |
A car is travelling along a flyover bridge which is a part of vertical circle of radius `10m`. At the highest point of it the normal reaction on the car is half of it’s weight, the speed of car isA. `7m//s`B. `10m//s`C. `14m//s`D. `20m//s` |
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Answer» Correct Answer - A Normal reaction, `N = mg - (m v^(2))/(r )` Given that `N = (mg)/(2)` |
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| 8. |
A vehicle is travelling with uniform speed along a concave road of radius of curvature `19.6m`. At lowest point of concave road is the normal reaction on the vehicle is three times its weight, the speed of vehicle isA. `4.9 m//s`B. `9.8 m//s`C. `14.7 m//s`D. `19.6 m//s` |
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Answer» Correct Answer - D Normal reaction, `N = mg+(m v^(2))/(r )`, Given `N = 3 mg` |
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| 9. |
A vehicle of mass 1000 kg is moving with a velocity of `15 ms^(-1)`.It is brought to rest by applying brakes and locking the wheels. If the slidding friction between the tyres and the road is 6000 N, then the distance moved by the vehicle before coming to rest isA. `37.5 m`B. `18.75m`C. `75m`D. `15 m` |
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Answer» Correct Answer - B `W = FS = Delta KE` |
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| 10. |
A projectile of mass `50 kg` is shot vertically upwards with n initial velocity of `100m//s`. After `5s`, it explodes into two fragments, one of (1st fragment) which having a mass of `20 kg` travels vertically up with a velovity of `150m//s (g = 10 m//s^(2))`. Bases on the above paragraph answer the following questions. What is the magnitude and direction of velocity of the 2nd fragments just after explosion isA. `(50)/(3)m//s` (up)B. `50m//s` (down)C. `50m//s`D. `(50)/(3)m//s` (down) |
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Answer» Correct Answer - D Velocity of projectile after `5s` is `= 50 m//s` Total momentum before explosion `=` Total momentum after explosion. `m xx 50 = m_(1)v_(1)+m_(2)v_(2)` `50 xx 50 = 20 xx 150 - 30 xx v_(2)` `v_(2) = (50)/(3)m//s` (velocity of 2nd fragment just after explosion) velocity of `2^(nd)` fragment `3s` after the explosion is `= (50)/(3) + 10 xx 3 = (140)/(3)`, momentum of `2^(nd)` fragement `3s` after the explosion is `(140)/(3) xx 30` Velocity of `1^(st)` one `3s` after the explosion is `= 150-10 xx 3 = 120 m//s` Sum of momentum of two fragment `3s` after the explosion `= 20 xx 120 - (30 xx 140)/(3)=2400-1400 = 1000 kg-m//s` |
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| 11. |
Wedges `B` and `C` are smooth and they are placed in contact as shown. Block `A` is placed on wedge `B` at a height `h` above ground. Block and the two wedges are all of same mass `m`. Neglect friction every where. The velocity of `A` when it has slide down to ground from wedge `C` isA. `0`B. `sqrt((gh)/(2))`C. `sqrt((gh)/(4))`D. `sqrt(gh)/(3)` |
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Answer» Correct Answer - A `mgh = ((1)/(2)mv^(2))2` `v = sqrt(gh)` From conservation of momentum `msqrt(gh) = 2mv_("common")` `v_("common") = sqrt(gh)/(2)` From conservation of energy `(1)/(2)mgh = mgh + (1)/(2)xx2mxx((gh)/(4))` `rArr h = (h)/(4)` |
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| 12. |
A body slides down a fixed curved track that is one quadrant of a circle of radius `R`, as in the figure. If there is no friction and the body starts from rest, its speed at the bottom of the track isA. `5g R`B. `sqrt(5 g R)`C. `sqrt(2g R)`D. `sqrt(gR)` |
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Answer» Correct Answer - C Loss of `P.E = "gain in K.E"` , `mgh = (1)/(2) m v^(2)` |
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| 13. |
A block of mass `1kg` slides down a curved track which forms one quadrant of a circle of radius `1m` as shown in figure. The speed of block at the bottom of track is `v=2 ms^(-1)`. The work done by the of friction is .A. `8J`B. `-8J`C. `4J`D. `-4J` |
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Answer» Correct Answer - B `W_("friction") = K.E_(f)-P.E_(i)` |
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| 14. |
A block of mass `4 kg` slides on a horizontal frictionless surface with a speed of `2m//s`. It is brought to rest in compressing a spring in its path. If the force constant of the spring is `400 N//m`, by how much the spring will be compressedA. `2 xx 10^(-2)m`B. `0.2 m`C. `20m`D. `200 m` |
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Answer» Correct Answer - B `K.E.` of block is converted into elastic `P.E.` in spring i.e., `(1)/(2) mv^(2) = (1)/(2)k x^(2)` |
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| 15. |
A block of mass `10 kg` slides down a rough slope which is inclined at `45^(0)` to the horizontal. The coeffficient of sliding friction is `0.30`. When the block has to slide `5m`, the work done on the block by the force of friction is nearlyA. `115 J`B. `-75 sqrt(2) J`C. `321.4 J`D. `-321.4 J` |
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Answer» Correct Answer - B `W = -f xx S = - mu mg cos theta xx S` |
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| 16. |
A block of mass `4 kg` is initially at rest on a horizontal frixtionless surface. A horizontal force `vec(F) = (3+x) hat(i)` newtons acts on it, when the block is at `x = 0`. The maximum kinetic energy of the block between `x = 0` and `x = 2 m` isA. `6 J`B. `8J`C. `9J`D. `10 J` |
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Answer» Correct Answer - B `Delta KE = W = underset(x_(1))overset(x_(2))int vec(F).vec(dx)` |
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| 17. |
A block of mass `4 kg` is initially at rest on a horizontal frictionless surface. A force `vec(F) = (2x+3x^(2))hat(i) N` acts horizontally on it. The maximum kinetic energy of the block between `x = 2m` and `x=4m` in joules isA. `40`B. `36`C. `68`D. `52` |
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Answer» Correct Answer - C `K.E = W = underset(x_(i))overset(x_(f))intf(x)dx` |
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| 18. |
If a force `vec(F) = (vec(i) + 2 vec(j)+vec(k)) N` acts on a body produces a displacement of `vec(S) = (4vec(i)+vec(j)+7 vec(k))m`, then the work done isA. `9J`B. `13 J`C. `5J`D. `1J` |
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Answer» Correct Answer - B `W = vec(F).vec(S)` |
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| 19. |
A body of mass `6kg` is under a force which causes displacement in it given by `S= (t^(2))/(4)` maters where `t` is time . The work done by the force in `2` sec isA. `12J`B. `9J`C. `6J`D. `3J` |
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Answer» Correct Answer - D `W = (1)/(2)m v^(2)`, `v = (d s)/(d t)` |
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| 20. |
A ball of mass `m = 1 kg` is hung vertically by a thread of length `1=1.50 m`. Upper end of the thread is attached to the ceiling of a trolley of mass `M = 4kg`. Initially, the trolley is stationary and it is free to move along horizontal rails with out friction. A shell of mass `m=1 kg`, moving horizontally with velocity `v_(0) = 6m//s` collides with thread starts to deflect towards right At the time a maximum deflection of the thread with vertical, the trolley will move with velocityA. `2m//s`B. `3 m//s`C. `1m//s`D. `4m//s` |
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Answer» Correct Answer - C When shell strikes the ball and gets stuck with it, combined body of mass `2m` starts to move to the right. Let velocity of the combined body (just after collision) be `v_(1)`. acoording to law of conservation of momentum, `(m+m) v_(1) = m v_(o)`, `v_(1) = (v_(0))/(2) = 3m//s` As soon as the combined body starts to move rightward, thread becomes inclined to the vertical. Horizontal component of its tension retards the combined body while trolley accelerates rightward due to the same component of tension. Inclination of thread with the vertical continues to increase till velocities of both (combined body and trolley) become identical or combined body comes to rest relative to the trolley. Let velocity at that instant of maximum inclnation of thread be `v`. Acoording to law of conservation of momentum, `(2m+M) v = 2mv , v = 1m//s` During collision of ball and shell, a part of energy is lost. but after that there is no lose of energy. Hence, after collision, kinetic energy lost is used up in increasing gravitational potential energy of the combined body. If maximum inclination of thread with the vertical is `theta`, then according to law of conservation of energy. `(1)/(2)(2m)v_(1)^(2)-(1)/(2)(2m+M)v^(2) = 2mgl(1- cos theta)` `cos theta = 0.8` (or) `theta = 37^(@)` |
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| 21. |
A man standing on the edge of the roof of a `20 m` tall building projects a ball of mass `100 gm` vertically up with a speed of `10 ms^(-1)`. The kinetic energy of the ball when it reaches the ground will be `[g = 10 ms^(-2)]`A. `5 J`B. `20 J`C. `25 J`D. Zero |
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Answer» Correct Answer - C `K.E_(f) = K.E_(i) + P.E_(i)` |
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| 22. |
A meter scale of mass `400 gm` is lying horizontally on the floor. If it is to be held vertically with one end touching the floor, the work to be done isA. `6 J`B. `4 J`C. `40 J`D. `2J` |
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Answer» Correct Answer - D `W = vec(F).vec(S) = FS_("com")` , where `F = mg` and `S_("com") = (L)/(2)` |
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| 23. |
A tennis ball has a mass of `56.7 gm` and is served by a player with speed of `180 kmph`. The work done in serving the ball is nearlyA. `710 J`B. `71 J`C. `918 J`D. `91.8 J` |
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Answer» Correct Answer - B `W = (1)/(2)m v_(i)^(2)` , |
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| 24. |
A uniform bar of length `12 L` and mass `48 m` is supported horizontally on two smooth tables as shown in the figure. A small moth (an insect) of mass `8m` is sitting on end `A` of the rod and a spider (an insect) of mass `16 m` is sitting on the other end `B`. Both the insects start moving towards each other along the rod with moth moving at speed `2v` and the spider at half of this speed. They meet at a point `P` on the rod and the spider eats the moth. After this the spider moves with a velocity `v//2` relative to the rod towards the end `A`. The spider takes negligible time in eating the insect. Also, let `v = L//T`, where `T` is a constant having value `4 sec`. The point `P` is atA. the centre of the rodB. the edge of the table supporting the end `B`C. close to the edge of the table supporting the end `A`D. none of the above |
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Answer» Correct Answer - B Displacement of moth `= (12L)/(3) xx 2 = 8L` `therefore` Point `P` is at the table supporting `B`. |
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| 25. |
A uniform bar of length `12 L` and mass `48 m` is supported horizontally on two smooth tables as shown in the figure. A small moth (an insect) of mass `8m` is sitting on end `A` of the rod and a spider (an insect) of mass `16 m` is sitting on the other end `B`. Both the insects start moving towards each other along the rod with moth moving at speed `2v` and the spider at half of this speed. They meet at a point `P` on the rod and the spider eats the moth. After this the spider moves with a velocity `v//2` relative to the rod towards the end `A`. The spider takes negligible time in eating the insect. Also, let `v = L//T`, where `T` is a constant having value `4 sec`. The speed of the bar after the spider eats up the moth and moves towards `A` isA. `v//2`B. `v`C. `v//6`D. `2v` |
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Answer» Correct Answer - C After spider eats the moth let speed of rod becomes `v_(0)` (towards right) `therefore` Speed of spider relative to rod `= (v)/(2)` (left ward) From com `24m ((v)/(2)-v_(0)) = 48 mv_(0) , v_(0) = v//6` |
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| 26. |
A uniform bar of length `12 L` and mass `48 m` is supported horizontally on two smooth tables as shown in the figure. A small moth (an insect) of mass `8m` is sitting on end `A` of the rod and a spider (an insect) of mass `16 m` is sitting on the other end `B`. Both the insects start moving towards each other along the rod with moth moving at speed `2v` and the spider at half of this speed. They meet at a point `P` on the rod and the spider eats the moth. After this the spider moves with a velocity `v//2` relative to the rod towards the end `A`. The spider takes negligible time in eating the insect. Also, let `v = L//T`, where `T` is a constant having value `4 sec`. Displacement of the rod by the time when the insects meet isA. `L//2`B. `L`C. `3L//4`D. zero |
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Answer» Correct Answer - C `P_("moth") = 8m(2v) = 16mv` , `P_("spider") =16mv` `P_("system") = 0` here rod does not move. |
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| 27. |
A `1 kg` mass at rest is subjected to an acceleration of `5m//s^(2)` and travels `40 m`. The average power during the motion isA. `40W`B. `8W`C. `50W`D. `200 W` |
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Answer» Correct Answer - C `P_("avg") = (W)/(t) = (FS)/(t) = (m a S)/(t)` , `v = sqrt(2 as), t = (v)/(a)` |
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| 28. |
An object is acted on by a retarding force of `10 N` and at a particular instant its kinetic energy is `6J`. The object will come to rest after it has travelled a distance ofA. `3//5`B. `5//3`C. `4m`D. `16m` |
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Answer» Correct Answer - A `Delta KE = W rArr KE_(i) = FS` |
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| 29. |
If the mass of a moving body decreased by one third of its initial mass and velocity is tripled, then the percentage change in its kinetic energy isA. `500%`B. `600%`C. `300%`D. `200%` |
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Answer» Correct Answer - A `K = (1)/(2)m v^(2) rArr (K_(1))/(K_(2)) = (m_(1)v_(1)^(2))/(m_(2)v_(2)^(2))` |
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| 30. |
A simple pendulum is swinging vertical plane. The ratio of its potential energy when it is making `45^(@)` and `90^(@)` with the vertical isA. `1:1`B. `1:(sqrt(2)+1)`C. `sqrt(2):1`D. `(sqrt(2)-1):sqrt(2)` |
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Answer» Correct Answer - D `U=mgl(1-cos theta)` |
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| 31. |
A steel wire can withstand a load up to `2940N`. A load of `150 kg` is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire dose not break when the load passs through the position of equilibrium, isA. `30^(@)`B. `60^(@)`C. `80^(@)`D. `85^(@)` |
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Answer» Correct Answer - B While crossing the equilibrium position `T = mg+(m v^(2))/(r ) = mg+(m)/(l)[2gl(1- cos theta)]` `= mg+2mg-2mg cos theta = mg[3-2 cos theta]` |
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| 32. |
An elevator can carry a maximum load of `1800 kg` (elevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force opposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?A. `59`B. `8`C. `22`D. `20` |
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Answer» Correct Answer - A `F = mg + "frictional force"`, `P = vec(F).vec(v)` |
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| 33. |
A body of mass `m` is rotated in a vertical circle of radius `R` by means of light string. If the velocity of body is `sqrt(gR)` while it is crossing highest point of vertical circle then the tension in the string at that instant isA. `2 mg`B. `mg`C. `(mg)/(2)`D. Zero |
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Answer» Correct Answer - D Tension at the highest point `T = (m v^(2))/(r )-mg` where `v = sqrt(gr)` |
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| 34. |
A body of mass `m` is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If `T_(1), T_(2)` be the tensions in the string when the body is crossing the highest and the lowest positions then the following relation is correctA. `T_(2)-T_(1) = 6mg`B. `T_(2)-T_(1) = 4mg`C. `T_(2)-T_(1) = 3mg`D. `T_(2)-T_(1) = 2mg` |
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Answer» Correct Answer - A `T_(1) = (m v_(1)^(2))/(r )-mg, T_(2) = (m v_(2)^(2))/(r )+mg` `v_(2)^(2)-v_(1)^(2) = 4gr, v_(1) = sqrt(g r)` |
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| 35. |
A spring of force constant `800 Nm^(-1)` is stretched initially by `5 cm`. The work done in stretching from `5 cm` to `15 cm` isA. `12.50 N-m`B. `18.75 N-m`C. `25.00 N-m`D. `6.25 N-m` |
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Answer» Correct Answer - C `W = (1)/(2)K(x_(2)^(2)-x_(1)^(2))` |
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| 36. |
A ball is dropped from the top of a tower. The ratio of work done by force of gravity in `1^(st), 2^(nd)`, and `3^(rd)` second of the motion of ball isA. `1:2:3`B. `1:4:16`C. `1:3:5`D. `1:9:5` |
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Answer» Correct Answer - C `W = FS rArr(W_(1))/(W_(2)) = (S_(1))/(S_(2)) = ((2n-1)_(1))/((2n-1)_(2))` `(because S = u+a(n-(1)/(2)))` |
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| 37. |
If the power of the motor of a water pump is `3kW`, then the volume of water in llitres that can be lifted to a height of `10 m` in one minute by the pump is `(g = 10 ms^(-2))`A. `1800`B. `180`C. `18000`D. `18` |
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Answer» Correct Answer - A `P = (W)/(t) = (mgh)/(t)` and `m = V rho` |
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| 38. |
A force `F` is applied on a lawn move at an angle of `60^(@)` with the horizontal. If it moves through a distance `x`, the work done by the force isA. `Fx//2`B. `F//2x`C. `2Fx`D. `2x//F` |
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Answer» Correct Answer - A `W = vec(F).vec(S) = FS cos theta` |
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| 39. |
A block is constrained to move along `x`axis under a force `F=-(2x)N`. Find the work done by the force when the block is displaced from `x=2 m` to `x = 4m`A. `12J`B. `8J`C. `-12J`D. `-8J` |
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Answer» Correct Answer - C `W = underset(x_(i))overset(x_(f))int f(x)dx` |
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| 40. |
Two identical ball of radii `r` are kept on a horizontal plane with their centres `d` distance apart. A third ball, identical to previous one, collide elastically with both the balls symmetrically as shown in the figure. If the third ball comes to rest after the collision, `d` should be A. `3r`B. `2sqrt(2r)`C. `(sqrt(2)+1)r`D. `(sqrt(2)+2)r` |
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Answer» Correct Answer - B `v_(2) = (4)/(2 cos theta), e = (v_(2)-v_(1))/(u_(1)-u_(2))` `rArr cos theta = (1)/(sqrt(2)) rArr d = 2sqrt(2)r` |
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| 41. |
The kinetic energy of a projectile at the highest point of its path is found to be `3//4^(th)` of its initial kinetic energy. If the body is projected from the ground, the angle of projection isA. `0^(@)`B. `30^(@)`C. `60^(@)`D. `40^(@)` |
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Answer» Correct Answer - B `K.E_("top") = (3)/(4)(K.E_(i)),(1)/(2)m(u cos theta)^(2) = (3)/(4) ((1)/(2)m u^(2))` |
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| 42. |
A weight of `5N` is moved up a frictionless inclined plane from `R` to `Q` as shown. What is the work done in joules?A. `15`B. `20`C. `25`D. `35` |
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Answer» Correct Answer - A `W = mgh` |
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| 43. |
A simple pendulum is oscillating with angular displacement `90^(@)` For what angle with vertical the acceleration of bob direction horizontal?A. `sin^(-1)((1)/(3))`B. `cos^(-1) ((1)/(3))`C. `sin^(-1)((1)/(sqrt(3)))`D. `cos^(-1)((1)/(sqrt(3)))` |
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Answer» Correct Answer - D `tan theta = (a_(c ))/(a_(t)) = (v^(2)//r)/(g sin theta)` but `v^(2) = 2 gh`, `h = r cos theta` |
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| 44. |
A `16 gm` mass is moving in the `+x` direction at `30 cm//s` while a `4 gm` is moving in the `-x` direction at `50 cm//s`. They collide head - on and stick together. Their common velocity after impact isA. `0.14 cm//s`B. `0.14 m//s`C. `0 ms^(-1)`D. `0.3 m//s` |
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Answer» Correct Answer - B `v = (m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))` |
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| 45. |
Two identical balls collide head on. The initial velocity of one is `0.75 ms^(-1)` , while that of the other is `-0.43 ms^(-1)`. If the collision is perfectly elastic, then their respective final velocities areA. `0.75ms^(-1) , -0.43 ms^(-2)`B. `-0.43 ms^(-1) , 0.75 ms^(-1)`C. `-0.75 ms^(-1) , 0.43 ms^(-1)`D. `0.43 ms^(-1) , 0.75 ms^(-1)` |
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Answer» Correct Answer - B `vec(v_(1)) = ((m_(1)-m_(2))/(m_(1)+m_(2)))vec(u_(1))+((2m_(2))/(m_(1)+m_(2)))vec(u_(2))` `vec(v_(2)) = ((2m_(1))/(m_(1)+m_(2)))vec(u_(1))+((m_(2)-m_(1))/(m_(1)+m_(2)))vec(u_(2))` |
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| 46. |
Two pendulum bobs of masses m and 2m collide head on elastically at the lowest point in their motion. If both the balls are released from a height H above the lowest point, to what heights do they rise for the first time after collision?A. `(25H)/(9)`B. `(H)/(9)`C. `(16 H)/(9)`D. `(H)/(4)` |
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Answer» Correct Answer - A `h_(1) = (v_(1)^(2))/(2g) = [(5)/(3)sqrt(2gH)]^(2)/(2g), h_(1) = (25)/(9)H` |
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| 47. |
If the kinetic energy of a body increases by `125%`, the percentage increases in its momentum isA. `50%`B. `62.5%`C. `250%`D. `200%` |
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Answer» Correct Answer - A `P = sqrt(2mKE)` `((P_(2)-P_(1))/(P_(1))) xx 100 = ((sqrt(KE_(2))-sqrt(KE_(1)))/(sqrt(KE_(1)))) xx 100` |
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| 48. |
A spring of spring constant `5 xx 10^(3) N//m` is stretched initially by 5 cm from the unstretched position. The work required to further stretch the spring by another 5 cm is .A. `6.25 Nm`B. `12.50 Nm`C. `18.75 Nm`D. `25 Nm` |
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Answer» Correct Answer - C `W = (1)/(2)K (x_(2)^(2)-x_(1)^(2))` |
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| 49. |
A woman weighing `63 kg` eats plum cake whose energy content is `9800` calories. If all this energy could be utilized by her, she can ascend a height ofA. `1m`B. `67m`C. `100 m`D. `42 m` |
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Answer» Correct Answer - B `W = JQ rArr mgh = JQ (1 cal = 4.2J)` |
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| 50. |
The friction coefficient between the horizontal surface and each of the block shown in the figure is `0.2`. The collision between the blocks is perfectly elastic. Find the separation between them when they come to rest. (Take `g=10m//s^2`). |
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Answer» Correct Answer - 5 Velocity of first block before collision `v_(1)^(2)=l^(2)-2(2)xx0.16=1-0.64, v_(1)=0.6m//s` By conservation of momentum `2xx0.6=2v_(1)+4v_(2)` also `v_(2)-v_(1)` for elastic collision It gives `v_(2)=0.4 m//s, v_(1)=-0.2 m//s` Now distance moved after collision `S_(2)=((0.4))/(2xx2)` and `S_(1)=((0.2)^(2))/(2xx2), s=s_(1)+s_(2)=0.05m = 5 cm`. |
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