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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.
| 401. |
In a ballistics demonstration, a police officer fires a bullet mass `50.0g` with speed `200ms^(-1)` on soft plywood of thickness 2.00cm. The bullet emerges only with `10%` of its initial kinetic energy. What is the emergent speed of the bullet ? |
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Answer» Here, `m=50.0g` `=(50)/(1000)kg=(1)/(20)kg` `v_(i)=200ms^(-1)` `:.` Initial K.E., `K_(i)=(1)/(2)mv_(i)^(2)` `=(1)/(2)xx(1)/(20)(200)^(2)=1000J` Final K.E., `K_(f)=10% (K_(i))` `=(10)/(100)xx1000J=100J` If `v_(f)` is emergent speed of the bullet, then `(1)/(2)mv_(f)^(2)=K_(f)` `v_(f)=sqrt((2K_(f))/(m))=sqrt((2xx100)/(1//20))=63.2m//s` Note that K.E. is reduced by `90%` but speed is reduced by nearly `68%`. |
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| 402. |
A bullet of mass 10 g moving with a velocity `300 ms^(-1)` strikes a wooden block and comes out from ot her side with a velocity of `200ms^(-1).` Find the work done by t he resistive force on the bullet. |
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Answer» `DeltaK=K_(f)-K_(i)=W` `W=1/2mv ^(2)mu^(2).m=10xx10^(-3)kg,u=300m//sv=200m//s` `W=K_(f)-K_(f)=1/2mv^(2)-1/2mu^(2)=1/2xx10xx10^(-3)xx(200^(2)-300^(2))` `=-5xx10^(-3)xx500x100=-250J` |
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| 403. |
A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass `10^(-2)` kg moving with a speed of `2 xx 10^(2) m s^(-1)`. The height to which the bob rises before swinging back is (Take `g = 10 m s^(-2)`)A. 0.2 mB. 0.6 mC. 8 mD. 1 m |
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Answer» Correct Answer - A Momentum of bullet =`10^(-2)xx2xx10^2 = 2 kg m s^(-1)` Let the combined velocity of the bob + bullet =v Momentum of bob + bullet =`(10^(-2) + 1)v=1.01 v` By conservation of momentum, 1.01 v =`2 kg m s^(-1)` or `v=2/1.01 = 1.98 m s^(-1)` By conservation of energy `1/2(M+m)v^2 =(M+m)gh` or `h=v^2/(2g)=((1.98)^2)/(2xx10)=0.196 ~=0.2` m |
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| 404. |
A pump is used to fill a tank 6m × 3m × 2m in half an hour. If the average height to which water to be lifted from the well is 110m, find the horsepower of the engine. |
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Answer» Volume of the tank = 6 × 3 × 2 = 36m3 ∴ mass of water to be lifted = V × d = 36 × 103 kg ∴ Work done = mg × h = 36 × 103 × 9.8 × 110 Power of the pump = \(\frac{w}{t}\) = \(\frac{36 \times 9.8\times1.1 \times 10^5}{30 \times60}\)= 0.216 × 105W =\(\frac{0.216\times 10^5}{30 \times746}\) = \(\frac{21600}{746}\) 28.9 HP. |
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| 405. |
State the law of conservation of energy. |
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Answer» Energy can neither be created nor be destroyed, but can only be converted from one form to another, i.e., The total energy of a closed system remains constant. |
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| 406. |
Fill in the blanks(a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ………. (b) The two factors related to potential energy to body are…………. (c) 1HP = ……… Watt |
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Answer» (a) 1 Joule (b) Position, Strain (c) 746W |
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| 407. |
State law of conservation of energy? |
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Answer» Energy can neither be created nor be destroyed. It can be transformed from one form to another. |
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| 408. |
A block of mass m is 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 `mu=(3)/(4)`. The minimum value of M required to move the block up the plane is (Neglect mass of string, mass of pulley and friction in pulley) A. `(3)/(5)m`B. `(4)/(5)m`C. `(6)/(5)m`D. `(3)/(2)m` |
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Answer» Correct Answer - A |
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| 409. |
A force F=kx (where k is a positive constant) is acting on a particle Work done: `{:(,"Column-1",," Column-2"),("(A)","in displacing the body from x=2 to x=4",,"(P) Negative"),("(B)","In displacing the body from x=-4 to x=-2",,"(Q) Positive"),("(C)","In displacing the body from x=-2 to x=+2",,"(R) Zero"):}` |
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Answer» Correct Answer - (A) Q, (B) P, (C ) R |
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| 410. |
A body is moved along a straight line by a machine delivering a power proportional to time `(P prop t)`. Then, match the following. |
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Answer» Correct Answer - (A) P, (B) Q, (C ) Q |
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| 411. |
A `4 kg` block is on a smooth horizontal table. The block is connected to a second block of mass `1 kg` by a massless flexible taut cord that passes over a frictionless pulley. The `1 kg` block is `1m` above the floor. The two block are released from rest. With what speed does the `1 kg` block hit the ground? . |
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Answer» Correct Answer - 2 |
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| 412. |
A `10-kg` block moves in a straight line under the action of a force that varies with position as shown in figure. How much work does the force do as the block moves from origin to `x=8m`? |
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Answer» The area under the curve with respect to x-axis gives the total work done by the force as the block moves from `x=0` to `x=8m`. Area=Work done `=20+1/2(2)(10)+0-1/2(5)(2)=25J` |
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| 413. |
A body moves from point A to B under the action of a force, varying in magnitude as shown in figure. Obtain the work done. Force is expressed in newton and displacement in meter. |
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Answer» Work done = area under F-s graph `W_(AB)=W_(AP)+W_(PQ)+W_(QR)+W_(RB)` `=10xx1xx1/2(10+15)xx1+1/2xx1xx15-1/2xx1xx15` `=22.5J` |
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| 414. |
The kinetic friction force acting on a sliding body moving in a straight line is constant, but it is non-conservative. Explain. |
| Answer» Kinetic friction depends on the velocity as it opposes the relative sliding. When the block slide towards right, let it experience, a kinetic friction f towards left at a point P. When it retraces its path, at the same point P, it experience as kinetic friction of same magnitude f but in opposite direction. Hence, work done by friction in to and fro journey will not cancel each other. As friction opposes the displacement, always the total work performed by it is non-zero. It means, the total frictional work done in a closed path depends upon the length of the path. Hence it is non-conservative force. The total work done by kinetic friction acting on two contacting surface is negative. However, the total work done by kinetic friction between two surfaces in contact does not depend on the reference frame. | |
| 415. |
A body is thrown on a rough surface such that the friction force acting on it is linearly varying with the distance travelled by it as `f=ax+b`. Find the work done by the friction on the box if before coming to rest the box travels a distance s. |
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Answer» As the force is acting in the direction opposite to the motion of the box, work done by this force must be negative. As force is not constant, we use `W=-underset0oversetsintintfdx=-underset0oversets int(ax+b)*dx=((ax^2)/(2)+bx)_0^s` `=-1/2as^2-bs` |
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| 416. |
A pendulum bob swings along a circular path on a smooth inclined plane as shown in figure, where `m=3kg`, `l=0.75m`, `theta=37^@`. At the lowest point of the circle the tension in the string is `T=274N`. Take `g=10ms^-2`. The speed of the bob at the highest point on the circle isA. (a) `sqrt46ms^-1`B. (b) `sqrt26ms^-1`C. (c) `sqrt52ms^-1`D. (d) `sqrt35ms^-1` |
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Answer» Correct Answer - A Let the speed at the highest point be `v_H`. `1/2mv_H^2+mg2lsin 37^@=1/2mv_L^2impliesv_H=sqrt(46)ms^-1` |
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| 417. |
Sand particles drop vertically at the rate of `2kgs^-1` on a conveyor belt moving horizontally with a velocity of `0.2ms^-1`. The extra force required isA. (a) `0.4W`B. (b) `0.08W`C. (c) `0.04W`D. (d) `0.2W` |
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Answer» Correct Answer - B Force required to keep the belt moving `=F` `F=v(dm)/(dt)=0.2xx2=0.4N` Extra power required: `P=Fv=0.4xx0.2=0.08W` Rate of change of kinetic energy is `1/2v^2((dm)/(dt))=1/2(0.2)^2xx2=0.04Js^-1` |
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| 418. |
Sand particles drop vertically at the rate of `2kgs^-1` on a conveyor belt moving horizontally with a velocity of `0.2ms^-1`. The extra force required to keep the belt moving isA. (a) `0.4N`B. (b) `0.08N`C. (c) `0.04N`D. (d) `0.2N` |
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Answer» Correct Answer - A Force required to keep the belt moving `=F` `F=v(dm)/(dt)=0.2xx2=0.4N` Extra power required: `P=Fv=0.4xx0.2=0.08W` Rate of change of kinetic energy is `1/2v^2((dm)/(dt))=1/2(0.2)^2xx2=0.04Js^-1` |
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| 419. |
A pump is required to lift `1000kg` of water per minute from a well `20m` deep and eject it at a rate of `20ms^-1`. a. How much work is done in lifting water? b. How much work is done in giving in KE? c. What HP(horsepower) engine is required for the purpose of lifting water? |
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Answer» Work done in lifting water=Gain in PE(potential energy) Work`=1000xxgxx20=1.96xx10^5Jmin^-1` Work done(per minute) in giving it KE `=1/2mv^2=1/2(1000)(20)^2=2xx10^5Jmin^-1` Power of the engine=Work done per second `=1/60(1.96+2)xx10^5J=6.6xx10^3W` Since `1HP=746W`, HP required `=8.85`. |
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| 420. |
Two identical `5kg` blocks are moving with same speed of `2ms^-1` towards each other along a frictionless horizontal surface. The two blocks collide, stick together, and come to rest. Consider the two blocks as a system. The works done by external and internal forces are, respectively,A. (a) 0, 0B. (b) 0, 20JC. (c) 0, -20JD. (d) 20J, -20J |
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Answer» Correct Answer - C `F_(ext)=0`, therefore, `W_(ext)=vecF_(ext)*vecs=0` By work-energy theorem: `W=K_F-K_I` `W_(ext)+W_(i nt)=0-(1/2mv^2+1/2mv^2)` `W_(i nt)=-mv^2=-5xx2^2=-20J` |
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| 421. |
Water falling from a `50m` high fall is to be used for generating electric energy. If `8xx10^5kg` in water falls per hour and half the gravitational potential energy can be converted into electric energy, how many `100W` lamps can be lit? |
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Answer» `h=50m`, `m=1.8xx10^5kg`, `P=100W` `PE=mgh=1.8xx10^5xx10xx50=900xx10^5Jh^-1` Because half the potential energy is converted into electricity Electrical energy `=1/2PE=450xx10^5Jh^-1` So, power in watt `(Js^-1)=(450xx10^5)//(3600)` The number of `100W` lamps that can be lit `=(450xx10^5)/(3600xx100)=125` |
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| 422. |
A man slowly pulls a bucket of water from a well of depth `h=20m`. The mass of the uniform rope and bucket full of water are `m=200g` and M `19.9kg`, respectively. Find the work done `(in kJ)` by the mean. |
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Answer» Correct Answer - `(4)` As the centre of gravity of rope comes up by `h//2` and that of bucket by h, required work would be `W=sqrt(etagr)+Mgh=4000J=4kJ` |
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| 423. |
Water is drawn from a well in a `5kg` drum of capacity `55L` by two ropes connected to the top of the drum. The linear mass density of each rope is `0.5kgm^-1`. The work done in lifting water to the ground from the surface of water in the well `20m` below is `[g=10ms^-2]`A. (a) `1.4xx10^4J`B. (b) `1.5xx10^4J`C. (c) `9.8xx10xx6J`D. (d) `18J` |
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Answer» Correct Answer - A Work done in lifting water and drum is `60xx10xx20J=12000J` Total mass of ropes `=40xx0.5kg=20kg` Work done in the case of ropes is `20xx10xx10=2000J` Total work done `=14000J` |
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| 424. |
A ladder of length `l` carrying a man of mass `m` at its end is attached to the basket of a balloon of mass M. The entire system is in equilibrium in the air. As the man climbs up the ladder into the balloon, the balloon descends by a height h. The potential energy of the manA. (a) Increases by `mg(l-h)`B. (b) Increases by `mgl`C. (c) Increases by `mgh`D. (d) Increases by `mg(2l-h)` |
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Answer» Correct Answer - A When the man climbs the ladder of length l, the balloon descends by height h. So net height gained by the man is `l-h`. Hence, gain in potential energy of the man is `mg(l-h)`. Work done by man: `W_1=mgl` Inclination in potential energy of balloon=work done by man -increase in potential energy of man, i.e., `mgl-mg(l-h)=mgh` |
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| 425. |
A `700N` marine in basic training climbs a `10.0m` vertical rope at a constant speed in `8.00s`. What is his power output? |
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Answer» `Power=W/l` `P=(mgh)/(t)=((700N)(10.0m))/(8.00s)=875W` |
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| 426. |
A rope ladder of length L is attached to a balloon of mass M. As the man of mass m climbs the ladder into the balloon basket, the balloon comes down by a vertical distance s. Then the increase in potential energy of man divided by the increase in potential energy of balloon is A. (a) `(L-s)/(s)`B. (b) `L/s`C. (c) `(s)/(L-s)`D. (d) `L-s` |
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Answer» Correct Answer - A Work done by man, `mgL=mg(L-s)+mgs` where `mg(L-s)` is the increase in potential energy of the man and `mgs` is the increase in potential energy of the balloon because the balloon would have been lifted up but for the climbing of the man. So Increase in PE of man/Increase in PE of balloon=`(mg(L-s))/(mgs)-(L-s)/(s)` |
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| 427. |
A body of mass `1kg` is taken from infinity to a point P. When the body reaches that point, it has a speed of `2ms^-1`. The work done by the conservative force is `-5J`. Which of the following is true (assuming non-conservative and pseudo-forces to be absent).A. (a) Work done by the applied force is `+7J`B. (b) The total energy possessed by the body at P is `+7J`C. (c) The potential energy possessed by the body at P is `+5J`D. (d) Work done by all forces together is equal to the change in kinetic energy. |
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Answer» Correct Answer - A::B::C::D `DeltaKE=1/2mv^2=1/2xx1xx2^2=2J` `W_(cons)=-DeltaU=-5JimpliesDeltaU=5J` `W_(ext)=DeltaU+DeltaKE=5+2=7J` |
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| 428. |
An athlete jumping vertically on a trampoline leaveles the surface with a velocity of `8.5 m//s` upwards. What maximum height does the reach?A. `10 m`B. `2.5 m`C. `5.0 m`D. `0.50 m` |
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Answer» Correct Answer - C Once the athlete leves the surface the tranpoline , only a conservative force (het weight) acts on her .Therefore the , mmechanical energy of the athlete - earth system is constent during her flight `k_(f) +U_(f) = k_(i) +U_(i))` .Taking the `y = 0` at the surface of the tranpoline.`U_(i) = mgv_(i)= 0` .Also , her speed when she rwaches maximum height ,zero , or `k_(f) = 0` This levels as with `U_(f) = k_(i)`, or` mgv_(max) = (1)/(2) mv_(i)^(2)` , which gives the maximum height as `y_(max) = (v_(i)^(2))/(2g)= ((10.0m//s)^(2))/(2(10.0m//s^(2))) = 5.0m` |
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| 429. |
The energy which an `e^(-)` e acquires when accelerated through a potential difference of 1 volt is calledA. 1 jouleB. 1 electron voltC. 1ErgD. 1 Watt. |
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Answer» Correct Answer - B |
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| 430. |
A stone is thrown at an angle of `45^(@)` to the horizontal with kinetic energy K. The kinetic energy at the highest point isA. `(K)/(2)`B. `(K)/(sqrt2)`C. KD. zero |
| Answer» Correct Answer - a | |
| 431. |
A spherical ball of mass `m_(1)` colliddes head on with another ball of mass `m_(2)` at rest . The collision is elastic . The fraction of kintic energy lost by `m_(1)` is `:`A. `(4m_1m_2)/((m_1+m_2)^2)`B. `m_1/(m_1+m_2)`C. `m_2/(m_1+m_2)`D. `(m_1m_2)/((m_1+m_2)^2)` |
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Answer» Correct Answer - A According to momentum conservation, we get `m_1v_(1i)=m_1v_(1f)+m_2v_(2f)`…(i) where `v_(1i)` is the initial velocity of spherical ball of mass `m_1` before collision and `v_(1f)` and `v_(2f)` are the final velocities of the balls of masses `m_1 and m_2` after collision. According to kinetic energy conservation, we get `1/2m_1v_(1i)^(2)=1/2m_1v_(1f)^(2)+1/2m_2v_(2f)^(2)` `m_1v_(1i)^(2)=m_1v_(1f)^(2)+m_2v_(2f)^(2)` ...(ii) From Eqs. (i) and (ii) , it follows that `m_1v_(1i)(v_(2f)-v_(1i))=m_1v_(1f)(v_(2f)-v_(1f))` or `v_(2f)(v_(1i)-v_(1f))=v_(1i)^2-v_(1f)^2=(v_(1i)-v_(1f))(v_(1i)+v_(1f))` `therefore v_(2f)=v_(1i)+v_(1f)` Substituting this in Eq.(i), we get `v_(1f)=((m_1-m_2))/(m_1+m_2)v_(1i)`...(iii) The initial kinetic energy of the mass `m_1` is `K_(1i)=1/2m_1v_(1i)^2` The final kinetic energy of the mass `m_1` is `K_(1f)=1/2m_1v_(1f)^2=1/2m_1((m_1-m_2)/(m_1+m_2))^2 v_(1i)^2`(Using (iii)) The fraction of kinetic energy lost by `m_1` is `f=(K_(1i)-K_(1f))/K_(1i)=(1/2m_1v_(1i)^2 -1/2m_1((m_1-m_2)/(m_1+m_2))^2 v_(1i)^2)/(1/2m_1v_(1i)^2)` `=1-((m_1+m_2)/(m_1+m_2))^2=(4m_1m_2)/((m_1+m_2)^2)` |
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| 432. |
A truck and a car moving with the same kinetic energy are brought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance?A. The truckB. The carC. Both will travel the same distance before coming to restD. Cannot be predicted |
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Answer» Correct Answer - C Work done = Force x distance = Change in kinetic energy. Both the truck and the car had same kinetic energy and hence same amount of work is needed to be done. As retarding force applied is same for both, therefore, both the truck and the car travel the same distance before coming to rest. |
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| 433. |
A lorry and a car moving with the same kinetic energy are brought to rest by the application of brakes which provides equal retarding forces. Which of them will come to rest in a shorter distance? |
| Answer» Both will come to rest after travelling the same distance. | |
| 434. |
If the force acting on a body is inversely proportional to its speed, then its kinetic energy isA. linearly related to timeB. inversely proportional to timeC. inversely proportional to the square of timeD. a constant |
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Answer» Correct Answer - A `Fprop1/v,F=C/v` Where C is a constant of proportionality. `implies ma=C/v " or" m(dv)/(dt)=C/v` `vdv=(Cdt)/m` Integrating both sides , we get `v^2/2 =(Ct)/m "or" 1/2 mv^2=Ct` or Kinetic energy, `K=1/2 mv^2=Ct "or" K prop t` |
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| 435. |
The amount of work done by a moving body depends on the (a) mass of the body (b) velocity (c) both (a) and (b)(d) time |
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Answer» (c) both (a) and (b) |
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| 436. |
A force of 600 N displaces an object through 2 m. Find the work done if the force and displacement are 1. Parallel 2. At right angles. |
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Answer» Work done is given by W = Fs cosθ 1. when F and s are parallel, θ = 0 and hence cosθ = 1 ∴ W = Fs = 600 × 2 = 1200 J. 2. when F and s are at right angles, θ = 90° and hence cosθ = 0 ∴ Work done, W = 0. |
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| 437. |
The kinetic energy of a body is given by(a) \(\frac{1}{2}\) mv2(b) ma(c) Fs(d) (v2 - u2)m |
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Answer» (a) \(\frac{1}{2}\)mv2 |
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| 438. |
The coefficient of restitution e for a perfectly elastic collision isA. 1B. 0C. `infty`D. `-1` |
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Answer» Correct Answer - A |
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| 439. |
Two perfectly elastic particles `A` and `B` of equal masses travelling along a line joining them with velocities `15 m//s` and `10m//s` respectively collide. Their velocities after the elastic collision will be (in m/s) respectivelyA. 0,25B. 5,20C. 10,15D. 20,5 |
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Answer» Correct Answer - C |
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| 440. |
Two perfectly elastic particles `A` and `B` of equal masses travelling along a line joining them with velocities `15 m//s` and `10m//s` respectively collide. Their velocities after the elastic collision will be (in m/s) respectivelyA. `0 and 25`B. `5 and 20`C. `10 and 15`D. `20 and 5` |
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Answer» Correct Answer - C In perfectly elastic collision, kinetic energy is conserved. So, velocities are not lost in collision. When masses are equal, velocities are interchanges. So velocities of `A` and `B` will be `10 m//s` and `15 m//s` respectively. |
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| 441. |
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is.A. constant and equal to `mg` in magnitude.B. constant and greater than `mg` in magnitude.C. variable but always greater than `mg`.D. at first greater than `mg`, and later becomes equal to `mg` |
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Answer» Correct Answer - D When the man gets straight up and sstand, reaction of ground on the man `=`mg. However, when he is squatting on the ground, reactio of ground is more than `mg`, as the man is to exert sone extra force on the ground to stand up . |
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| 442. |
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,A. the two magnetic forces are equal and opposite so they produce no net effect.B. the magnetic forces do no work on each particleC. the magnetic forces do equal and opposite (but non-zero) work on each particle.D. the magnetic forces are necessarily negligible. |
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Answer» Correct Answer - B As the magnetic forces due to motion of electron and proton act in a direction perpendicular to the direction of motion, no work is done by these forces That is why one ignores the magnetic force of one particle on another |
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| 443. |
A ball of mass m and density `rho` is immersed in a liquid of density `3 rho` at a depth `h` and released. To what height will the ball jump up above the surface of liqud ? (neglect the reistance of water and air).A. hB. `h//2`C. 2hD. 3h |
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Answer» Correct Answer - C |
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| 444. |
Select the correct option(s).A. (a) A single external force acting on a particle necessarily changes its momentum and kinetic energy.B. (b) A single external force acting on a particle necessarily changes its momentum.C. (c) The work-energy theorem is valid for all types of forces: internal, external, conservative as well as non-conservative.D. (d) The kinetic energy of the system can be increased without applying any external force on the system. |
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Answer» Correct Answer - B::C::D Single force on a particle may change the magnitude and or direction of velocity, so momentum changes necessarily but kinetic energy may or may not change. From work-energy theorem, `DeltaK=W_(ext)+W_(i nt)+W_(conservative) +W_(NC)` Kinetic energy can be increased due to work done by internal forces also. |
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| 445. |
An object of mass `m` is tied to a string of length `l` and a variable force F is applied on it which brings the string gradually at angle thit `theta` with the vertical. Find the work done by the force `F` . |
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Answer» Correct Answer - A::C In this case, three forces are acting on the object: 1. tension (T) 2. weight (mg) and 3. applied force (F) Using work-enegy theorem. `W _ (n et)=DeltaKE` or `W_(T) + W_(mg) + W_(F)=0` ...(i) as `DeltaKE=0` because `K_(i) = K_(f) = 0` Further, `W_T=0,` . as tension is always perpendicular to displacement. `W_(mg) =-mgh` or `w_(mg)=-mgl(1-costheta)` Substituting these values in Eq. (i), we get `W_(F) = mgl(1-costheta)`. |
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| 446. |
A particle of mass 10 g moves along a circle of radius `6.4` cm with a constant tangential acceleration. What is the magnitude of this acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to `8xx10^(-4)`J by the end of the second revolution after the beginning of the motion?A. `0.15" ms"^(-2)`B. `0.18" ms"^(-2)`C. `0.2" ms"^(-2)`D. `0.1" ms"^(-2)` |
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Answer» Correct Answer - D (d) `W_("Tengential")=DeltaKE=K_(f)-K_(i)=K_(f)-0` `(ma_(T))xx5=8xx10^(-4)` `a_(T)=(8xx10^(-4))/(10^(-2)(2pirxx2))=(2xx10^(-2))/(pixx2pixx10^(-2))=0.1 ms^(-2)` |
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| 447. |
Asseration : A wooden block is floating in a liquid as shown in figure, in vertical direction equilibrium of block stable. . Reason : When depressed in downward direction is starts oscillating.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. If Assertion is true, but the Reason is falseD. If Assertion is false but the Reason is true. |
| Answer» Correct Answer - A | |
| 448. |
A force `F = -K(y hatI + x hatj) (where `K` is a posive constant ) acts on a particle moving in the `xy` plane . Starting form the original , the partical is taken along in the positive `x` axis to the point `(x,0)` and then partical to they axis the point `(x,0)` . The total work done by the force `F` on the particls isA. `- 2 Ka^(2)`B. ` 2 Ka^(2)`C. `- Ka^(2)`D. `Ka^(2)` |
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Answer» Correct Answer - C While moving from `(0.0)` to `(a.0)` Along positive x-axis , `y = 0 :. vecF = - kx hatj` i.e. , force is in negative y-direction while dicplacement is in positive direction. `:. W_(1) = 0` Because force is perpendicular to displacement . Then partical moves from `(a.0)` to `(a.a)` during this `vec F = - k(y hati + a hatj)` The first componenet of force, `- ky hatj` will not contribitute any work because this component is along negative x- direction `(- hati)` while displacement is in positive y- direction `(a.0)` to `(a.a)` . The second component of force i.e. `- ka hatj` will perform negative work `:. W_(2) = (- ka hatj) (a hatj) = (-ka) (a) = - ka^(2)` So net work done on the partical `W = W_(1) + W_(2)` `= 0 + (-ka^(2)) = - ka^(2)` |
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| 449. |
if a number of force act on a body and the body is in state or dynamic force equalibrium , then .A. work done by any individual force mucst be zeroB. Net work done by all the force is `+ve`C. Net work done by all the force is `-ve`D. Net work done by all the force is zero |
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Answer» Correct Answer - D When a number of force act on a body and the body is in state or dymanic equilibrium (it mean `F_(max) = 0)`, then individual force may do the work , but net work done by all the force will be zero [Note, In such a situation, work done by some forces is positive while by others is equally negative, so that net work becomes zero.]`. |
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| 450. |
Does the work done in moving a body depend on how fast or how slow the body is moved? |
| Answer» No, time is not involved, in work or energy. Time is involved only in power. | |