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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.
| 1001. |
A person trying to lose weight (dieter ) lifts a 10 kg mass through 0.5m, 1000 times, A ssume that the potential energy lost each time she lowers the mass is dissipated (a) How much work does she does against the gravitational force ? (b) Fat supplies `3.8xx10^(7) J` of energy per kilogram which is converted to mechanical energy with a `20%` efficiency rate. How much fat will the dieter use up ? |
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Answer» Here, `m=10kg, h=0.5m, n=1000` (a) Work done against gravitational force `W=n(mgh)=1000xx(10xx9.8xx0.5)=49000J.` (b) Mechanical energy supplied by 1 kg of fat `=3.8 xx10^(7)xx(20)/(100)=0.76xx10^(7)J//kg` `:.` Fat used up by the dieter `=(1kg)/(0.76xx10^(7))xx49000=6.45xx10^(-3)kg` |
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| 1002. |
An elevator weighing 500 kg is to be lifted up at a constant velociyt of 0.20 m/s. What would be the minimum horsepower of the motor to be used?A. `10.30 hp`B. `5.15 hp`C. 2.62 hpD. 1.31 hp |
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Answer» Correct Answer - D (d) Given, mass of elevator=500 kg velocity=0.20 `ms^(-1)` Weight of elevator`=500xx9.8=F` Now,power, `P=Fv=500xx9.8xx0.20=980W` Therefore, hp-rating of motor`=(1)/(746)xx980=1.31hp` |
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| 1003. |
Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h. |
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Answer» Here, `F=?, m=500kg, s=25m, v=0, u=72km//h=(72xx1000)/(60xx60)m//s=20m//s` According to word energy principle, `W =`change of K.E. of the body `Fs=(1)/(2)m(v^(2)-u^(2))` `Fxx25=(1)/(2)xx500(0-20^(2))=-(500xx400)/(2)` `F=(500xx200)/(25)=-4000N :.` Average frictional force `=400N` |
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| 1004. |
A car weighing 1400 kg is moving at speed of 54 km/h up a hill when the motor stops. If it is just able to read the destination which is at a height of 10 m above the point calculte the work done against friction (negative of the work done by the friction).A. `17500 J`B. `12500 J`C. `25000 J`D. `50000 J` |
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Answer» Correct Answer - A `v=54(km)/(h)=54xx(5)/(18)=15(m)/(s)` `(1)/(2)mv^2=mgh+W_f` `(1)/(2)xx1400(15)^2=1400xx10xx10+W_f` `W_f=17500J` |
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| 1005. |
In the arrangement show in figure, the track is frictionless until the body reaches the higher level. A frictional force stops the body in a distance d from A. The initial speed of the body `v_(0)` is 6 m/s the height h is 1 m and the coefficient of kinetic friction is 0.2. The speed of body at the higher level A isA. 3 m/sB. 4 m/sC. 5 m/sD. 4.25 m/s |
| Answer» Correct Answer - B | |
| 1006. |
If the earth stops rotating, the apparent value of g on its surface willA. Increases everywhereB. Decrease everywhereC. Remain the same everywhereD. Increase at some places and remain the same at some other places |
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Answer» Correct Answer - D |
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| 1007. |
A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional toA. `t^(1//2)`B. `t^(3//4)`C. `t^(3//2)`D. `t^(2)` |
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Answer» Correct Answer - C |
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| 1008. |
The velocity-time graph of a particle moving in a straight line is shown in figure. The mass of the particle is `2kg`. Work done by all the forces acting on the particle in time interval between `t=0` to `t=10s` is A. (a) `300J`B. (b) `-300J`C. (c) `400J`D. (d) `-400J` |
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Answer» Correct Answer - A From work-energy theorem, `W=DeltaKE=K_f-K_i=1/2m(v_f^2-v_i^2)` `=1/2xx2[(-20)^2-(10)^2]=300J` |
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| 1009. |
A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional toA. (a) `sqrtt`B. (b) `t^(3//4)`C. (c) `t^(3//2)`D. (d) `t^2` |
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Answer» Correct Answer - C Let us assume that the displacement of the body is directly proportional to `t^n`, i.e., `s=Kt^n`, `v=(ds)/(dt)=Knt^(n-1)` and `a=(dv)/(dt)=Kn(n-1)t^(n-2)` Force `F=ma=mKn(n-1)t^(n-2)` Power, `P=Fv=[mKn(n-1)t^(n-2)][Knt^(n-1)]` `=mKn^2(n-1)t^(2n-3)` As power is constant, i.e. independent of time, hence `2n-3=0` or `n=3/2` or `spropt^(3/2)` |
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| 1010. |
a. A `2-kg` situated on a smooth fixed incline is connected to a spring of negligible mass, with spring constant `k=100Nm^-1`, via a frictionless pulley. The block is released from rest when the spring is unstretched. How far does the block moves down the incline before coming (momentarily) to rest? What is its acceleration at its lower point? b. The experiment is repeated on a rough incline. If the block is observed to move `0.20m` down along the incline before it comes to instantaneous rest, calculate the coefficient of kinetic friction. |
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Answer» Correct Answer - a. `24m`; `6ms^-2`, b. `1/8` (a) At the extreme position blocks stops. Applying work-energy theorem, we get `mg sin 37^@=1/2ks^2` `2xx10xxsxx3/5=1/2xx100xxs^2` On solving `s=0.24m` Acceleration at its lowest point, `a=(ks-mg sin 37^@)/(m)` `=(100xx0.24-2xx10xx3/5)/(2)=6ms^-2` (b) `W_g+W_(f riction)+W_(spri ng)=DeltaKE` `mg sin 37^@+mumgcos37^@xxs=1/2ks^2` `mgsin37^@-1/2ks=mumgcos 37^@` `2xx10xx3/5-1/2xx100xxs=muxx2xx10xx4/5` Gives `s=0.020m` `mu=(12-50s)/(16)impliesmu=1/8` |
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| 1011. |
In the position shown in figure, the spring is at its natural length. The block of mass `m` is given a velocity `v_0` towards the vertical support at `t=0`. The coefficient of friction between the block and the surface is given by `mu=alphax`, where `alpha` is a positive constant and x is the position of the block from its starting position. The block comes to rest for the first time at x, which is A. (a) `v_0sqrt((m)/(k+alphamg))`B. (b) `v_0sqrt(m/k)`C. (c) `v_0sqrt((m)/(alphag))`D. (d) None of these |
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Answer» Correct Answer - A According to the work-energy theorem, `1/2mv_0^2=mg alpha underset0oversetx int dx+1/2kx^2` Solving we get, `x=v_0sqrt((m)/(k+alphamg))` |
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| 1012. |
A moving railway compartment has a spring of constant k fixed to its front wall. A boy stretches this spring by distance x and in the mean time the compartment moves by a distance s. The work done by boy w.r.t. earth is A. (a) `1/2kx^2`B. (b) `1/2(kx)(s+x)`C. (c) `1/2kxs`D. (d) `1/2kx(s+x+s)` |
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Answer» Correct Answer - A Total work done in equal to the energy stored in the spring. |
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| 1013. |
A block of mass `m` has initial velocity `u` having direction towards `+x` axis. The block stops after covering a distnce S causing similar extension in the spring of constant K holding it. If `mu` is the kinetic friction between the block and the surface on which it was moving, the distance S is given A. (a) `1/Kmu^2m^2g^2`B. (b) `1/K(mKu^2-mu^2m^2g^2)^(1//2)`C. (c) `1/K(mu^2m^2g^2+mKmu^2+mumg)^(1//2)`D. (d) `(-mumg+sqrt(mu^2m^2g^2+m u^2k))/(k)` |
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Answer» Correct Answer - D By the work-energy theorem, `1/2m u^2=mumgS+1/2kS^2` i.e., `S^2+(2mumgS)/(k)-(mu^2)/(k)=0` `impliesS=((-2mumg)/(k)+sqrt((4mu^2m^2g^2))/(k^2)(4m u^2)/(k))/(2)` `=(-mumg+sqrt(mu^2m^2g^2+m u^2k))/(k)` |
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| 1014. |
Two springs have force constants `K_(1)` and `K_(2)` , where `K_(1)gtK_(2)`. On which spring, more work is downe if (i) they are stretched by the same force? (ii) they are stretched by the same amount? |
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Answer» (i) Under the action of same force F, let the springs be stretched by `x_(1)` and `x_(2)` respectively. `:.F=K_(1)x_(1)=K_(2)x_(2),:. (x_(1))/(x_(2))=(K_(2))/(K_(1))` `(W_(1))/(W_(2))=((1)/(2)K_(1)x_(1)^(2))/((1)/(2)K_(2)x_(2)^(2))=(K_(1))/(K_(2))((x_(1))/(x_(2)))^(2)=(K_(1))/(K_(2))((K_(2))/(K_(1)))^(2)=(K_(2))/(K_(1))` ltBRgt As `K_(1)gtK_(2)` `:. W_(2)gtW_(1)` (ii) When the two springs are stretched by the same distance, say x, then `(W_(1))/(W_(2))=((1)/(2)K_(1)x^(2))/((1)/(2)K_(2)x^(2))=(K_(1))/(K_(2))` As `K_(1)gtK_(2)` therefore `W_(1)gtW_(2)` . |
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| 1015. |
(1) A scissor is a_________ multiplier (2) 1 kWh =________ J. |
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Answer» A scissor is a force multiplier because the effort applied is less than the load. 1kWh = 1kilowatt x 1 hour = 1000Js- x 3600s = 3.6 x 106J |
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| 1016. |
A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? |
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Answer» Let `m` be the mass of bullet and `upsilon` be its initial velocity. If `F` is constant resistance to motion, then `(i)` Work done `=` loss in KE `=` force `xx` distance `=(1)/(2) m(upsilon^(2)-(upsilon^(2))/( 4))=Fxx3` or `(1)/(2) m ( (3 upsilon^(2))/(4))=Fxxs ....(i)` `(ii)` Work done `=` loss in KE `=` force `xx` distance `=(1)/(2) m ( (upsilon^(2))/( 4)-0)=Fxxs...(ii)` Dividing `(ii)` and `(i)`, we get `(1)/(3) =(s)/(3)` `:. s = 1 cm` |
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| 1017. |
A force `(4hati+hatj-2hatk)` N acting on a body maintains its velocity at `(2hati+2hatj+3hatk) m s^(-1)`. The power exerted isA. 4 WB. 6 WC. 2 WD. 8 W |
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Answer» Correct Answer - B Force, `vecF=(4hati+2hatj-2hatk)N` Velocity, `vecv=(2hati+2hatj+3hatk)m s^(-1)` Power, `P=vecF.vecv =(4hati+2hatj-2hatk).(2hati+2hatj+3hatk)` = (8 + 4-6) W =6 W |
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| 1018. |
Work done by a force `F=(hati+2hatj+3hatk)` acting on a particle in displacing it from the point `r_(1)=hati+hatj+hatk` to the point `r_(2)=hati-hatj+hat2k` isA. `-3 J`B. `-1 J`C. zeroD. 2 J |
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Answer» Correct Answer - B (b) `W=F.s=F.(s_(2)-s_(1))` `=(hati+2hatj+3hatk).[(hati-hatj+2hatk)-(hati+hatj+hatk)]` `=(hati+2hatj+3hatk).(-2hatj+hatk)=-4+3=-1J` |
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| 1019. |
Assertion: Work done by the force of friction in moving a body around a closed loop is zero. Reason: Work done does not depend upon the nature of force.A. If both assertion and reason are true and reason is a true explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion is and reason are false. |
| Answer» Work done during the motion of a body over a closed path depends upon whether the force doing work is conservative or nonr-conservative in nature. For conservative forces like gravitational force like frictional force some work is done. | |
| 1020. |
A particle moved from position `vec r_(1) = 3 hati + 2 hatj - 6 hatk `to position`vecr_(2) = 14 hati + 13 hatj +9 hatk` undre the action of a force `(4 hati + hatj +3 hatk)` newtons . Find the work done .A. `10J`B. `100 J`C. `0.01 J`D. `1J` |
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Answer» Correct Answer - B displacement : `vec s = vec r_(2) - vec r_(1)`. Now find `W = vec F. vecs` |
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| 1021. |
A ball is dropped from height 10 m . Ball is embedded in sand 1 m and stops, thenA. Only momentum remains conservedB. Only kinetic energy remains conservedC. Both momentum and K.E. are conservedD. Neither K.E. nor momentum is conserved |
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Answer» Correct Answer - A |
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| 1022. |
A force `vecF = 2 x hati + 2 hatj + 3z^(2)hatk N` is acting on a particle .Find the work done by this force in displacing the body from `(1, 2, 3) m` to `(3,6,1)m`A. `1-0J`B. `100 J`C. `10 J`D. `1J` |
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Answer» Correct Answer - A `W = int _(x 1)^(x2) F_(x) dx + int _(y 1)^(y2) F_(y) dy + int _(z 1)^(z2) F_(z) dz` ` = int _(1)^(2) 2xdx + int _(2)^(6) 2 dy + int _(3)^(1) 3z^(2) dz = - 10J` |
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| 1023. |
An open knife of mass m is dropped from a height h on a wooden floor. If the blade penetrates up to the depth d into the wood. The average resistance offered by the wood to the knife edge is to the depth d into the wood, the average resistance offered by the wood to the knife edge is .A. `mg (1 + (h)/(d))`B. `mg (1 + (h)/(d))^(2)`C. `mg(1-h/d)`D. `mg (1 + /(h))` |
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Answer» Correct Answer - A Decerase in potential energy =Work done against friction `:. Mg (h + d) = F.d` here `F =` average resistance `rArr F =mg(1 + (h)/(d))` |
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| 1024. |
A bag of sand of mass M is suspended by a string. A bullet of mass m is fired at it with velocity v and gets embedded into it. The loss of kinetic energy in this process isA. `(1)/(2)mv^(2)`B. `(1)/(2)mv^(2)xx(1)/(M+m)`C. `(1)/(2)mv^(2)xx(M)/(m)`D. `(1)/(2)mv^(2)((M)/(M+m))` |
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Answer» Correct Answer - D |
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| 1025. |
A bullet hits and gets embedded in a solid block resting on a frictionless surface. In this process, which of the following is correct ?A. Only momentum is conservedB. Only kinetic energy is conservedC. Neither momentum nor kinetic energy is conservedD. Both, momentum and kinetic energy are conserved |
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Answer» Correct Answer - A As the bullet gets embedded in the solid block, whole of its kinetic energy gets lost. This will be perfectly inelastic collision in which only momentum is conserved. |
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| 1026. |
A force `(vecF) = 3 hati + c hatj + 2 hatk` acting on a partical causes a displacement: `(vecs) = - 4 hati + 2 hatj + 3 hatk` in its own direction . If the work done is `6 j`, then the value of `c` is |
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Answer» `W = (3 hati + c hatj + 2 hatk) . (- 4 hati + 2 hatj + 3 hatk) = 6` joule `W = - 12 + 2c + 6 = 6 implies c = 6` |
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| 1027. |
An open knife of mass m is dropped from a height h on a wooden floor. If the blade penetrates up to the depth d into the wood. The average resistance offered by the wood to the knife edge is .A. `mg`B. `mg(1-(h)/(d))`C. `mg(1+(h)/(d))`D. `mg(1+(h)/(d))^(2)` |
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Answer» Correct Answer - C |
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| 1028. |
It is easier to draw up a wooden block along an inclined plane than to haul it vertically, principally becauseA. The friction is reducedB. The mass becomes smallerC. Only a part of the weight has to be overcomeD. ‘ g ’ becomes smaller |
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Answer» Correct Answer - C |
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| 1029. |
A bullet hits and gets embedded in a solid block resting on a frictionless surface. In this process, which of the following is correct ?A. Momentum and kinetic energyB. Kinetic energy aloneC. Momentum aloneD. Neither momentum nor kinetic energy |
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Answer» Correct Answer - C |
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| 1030. |
The work done in pulling up a block of wood weighing 2 kN for a length of 10 m on a smooth plane inclined at an angle of 15° with the horizontal isA. 4.36 kJB. 5.17 kJC. 8.91 kJD. 9.82 kJ |
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Answer» Correct Answer - B |
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| 1031. |
A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0 m . How much energy is dissipated by air drag acting on the ball during the ascentA. 19.6 JouleB. 4.9 JouleC. 10 JouleD. 9.8 Joule |
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Answer» Correct Answer - D |
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| 1032. |
A body of mass 2kg slides down a curved track which is quadrant of a circle of radius 1 metre . All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track isA. 4.43m/secB. 2m/secC. 0.5m/secD. 19.6m/ec |
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Answer» Correct Answer - A |
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| 1033. |
A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by himA. zeroB. positiveC. negativeD. Nothing can be said |
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Answer» Correct Answer - A |
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| 1034. |
A force `vecF=3hati+chatj+2hatk` acting on a particle causes a displacement `vecd=-4hati+2hatj-3hatk`. If the work done is 6 J. then the value of c will be |
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Answer» Correct Answer - C |
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| 1035. |
A force `vecF=(5hati+3hatj)N` is applied over a particle which displaces it from its original position to the point `vecs=S(2hati-1hatj)m`. The work done on the particle isA. `-7`B. `+7`C. `+10`D. `+13` |
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Answer» Correct Answer - B |
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| 1036. |
A force `vecF=6hati+2hatj-3hatk` acts on a particle and produces a displacement of `vecs=2hati-3hatj+xhatk`. If the work done is zero, the value of x isA. `-2`B. `1//2`C. `6`D. `2` |
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Answer» Correct Answer - D |
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| 1037. |
If force `(vecF)=4hati+4hatj` and displacement `(vecs)=3hati+6hatk` then the work done isA. `4xx6` unitB. `6xx3` unitC. `5xx6` unitD. `4xx3` unit |
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Answer» Correct Answer - D |
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| 1038. |
A force `vecF=5hati+6hatj+4hatk` acting on a body, produces a displacement `vecS=6hati-5hatk`. Work done by the force isA. 18 unitsB. 15 unitsC. 12 unitsD. 10 units |
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Answer» Correct Answer - D |
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| 1039. |
A particle of mass m is tied to light string and rotated with a speed v along a circular path of radius r. If T=tension in the string and `mg=` gravitational force on the particle then actual forces acting on the particle areA. mg and T onlyB. mg, T and an additional forces of `mv^(2)//r` directed inwardsC. mg, T and an additional forces of `mv^(2)//r`directed outwardsD. Only a foce `mv^(2)//r` directed outward |
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Answer» Correct Answer - A |
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| 1040. |
The maximum tension an inextensible rope of mass density 0.1 kg `m^(-1)` can bear is 40 N Length of rope is 2m. The maximum angular velocity with which it can be rotated horizontally in a circular path on a frictionless table is :A. `10 sqrt(2) rad//s`B. `18 rad//s`C. `16 rad//s`D. `15 rad//s` |
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Answer» Correct Answer - A |
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| 1041. |
A force `(3hati+4hatj)` newton acts on a boby and displaces it by `(3hati+4hatj)` metre. The work done by the force isA. 10 JB. 12 JC. 16 JD. 25 J |
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Answer» Correct Answer - D (d) Given, Force `F=(3hati+4hatJ)N` and Displacement, `" "s=(3hati+4hatj) m` `:. "Work done", W=F.s=(3hati+4hatj).(3hati+4hatj)` W=9+16 `implies" "`W=25 J |
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| 1042. |
A block of mass `m=1kg` moving on a horizontal surface with speed `v_(i)=2ms^(-1)` enters a rough patch ranging from `x0.10m to x=2.01m`. The retarding force `F_(r)` on the block in this range ins inversely proportional to x over this range `F_(r)=-(k)/(x) fo r 0.1 lt xlt 2.01m` `=0` for `lt 0.1m` and `x gt 2.01m` where `k=0.5J`. What is the final K.E. and speed `v_(f)` of the block as it crosses the patch? |
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Answer» Here, `m=1kg, v_(i)=2ms^(-1),k=0.5J` Initial K.E., `K_(i)=(1)/(2)mv_(i)^(2)=(1)/(2)xx1(2)^(2)=2J` Work done against friction `W=int_(x=0.1m)^(x=2.01m)F_(r), dx=int_(x=0.1m)^(x=2.01m)-(k)/(x)dx` `=-0.5[log_(e)x]_(x=01m)^(x=2.01m)` `=-0.5"log"_(e)(2.01)/(0.1)` `W=-0.5xx2.303log_(10)20.10` `=-0.5xx2.303xx1.303=-1.5J` `:.` Final K.E., `K_(f)=K_(i)+W=2.0-1.5J` `=0.5J` `v_(f)=sqrt((2k_(f))/(m))=sqrt((2xx0.5)/(1))=1ms^(-1)` |
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| 1043. |
A small body of mass m moving with velocity `v_(0)` on rough horizontal surface, finally stops due to friction. Find, the mean power developed by the friction force during the motion of the body, if the frictional coefficient `mu`=0.27,m=1 kg and `v_(0)=1.5 ms^(-1)`. |
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Answer» The retardation due to friction is, `a=("force of friction")/("mass")=(mu mg)/(m)=-mug` Further,`" " v_(0)`=at Therefore, `" "t=(v_(0))/(a)=(v_(0))/(mug)`………..(i) From work energy theorem, work done by force of friction =change in kinetic energy or `" " W=(1)/(2)mv_(0)^(2)`..............(ii) Mean power`=(W)/(t)` From Eqs. (i) and (ii), we get `P_(mean)=(1)/(2) mu mgv_(0)` Substituting the values, we have `P_(mean)=(1)/(2)xx0.27xx1.0xx9.8xx1.5~~2 W` |
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| 1044. |
A force of constant magnitude `F_(0)` is applied in the tangential direction as shown in the figure. Assume that the bob is at its lowest point initially. A. Speed of bob at `theta=60^(@) " is "sqrt((2L)/(m)[(F0pi)/(3)+(mg)/(2)])`B. Speed of bob at `theta=60^(@) " is "sqrt((2L)/(m)[(F0pi)/(3)-(mg)/(2)])`C. Tension in thread at `theta=60^(@) " is "[(2F0pi)/(3)-(mg)/(2)]`D. Tension in thread at `theta=60^(@) " is "[(2F0pi)/(3)+(mg)/(2)]` |
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Answer» Correct Answer - B::C |
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| 1045. |
A block of mass m is placed on a circular track and then it is given a velocity `upsilon` vertically downwards at position A on track. If block moves on track with constant speed, then A. Coefficient of friction between block and circular track as function of angle `theta` is `mu=(sintheta)/(cottheta+v^(2)/(Rg))`B. Coefficient of friction between block and circular track as function of angle `theta` is `mu=(sintheta)/(costheta+v^(2)/(Rg))`C. Instantaneous power due to friction is `-mgv sin theta`D. Work done from A to C by friction on block will be `-mgR` |
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Answer» Correct Answer - B::C::D |
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| 1046. |
A particle moves move on the rough horizontal ground with some initial velocity `V_(0)`. If `(3)/(4)` of its kinetic enegry lost due to friction in time `t_(0)`. The coefficient of friction between the particle and the ground is.A. `(v_(0))/(2"gt"_(0))`B. `(v_(0))/(4"gt"_(0))`C. `(3v_(0))/(4"gt"_(0))`D. `(v_(0))/("gt"_(0))` |
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Answer» Correct Answer - A (a) `(3)/(4)`th of KE is lost. Hence left KE is `(1)/(4)`th or, `v^(2)=(v_(0)^(2))/(4)` `:. " " v=(v_(0))/(2)=v_(0)-at_(0)=v_(0)-mu" gt"_(0)` or `" " mu=(v_(0))/(2"gt"_(0))` |
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| 1047. |
A bead slides on a fixed frictionless wire bent into a horizontal semicircle of radius `R_(0)` as shown in figure. In addition to any normal forces exerted by the wire, the bead is subjected to an external force that points directly away from origin and depends on distance r from the origin according to the formula `F=F_(0)((r )/(R_(0)))^(2)hat(r)` A. Given force is a central forceB. Given force is a conservation forceC. Work done by external force as bead leaves the track (starting from origin) is `(8F_(0)R_(0))/(3)`D. Speed v of bead as it leaves the wire at P is `sqrt(v_(0)^(2)+(18F_(0)R_(0))/(3m))` |
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Answer» Correct Answer - A::B::C |
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| 1048. |
A track (ACB) is in the shape of an arc of a circle. It is held fixed in vertical plane with its radius OA horizontal. A small block is released on the inner surface of the track from point A. It slides without friction and leaves the track at B. What should be value of q so that the block travels the largest horizontal distance by the time it returns to the horizontal plane passing through B? |
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Answer» Correct Answer - `theta=tan^(-1)(1/(sqrt(2)))` |
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| 1049. |
A particle moves move on the rough horizontal ground with some initial velocity `V_(0)`. If `(3)/(4)` of its kinetic enegry lost due to friction in time `t_(0)`. The coefficient of friction between the particle and the ground is.A. `(v_(0))/(2g t_(0))`B. `(v_(0))/(4g t_(0))`C. `(3v_(0))/(4g t_(0))`D. `(v_(0))/(g t_(0))` |
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Answer» Correct Answer - A |
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| 1050. |
Two particles `1` and `2` are allowed to descend on the two frictionless chord `OA` and `OB` of a vertical circle, at the same instant from point `O`. The ratio of the velocities of the particles `1` and `2` respectively, when they reach on the circumference will be (OB is the diameter).A. `(1)/(4)`B. `(1)/(2)`C. 1D. `(1)/(2sqrt(2))` |
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Answer» Correct Answer - B |
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