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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.
| 1251. |
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 magnitudeB. constant and greater than mg in magnitudeC. variable but always greater than mgD. at first greater than mg and later becomes eual to mg |
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Answer» Correct Answer - D (d) When the man is squatting on the ground he is tilted somewhat, hence he also has to balance frictional force besides his weight in this case. R=reactional froce`="friction" +mg rArr R gt mg` When the man gets straight up in that case friction ~~ 0 `rArr" " "Rectional force" ~~mg` |
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| 1252. |
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 magnitudeB. constant and greater than mg in magnitudeC. variable but always greater than mgD. at first greater than mg and later becomes equal to mg |
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Answer» (d) When the man is squatting on the ground he is tilted somewhat, hence he also has to balance frictional force besides his weight in this case. R=reactional force=friction+mg `Rightarrow Rgtmg` When the man gets straight up in that case friction=0 Reactional force=mg |
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| 1253. |
A proton is kept at rest. A positively charged particle is released from rest at a distance `d` in its field. Consider two experiments, one ini which the charged particle is also a proton and in another, a position. In the same time `t`, the work done on the two moving charged particles isA. same as the same force law is involved in the two experiments.B. less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.C. more for the case of a positron, as the positron moves away a larger distance.D. same as the work done by charged particle on the stationary proton. |
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Answer» Correct Answer - C Force between two protons `=` froce between a proton and a positron. As positron is much lighter than proton, it moves away through much larger distance compared to proton. A work done `=` force `xx` distance, therefore in the same time`t`, work done in case of positron is more than that in case of proton. |
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| 1254. |
Which one of the following is a non-conservative force ?A. Force of frictionB. Magnetic forceC. Gravitational forceD. Electrostatic force |
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Answer» Correct Answer - A Among the given forces, force of friction is a non-conservative force whereas all other forces are conservative forces. |
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| 1255. |
Which of the following statements are incorrect?A. If there were on friction, work needed to be done to move a body up an inclined plane will be zero.B. If there were no friction, the moving vechicles could not be stopped even by locking the breakesC. As the angle of inclination is increased, the normal reaction on the body placed on it increasesD. A duster weighing 0.5 N is pressed against a vertical board with a force of 11 N If the coefficient of friction is 0.5 the work done in rubbing it upwards through a distance of 10 cm is 0.55 J |
| Answer» Correct Answer - A::C::D | |
| 1256. |
Which of the following statements is incorrect?A. No work is done if the displacement is perpendicular to the direction of the applied force.B. If the angle between the force and displacement Vectors is obtuse, then the work done is negativeC. Frictional force is a non-conservativeD. All the central forces are non-conservative |
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Answer» Correct Answer - D All the central forces are conservative. All other statements are correct |
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| 1257. |
Which of the following statements is incorrect?A. Kinetic energy may be zero, positive or negativeB. Power, energy and work are all scalarsC. Potential energy may be zero, positive or negativeD. Ballistic pendulum is a device for measuring the speed of bullets |
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Answer» Correct Answer - A If an object of mass m moving with velocity `vecv`, its kinetic energy K is given by `K=1/2 mvecv.vecv=1/2 mv^2` As mass m and `v^2(vecv.vecv)` always positive, therefore kinetic energy is always positive. The kinetic energy can never be negative. |
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| 1258. |
The work done by a body against friction always results inA. loss of kinetic energyB. loss of potential energyC. gain of kinetic energyD. gain of potential energy |
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Answer» Correct Answer - A Friction always opposes motion. A body does work against friction at the expense of its kinetic energy. Hence, the work done by a body against friction always results in loss of kinetic energy. |
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| 1259. |
the work done against force of friction isA. 8.7 JB. 10.7 JC. 7.8 JD. 12.7 J |
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Answer» Correct Answer - A Work against friction is `W_f=fd=muNd=mumgcosthetad " " (therefore N=mgcos theta)` `=0.1xx1 kg xx 10 m s^(-2) xx cos 30^@ xx 10 m` =8.7 J |
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| 1260. |
the work done by applied force isA. 10 JB. 50 JC. 100 JD. 150 J |
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Answer» Correct Answer - C Work done by applied force is W=Fd= 10 N x 10 m =100 J |
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| 1261. |
the work done by friction in 10 s isA. 200JB. `-200 J`C. 600 JD. `-600 J` |
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Answer» Correct Answer - B The frictional force and the displacement are in opposite direction, therefore the angle between frictional force and displacement is `theta=180^@`. Thus, work done by the friction is `W_f=fd cos 180^@ =(2 N) (100 m) cos 180^@=-200 J` |
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| 1262. |
A block of mass m is placed at the top of a smooth wedge ABC. The wedge is rotated about an axis passing through C as shown in the figure - 3.81. The minimum value of angular speed `omega` such that the block does not slip on the wedge is : A. `(sqrt((g sintheta)/l))sectheta`B. `(sqrt(g/l))costheta`C. `(sqrt(g/(lcostheta)))costheta`D. `sqrt((gsintheta)/l)` |
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Answer» Correct Answer - A |
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| 1263. |
A flexible chain of length `L=20 sqrt(2)` m and weight W=10 kg in initially placed at rest on a smooth frictionless wedge surface ABC. It is given a slight jerk on one side so that it will start sliding on the side. Find the speed of the chain when its one end will leave the vertex of the wedge `(Take g=10m//s^(2))` A. `10 sqrt(2)m//s`B. `10 m//s`C. `4 m//s`D. `(10sqrt(2))^(1//2)` |
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Answer» Correct Answer - B |
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| 1264. |
Unit of work done (a) Nm (b) joule (c) either a or b (d) none |
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Answer» (c) either a or b |
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| 1265. |
Assertion: Work done by the friction or viscous force on a moving body in negative. Reason: Work done is a scalar quantity which cannot be negative like mass.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true not but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false. |
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Answer» Correct Answer - C Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities |
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| 1266. |
Assertion: Work done by the friction or viscous force on a moving body in negative. Reason: Work done is a scalar quantity which cannot be negative like mass.A. If both assertion and reason are true and reason is the correct 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 assertion and reason are false. |
| Answer» Work done is a scalar quantity. It can be positive or negative unlike mass nad kinetic energy which are positive scalar quantities. | |
| 1267. |
A small ball is given some velocity at point A towards right so that it moves on the semicircular track and does not leave contact up to the highest point B. After leaving the highest point B, it falls at the top of a building of height R and width `x(xlt lt 2R)`. (All the surfaces are frictionless). The velocity given to the ball at point A so that it may hit the top of the building isA. (a) `sqrt(4gR)`B. (b) `sqrt(2gR)`C. (c) `sqrt(gR)`D. (d) `sqrt(6gR)` |
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Answer» Correct Answer - D Let t be the time taken by the ball from B to C. `R=1/2"gt"^2` `t=sqrt((2R)/(g))` `2R=vsqrt((2R)/(g))` `v=sqrt(2gR)` Applying energy conservation at point A and B, `1/2mv_0^2=1/2mv^2+mg2R` `impliesv_0=sqrt(6gR)` |
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| 1268. |
A srping lies along the x-axis attached to a wall at one end and a block at the other end. The block rests on a friction less surface at `x=0`. A force of constant magnitude F is applied to the block that begins to compress the spring, until the block comes to a maximum displacement `x_(max)`. During the first half of the motion, applied force transfers more energy to theA. (a) Kinetic energyB. (b) Potential energyC. (c) Equal to bothD. (d) Depends upon mass of the block |
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Answer» Correct Answer - C From beginning to end of motion: `DeltaKE=0` `x=(2F)/(K)` (from work-energy theorem) First half corresponds to `0lexle(F/K)` Work done by force during first half: `W_1=Fx/2=F^2/K` Energy in spring `U_1=1/2K(x/2)^2=(F^2)/(2K)` |
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| 1269. |
A heavy body of mass `25kg` is to be dragged along a horizontal plane (`mu` = `(1)/(sqrt(3))`. The least force required is `(1 kgf = 9.8 N)`A. 25 kgfB. 2.5 kgfC. 12.5 kgfD. 50 kgf |
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Answer» Correct Answer - C |
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| 1270. |
A single conservative force `F(x)` acts on a `1.0-kg` particle that moves along the x-axis. The potential energy `U(x)` is given by `U(x)=20+(x-2)^2` where x is in meters. At `x=5.0m`, the particle has a kinetic energy of `20J`. What is the mechanical energy of a system?A. (a) `35J`B. (b) `64J`C. (c) `86J`D. (d) `49J` |
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Answer» Correct Answer - D At ` x=5m`, `U=20+(5-2)^2=29J`, `K=20J` Mechanical energy `=E=U+K=29+20=49J` |
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| 1271. |
A single conservative force `F(x)` acts on a `1.0-kg` particle that moves along the x-axis. The potential energy `U(x)` is given by `U(x)=20+(x-2)^2` where x is in meters. At `x=5.0m`, the particle has a kinetic energy of `20J`. The maximum and minimum values of x, respectively, areA. (a) `7.38m, -3.38m`B. (b) `6.38m, -4.38m`C. (c) `7.38m, -2.83m`D. (d) `6.38m, -2.38m` |
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Answer» Correct Answer - A The greatest or least value of x will be at the point where K is zero and U is maximum. `U_(max)=49Jimplies20+(x-2)^2=49implies(x-2)^2=29` `impliesx-2=+-sqrt(29)impliesx=-3.38m, 7.38m` |
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| 1272. |
The potential energy of configuration changes in x and y directions as `U=kxy`, where k is a positive constant. Find the force acting on the particle of the system as the function of x and y. |
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Answer» Substituting `U=kxy` in the expression `F=-(delU)/(delx)hati-(delU)/(dely)hatj-(delU)/(delz)hatk` we have `oversetrarrF=-k(yhati+xhatj)` |
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| 1273. |
The potential energy of a particle in a certain field has the form `U=(a//r^2)-(b//r)` , where a and b are positive constants and r is the distance from the centre of the field. Find the value of `r_0` corresponding to the equilibrium position of the particle, examine whether this position is stable. |
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Answer» `U(r)=a/r^2-b/r` Force `=-(dU)/(dr)=-((-2a)/(r^3)+b/r^2)=-((br-2a))/(r^3)` At equilibrium, `F=-(dU)/(dr)=0` Hence, `br-2a=0`, at equilibrium. `r=r_0=(2a)/(b)` corresponds to equilibrium. At stable equilibrium, the potential energy is minimum and at unstable equilibrium, it is maximum. For minimum potential energy, `du//dr=0` and `d^2U//dr^2gt0` at `r=r_0`. Let us investigate the second derivative. `(d^2U)/(dr^2)=(d)/(dr)((dU)/(dr))=(d)/(dr)((-2a)/(r^3)+(b)/(r^2))=(6a)/(r^4)-(2b)/(r^3)` At `r=r_0=(2a)/(b)implies(d^2U)/(dr^2)=(6a-2br_0)/(r_0^4)=(2a)/(r_0^4)gt0` Hence, the potential energy function `U(r)` has a minimum value at `r_0=2a//b`. The system has a stable equilibrium at minimum potential energy state. |
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| 1274. |
The displacement of a particle of mass `1kg` on a horizontal smooth surface is a function of time given by `x=1/3t^3`. Find out the work done by the external agent for the first one second. |
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Answer» Given that `x=1/3t^3`. The velocity of the particle at any instant t is `v=(dx)/(dt)=t^2` The acceleration of the particle at any instant t is `a=(d^2x)/(dt^2)=2t` Therefore, work done by the force imposed is `W=intFdx=intF(dx)/(dt)*dt=intm(a)((dx)/(dt))(dt)` Putting the values of `m=1kg`, `(dx)/(dt)`, and a, we obtain `W=underset0overset(t=1)int(1)(t)^2(2t)dt=2underset0overset1intt^3dt=2(t^4/4)_0^1=0.5J` |
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| 1275. |
Give explanations for the following Why well-off Londoners supported the need to build housing for the poor in the nineteenth century. |
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Answer» Well-off Londoners supported the need to build housing for the poor in the nineteenth century on account of three reasons: one-room houses of the poor came to be seen as the breeding ground of diseases, and hence, a threat to public health; fire hazards became a worry in these over-crowded, badly ventilated, unhygienic homes; lastly, there was a widespread fear of social disorder, especially after the 1917 Russian Revolution. Housing schemes were undertaken to avoid a rebellion by the poor. |
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| 1276. |
A car travelling on a smooth road passes through a curved portion of the road in the form of an arc of circle of radius `10m`. If the mass of car is `120kg`, find the reaction (in kN) on car at lowest point P where its speed is `20ms^-1`. |
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Answer» Correct Answer - `(6)` `N=mg+mv^2//r=120[10+20^2//10]=6,000N=6kN` |
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| 1277. |
The speed v reached by a car of mass m in travelling a distance x, driven with constant power P, is given byA. (a) `v=(3xP)/(m)`B. (b) `v=((3xP)/(m))^(1//2)`C. (c) `v=((3xP)/(m))^(1//3)`D. (d) `v=((3xP)/(m))^2` |
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Answer» Correct Answer - C `P=Fv=m(dv)/(dt)v` or `v(dv)/(dt)=P/m` or `v(dv)/(dx)(dx)/(dt)=P/m` or `v^2(dv)/(dx)=P/m` or `v^2dv=P/mdx` On integration, we get `v^3/3=(Px)/(m)` or `v=((3xP)/(m))^(1//3)` |
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| 1278. |
An elevator can carry a maximum load of `1800 kg` (celevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force oppposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?A. `22 kW`B. `44 kW`C. `66 kW`D. `88 kW` |
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Answer» Correct Answer - B Here, `m = 1800 kg` Frictional force `f = 4000 N` Uniform speed, `v = 2ms^(-1)` Downward force on elevator is `F = mg + f` `= (1800 kg xx 10 ms^(-2)) + 4000 N = 22000N` The motor must supply enough power to balanced this force hence, `P = Fv = (22000 N) (2 ms^(-1)` `= 44000 W = 44 xx 10^(3)W = 44 kW`. |
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| 1279. |
A ball is projected with an velocity`V_(0)` at an angle of elevation `30^(@)`. Mark of the correct statement.A. Kinetic energy will be zero at highest point of the trajectory.B. Vertical component of momentum will be conservedC. Horizontal component of momentum will be conservedD. Gravitational potential energy will be minimum at the highest pointof the trajectory |
| Answer» Since horizontal component of vertical is constant hence momentum is constant. | |
| 1280. |
The potential energy function for the force between two atoms in a diatomic molecule is approximate given by `U(r) = (a)/(r^(12)) - (b)/(r^(6))`, where `a` and `b` are constants and `r` is the distance between the atoms. If the dissociation energy of the molecule is `D = [U (r = oo)- U_("at equilibrium")],D` is |
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Answer» Correct Answer - `[(6v_(0))/a{2(a/r)^(13)-(9/r)^(7)},2^(1//6)a]` |
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| 1281. |
A constant power `P` is applied to a particle of mass `m`. The distance traveled by the particle when its velocity increases from `v_(1)` to `v_(2)` is (neglect friction):A. `(m)/(3P) (v_(2)^(3) - v_(1)^(3))`B. `(m)/(3P) (v_(2) - v_(1))`C. `(3P)/(m) (v_(2)^(2) - v_(1)^(2))`D. `(m)/(3P) (v_(2)^(2) - v_(1)^(2))` |
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Answer» Correct Answer - A `P=F_(v) =mav rArr a=(P)/(mv)` `rArr v(dv)/(ds) =(P)/(mv) rArr v^(2) dv =(P)/(m) ds` `rArr (P)/(m) int_(0)^(s) ds =int_(v1)^(v2)v^(2) dv rArr (P)/(m) s =(1)/(3) (v_(2)^(3) -v_(1)^(3))` `rArr s=(m)/(3P) (v_(2)^(3) -v_(1)^(2))`. |
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| 1282. |
The potential energy function for the force between two atoms in a diatomic molecule is approximate given by `U(r) = (a)/(r^(12)) - (b)/(r^(6))`, where `a` and `b` are constants and `r` is the distance between the atoms. If the dissociation energy of the molecule is `D = [U (r = oo)- U_("at equilibrium")],D` isA. `(b^2)/(6a)`B. `(b^2)/(2a)`C. `(b^2)/(12a)`D. `(b^2)/(4a)` |
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Answer» Correct Answer - D `U=(a)/(x^(12))-(b)/(x^(6))=ax^(-12)-bx^(-6)` `F=-(dU)/(dx)=-[a(-12)x^(-13)-b(-6)x^-7]=0` (for equilibrium) `x=((2a)/(b))^((1)/(6))` At `xinfty`,`U(infty)=0` `x=((2a)/(b))^((1)/(6))` `U_(at eq)=(axxb^2)/(4a^2)-(bxxb)/(2a)=(b^2)/(4a)-(b^2)/(2a)=-(b^2)/(4a)` `U_(infty)-U_(at eq)=(b^2)/(4a)` |
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| 1283. |
A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation `v=asqrtx`, wher ea is a constant. Find the total work done by all the forces during a displacement from `x=0 to x=d`. |
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Answer» Work done by all forces, `W=DeltaKE=(1)/(2)mv_(2)^(2)-(1)/(2)mv_(1)^(2)` Here,`v_(1)=asqrt0=0, v_(2)=asqrtd`, so, `W=(1)/(2)ma^(2)d-0=(1)/(2)ma^(2)d` |
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| 1284. |
A constant power `P` is applied to a particle of mass `m`. The distance traveled by the particle when its velocity increases from `v_(1)` to `v_(2)` is (neglect friction):A. `(3P)/m(v_(2)^(2)-v_(1)^(2))`B. `m/(3P)(v_(2)-v_(1))`C. `m/(3P)(v_(2)^(3)-v_(1)^(3))`D. `m/(3P)(v_(2)^(2)-v_(1)^(2))` |
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Answer» Correct Answer - C |
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| 1285. |
A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation `v=asqrtx`, wher ea is a constant. Find the total work done by all the forces during a displacement from `x=0 to x=d`.A. `(mad)/(2)`B. `(mad^2)/(d)`C. `(ma^2d)/(2)`D. `(m^2ad)/(2)` |
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Answer» Correct Answer - C `v=asqrtx` `x=0`,`v_1=0` `x=d`,`v_2=asqrtd` `W_(1rarr2)=K_2-K_1=(1)/(2)m(v_2^2-v_1^2)` `=(1)/(2)m(ad-0)=(1)/(2)ma^2d` |
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| 1286. |
A particle of mass `m` moves on a straight line with its velocity varying with the distance traveled. Find the total work done by all the forces during a displacement `x=0` to `x=d` if the velocity is equal to (a) `v=lamdasqrtx` and (b) `v=lamdax` is constant. |
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Answer» We know `W_(t otal)=K_2-K_1` (a) `x=0`, `v_1=0`, `K_1=0` `x=d`,`v_2=lamdasqrtd`,`K_2=(1)/(2)mv_2^2=(1)/(2)=(1)/(2)mlamda^2d` `W_(t otal)=K_2-K_1=(1)/(2)mlamda^2d` (b) `x=0`,`v_1=0`,`K_1=0` `x=d`,`v_2=lamdad`,`K_2=(1)/(2)mv_2^2=(1)/(2)=(1)/(2)mlamda^2d^2` `W_(t otal)=K_2-K_1=(1)/(2)mlamda^2d^2` |
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| 1287. |
Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .A. `1600 J`B. `160 J`C. `16 J`D. `1.6 J` |
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Answer» Correct Answer - C `v=(dx)/(dt) =t^(2)` `W =1/2mv^(2) =1/2mt^(4)` `1/2 xx 2 xx (2)^(4) =16 J` |
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| 1288. |
The potential energy of a particle of mass 5 kg moving in xy-plane is given as `U=(7x + 24y)` joule, x and y being in metre. Initially at `t=0`, the particle is at the origin `(0,0)` moving with velovity of `(8.6hati+23.2hatj) ms^(1)`, ThenA. The velocity of the particle at `t=4s`, `5ms^(-1)`B. The acceleration of the particle is `5ms^(-2)`.C. The direction of motion of the particle initially (at t=0) is right angles to the direction of acceleration.D. The path of the particle is circle. |
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Answer» Correct Answer - A::B `F=-[(del)/(delX)hati + (delU)/(dely)hatj]=(7hati-24hatj) N` `a=F/m =(7/5hati - (24)/5hatj)m//s` `|a|=ssqrt((7/5)^(2) + (24/5)^(2) =5 m//s^(2))` Since, `a="constant"`, we can apply, `v=u+at` `=(8.6hati + 23.2hatj) + (-7/5hati-(24)/5hatj)(4)` `=(3hati +4hatj)m//s` `|v|=sqrt((3)^(2) + (4)^(2)=5m//s)` |
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| 1289. |
A block is constrained to move along x-axis under a forc `F=4/(x^(2))(xne0)`. Here, F is in newton and x in metre. Find the work done by this force when the block is displaced from `x=4` m to `x=2m`. |
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Answer» Correct Answer - A `W = int_(4)^(2)Fdx = int_(4)^(2)(4)/(x^(2))dx` `= -4[(1)/(x)]_(4)^(2) = -4[(1)/(2) - (1)/(4)] = -1J` |
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| 1290. |
A force `F=(2+x)` acts on a particle in x-direction where F is in newton and x in metre. Find the work done by this force during a displacement form 1. 0 m to x = 2.0 m. |
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Answer» Correct Answer - C As the force is variable, we shall find the work done in a small dispacent from x to x + dx and then intrgrate it to find the total work done in thes small displacement is. `dW=F dx=(2+x)dx` Thus, `W=int_(2.0)^(1.0)dW = int_(1.0)^(2.0)(2+x)dx` `=[2xxx2/2]_(1.0)^(2.0)=3.5J`. |
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| 1291. |
Work done when a force `F=(hati + 2hat(j) + 3hatk) N` acting on a particle takes it from the point `r_(1) =(hati + hatk)` the point `r_(2) =(hati -hatj + 2hatk)` is .A. `3 J`B. `1J`C. `zero`D. `2J` |
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Answer» Correct Answer - B `W =F.S =F.(r_(f)-r_(i))`. |
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| 1292. |
A force (F) acting on a particle varoes work done by a particle varies with the with the position x as shown in figure . Find the work done by by force in displacing the particle from . (a) `x =-2m` to `x=0` (b) `x=0` to `x =2m.`. |
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Answer» Correct Answer - A (a) From x=-2 m to x=0, displacement of the particle is along positive x-direction while force acting on the particle is along negative `x`-direction Therefore, work don eis negative and given by the area under F-x graph with projevtion along x-axis. `:. W = -(1)/(2)(2)(10) = -10J` (b) From `x=0` to `x=2m`, displacement of particle and force acting on the particle both are along positive x-direction. Therefore, work done is positive and given by the area under F-x graph, or `W=1/2(2)(10)=10 J`, |
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| 1293. |
A force`F=(3thati + 5hatj)N` acts on a body due to which its displacement varies as `S=(2t^(2)hat-5hatj)m`. Work done by these force in `2 s` is .A. `32 J`B. `24 J`C. `46 J`D. `20 J` |
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Answer» Correct Answer - B `v=(dS)/(dt) =(4t)hati` `P=F.v =12 t^(2)` `:. W =int_(0)^(2) Pdt =int_(0)^(2)(12) t^(2)dt` =24 J`. |
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| 1294. |
Displacement of a particle of mass 2 kg varies with time as `s=(2t^(2)-2t + 10)m`. Find total work done on the particle in a time interval from `t=0` to `t=2s`. |
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Answer» Correct Answer - B::C `v=(ds)/(dt) =(4t-2)` `W_("all") =DeltaK=K_(f)-K_(i)=K_(2s)-K_(0s) `=1/2m(v_(f)^(2)-v_(i)^(2)) `=1/2 xx 2[(4 xx 2-2)^(2)-(4 xx 0-2)^(2)]` `=32J` . |
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| 1295. |
A block of mass 1kg start moving with constant acceleration `a=4m//s_(2)` Find. (a) average power of the net force in time inteval from `t=0` to `t=2s`, (b) instantaneous power of the net force at `t=4 s`. |
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Answer» Correct Answer - A::B::D (a) `P_(av)=W/t =((1)/(2)mv^(2))/t =((1)/(2)m(at)^(2))/t` `=1/2ma^(2) t =1/2 xx 1 xx (4)^(2) (2)` `=16W` (b) `P_(i) =Fv=(ma)(at)` `=mg^(2)t=(1) (4)^(2) (4)` `64W`. |
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| 1296. |
Force acting on a particle varies with displacement as shown is Fig. Find the work done by this force on the particle from `x=-4 m` to `x = + 4m. . |
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Answer» Correct Answer - C W=area under F-x graph From `X=-4` to `X =-2` `F=-ve` , (from graph) `S=+ve` `:. W_(1) =-1/2 xx 10 =10J` From `-2` to `4` `F = + ve` `S = + ve` `:. W_(2) = + 1/2(6 + 2) (10)` `= + 40J` `:. W_(T) =W_(1) + W_(2) =30J` |
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| 1297. |
A 5 kg mass is raised distance of `4m` by a vertical force of 80 N. Find the final kinetic energy of the mass if it was originally at rest.`g =10 m//s^(2)`. |
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Answer» Correct Answer - A::B `W_(F) =FS cos 0^@ =80xx4xx1 =320J` `W_(mg)=(mg)(S) cos 180^@` `=(50)(4)(-1)` `=-200J` `K_(f)=W_(All)=120J`. |
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| 1298. |
A body of mass `2 kg` makes an elastic head - on collision another body at rest and continues to move in the original direction with one fourth of its original speed . The mass of the second body which collides with the first body isA. 2 kgB. 1.2 kgC. 3 kgD. 1.5 kg |
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Answer» Correct Answer - B |
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| 1299. |
A rock of mass `m` is dropped to the ground from a height `h`. A second rock, with mass `2m`, is dropped from the same height. When the second rock strikes the ground, what is its kinetic energy? (a) Twice that of the first rock, (b) four times that of the first rock, (c) same as that of the first rock, (d) half as much as that of the first rock, (e) impossible to determine. |
| Answer» a. The more massive rock has twice as much gravitational potential energy associated with it compared with that of the lighter rock. Because mechanical energy of an isolated system is conserved, the more massive rock will arrive at the ground with twice as much kinetic energy as the lighter rock. | |
| 1300. |
Constant the following two statements : A. Liner momentum of a system of particles is zero B . Kinetic energy of a system of particls is zero .ThenA. A does not impty `B` and `B` does not imply `A`B. A implies`B` and `B` does not imply `A`C. A does not impty `B` and `B` implies`A`D. A impies `B` and `B` implies`A` |
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Answer» Correct Answer - C (c ) Kinetic energy of a system of particle is zero only when the speed of each particle is zero . And if speed of each particle is zero , the linear momentum of the system of partuicle has to be zero . Also the linear momentum of the system may be zero even when the particle are moving . this is became linear momentum is a vector quantity in this case the kinetic energy of the particle will not be zero :. A does not imply `B` but `b` imlies `A` . |
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