InterviewSolution
Saved Bookmarks
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.
| 551. |
A man carries a load of 5 kg on his head through a distance of 10 m. the minimum amount of work is done when heA. Move it up an inclined planeB. Moves it down an inclined planeC. Moves it over a horizontal surfaceD. Lifts it vertically upwards. |
| Answer» Correct Answer - C | |
| 552. |
Displacement time graph of a particle moving in a straight line is as shown in figure. Select the correct alternative (s) : .A. Work done by all the forces in region OA and BC is positive.B. Work done by all the forces in region AB is zero.C. Work done by all the forces in region BC is negative.D. Work done by all the forces in region OA is negative. |
|
Answer» Correct Answer - B::C |
|
| 553. |
The potential energy of a partical of mass `2 kg` moving in `y-z` plane is given by `U=(-3y+4z)J` where `y` and `z` are in metre. The magnitude of force ( in newton ) on the particle is `:` |
|
Answer» Here, `U=(-3 y + 4 z)J` As `vec(F)=-(deltau)/( deltay)hat(j)- (deltau)/( deltaz) hat(k)` `:. vec(F) 3 hat (j) - 4 hat(k)` `|vec(F)|=sqrt(3^(2)+(-4)^(2))=sqrt(9+16)=5N` |
|
| 554. |
A bucket of mass `4//3` is tied to a string and is lowered at a constant acceleration of `g//4`. The magnitude of work done (in joule ) by the string in lowering the bucket by `10 cm` would be `:` |
|
Answer» Here, `m=(4)/(3) kg, a =(g)/(a) , ` downwards. If `T` is tension in the string, then `T=m(g-a)=(4)/(3)(g-(g)/(4))=(4)/(3).(3g)/(4)=g` Magnitude of work done by string, `W=Txxs=gxx(10)/(100)=10 xx (10)/(100)=1J` |
|
| 555. |
A block of mass m tied to a string is lowered by a distance d, at a constant acceleration of `g//3`. The work done by the string isA. `2/3mgh`B. `(-mgh)/(3)`C. `mgh`D. `4/3mgh` |
| Answer» Correct Answer - D | |
| 556. |
When work done by force of gravity is negative, thenA. PE increasesB. KE increasesC. PE remains constantD. PE decrease |
|
Answer» Correct Answer - A (a) When a body is moved upward, work done by gravity is negative but potential energy increases. |
|
| 557. |
A constant power P is applied on a particle of mass m. find kintic energy, velocity and displacement of particle as function of time t. |
|
Answer» Correct Answer - B::C (i) `W =Pt =1/2mv^(2)` `(ii) v=sqrt(2pt)/m` (iii) integrating the velocity, we will get displacement `:. S=sqrt((2p)/(m))(t^(3//2)/(3//2)) =sqrt(8p)/(9m)t^(3//2)`. |
|
| 558. |
A particle moves along the x-axis from `x=0` to `x=5m` under the influnce of a given by `F =7-2x + 3x^(2)`. The instantaneous power applied to the particle is.A. `360 J`B. `85 J`C. `185 J`D. `135 J` |
|
Answer» Correct Answer - D `W =int_(0)^(5)Fdx =int_(0)^(5)(7-2x + 3x^(2)) dx` `=135 J` |
|
| 559. |
Assertion : The work done by the spring force in a cyclic process is zero. Reason : Spring force is a conservative force.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. |
|
Answer» Correct Answer - A In case of spring force, the work done only depends on the initial and final positions. So spring force is a conservative force and in cyclic process the initial and final positions are same, hence work done is zero. |
|
| 560. |
Assertion: Universe as a whole may be viewed an isolted system. Reason: Total energy of an isolated system remain constant or stretched.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. |
|
Answer» Correct Answer - B Universe as a whole may be viewed as an isolated system. The total energy of the universe is constant. If one part of the universe loses energy, another part must gain an equal amount of energy |
|
| 561. |
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 jouleB. 160 jouleC. 16 jouleD. 1.6 joule |
|
Answer» Correct Answer - C |
|
| 562. |
The position of a particle moving on a stringht line under rthe action of a force is given as `x=50t-5t^(2)` Here, x is in metre and t is in second. If mass of the particle is 2 kg, the work done by the force acting on the particle in first 5 s is.A. 2500 JB. `-2500J`C. 5000 JD. `-5000J` |
| Answer» Correct Answer - B | |
| 563. |
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 JB. 160 JC. 16 JD. 1.6 J |
|
Answer» Correct Answer - C (c ) Speed of the body, `v=(dx)/(dt)=(d)/(dt)((t^(3))/(3))=t^(2)` At`" "` t=0, v=0, At`" " t=0 s, v=4 ms^(-1)` From work energy theorem, W=change in kinetic energy `K_(f)-K_(i)` `=(1)/(2)m(v_(f)^(2)-v_(i)^(2))=(1)/(2)xx2xx(16-0)=16J` |
|
| 564. |
A particle of mass `m` moving with velocity `v` strikes a stationary particle of mass `2 m` and sticks to it. The speed of the system will be.A. v/2B. 2vC. v/3D. 3v |
|
Answer» Correct Answer - C |
|
| 565. |
A rod mass (M) hinged at (O) is kept in equilibrium with a spring of stiffness (k) as shown in figure. The potential energy stored in the spring is . A. `(mg)^(2)/(4k)`B. `(mg)^(2)/(2k)`C. `(mg)^(2)/(8k)`D. `(mg)^(2)/(k)`. |
|
Answer» Correct Answer - C `sum ("Moment about" O)=0` `:. (kx) l = mg (l/2)` or `x=(mg)/(2k)` `U=(1)/(2)kx^(2) =(mg)^(2)/(8k)` |
|
| 566. |
System shown in figure is in equilibrium, find the magnitude of net change in the string tension between two masses just after, when one of the springs in cut, Mass of both the blocks is same and equal to m spring constant of both springs is k. .A. `(mg)/(2)`B. `(mg)/(4)`C. `(mg)/(3)`D. `(3mg)/(2)` |
|
Answer» Correct Answer - A `T_(i)=gh`......(i) `2kx =2mg` `:. kx =mg` One `kx` force (acting is upward directon) is suddenly removed. So, net downward force on system will be `kx` or `mg`. Threrfore, net downward acceleration of system, `a=(mg)/(2m) =g/2` Free body diagram of lower block gives the equation, `mg-T_(f) =ma=(mg)/(2)` `:.T_(f)=(mg)/(2)` From these two equation, we get `DeltaT =(mg)/(2)`. |
|
| 567. |
Fqual net forces ant on two different block (A) and (B) masses (m) and 4(m) respectively For same displacement, identify the correct statement.A. Their kinetic energies are in the ratio `(K_(A))/(K_(B)) = (1)/(4)`B. Their speeds are in the ration `(v_(A))/(v_(B)) = (1)/(1)`C. Work done on the block are in the ratio `(W_(A))/(W_(B)) = (1)/(1)`D. All of the above |
|
Answer» Correct Answer - C `W =FS` `F` and `S`are same. Therefore, `W_(A)/K_(B)=1/1` From work energy theorem, `K_(A)/(K_(B))=W_(A)/(W_(B))=(1)/(1)` `1/2 m_(A)v_(A)^(2) =1/2 m_(B) v_(B)^(2)` `:. (v_(A))/v_(B) =sqrt((m_(B))/(m_(A))) =2/1`. |
|
| 568. |
Asseration : Power of a constant force is also constant. Reason : Net constant force will always produce a constant acceleration.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 - D `F = "constant"` `:. a = (F)/(m) = "constant"` `v = at = (F)/(m) t` `P = F.v = (F^(2))/(m)t` `P = f.v=(F^(2))/(m)t` `:. P ne "constant"` But `P prop t` |
|
| 569. |
Identify, which of th following energies can be positive (or zero) only ?A. Kinetic energyB. Potential energyC. Mechanical energyD. Both kinetic and mechanical energy |
| Answer» Correct Answer - A | |
| 570. |
If stretch in a spring of force constant k is doubled, calculate (a) ration of final to initial force in the spring. (b) ratio of elastic energies stored in the two cases. (c) work done in changing to the state of double stretch. |
|
Answer» (a) Foa a given spring, `F=kx` `:. (F_(2))/(F_(1))=(kx_(2))/(kx_(1))=(2x)/(x)=2` (b) For a given spring, `U=(1)/(2)kx^(2),` `(U_(2))/(U_(1)) =((1)/(2)kx_(2)^(2))/((1)/(2)kx_(1)^(2))=((2x)^(2))/(x^(2))=4` (c) As work done in stretching the spring is stored in the spring in the form of elastic potential energy of the spring, therefore, `W=U_(2)-U_(1)=(1)/(2)kx_(2)^(2)-(1)/(2)kx_(1)^(2)` `=(1)/(2)k[(2x)^(2)-x^(2)]=(3)/(2)kx^(2)` |
|
| 571. |
Two collies lift some load from the road to the roof of a bus. One of them takes 1 min. and the other takes two min. to do the same job. Who has done more work and whose power is more ? |
| Answer» As `W=Fxxs`, therefore in lifting the same load (F) through the same distance (s), work done by both the coolie is the same. As power `=` work`//` time, therefore, power of first coolie is more than the power of second coolie. | |
| 572. |
The potential energy of two atoms separated by a distance x is give by `U=-A//x^(0)`, where A is a positive constant. What is the force exerted by one atom on another atom ? |
|
Answer» Here, `U=A//x^(6), F=?` From `F_(x)=-(partial)/(partialx)(U)` `=-(partial)/(partialx)((-A)/(x^(6)))=A((-6)/(x^(7)))` `:.` force exerted by one atom on another atom `=-6A//x^(7)` |
|
| 573. |
Match the column I with column II. A. A-p,B-q,C-r,D-sB. A-q,B-r,C-s,D-pC. A-s,B-r,C-q,D-pD. A-s,B-p,C-q,D-r |
|
Answer» Correct Answer - D When a body does some work against friction, its kinetic energy decreases. `therefore` A-s Work done by a body is independent of time `therefore` B-p Power of a body (P=W/t) varies inversely as time. C-q When work done over a closed path is zero, force must be conservative, `therefore` D-r |
|
| 574. |
A man of mass `m` , standing at the bottom of the staircase of height `L` climbs it and stands at its top .A. Work done by all forces on man is equal to the rise in potentail energy mgLB. Work done by all forces on man is zeroC. Work done by the gravitational force on man is mgLD. The reaction force from a step does not do work because the point of application of the force does not move while the force exists |
|
Answer» (b,d) When a man of mass m climbs up the staircase of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic energy is almost zero. Hence, total work done =-mgL+mgL=0 As the point of application of the contact forces does not move hence work done by reasction forces will be zero. NOTE: Here work done by friction will also be zero as there is no dissipation or rubbing is involved. |
|
| 575. |
When an air bubble rises in water, what happens to its potential energy / |
| Answer» Potential energy of air bubble decreases, because work is done by upthrust on the bubble. | |
| 576. |
A particle of rest mass `m_(0)` moves with a spess `c//2` . Calculate its (i) mass (ii) momentum (iii) total energy (iv)K.E. |
|
Answer» (i) From `m=(m_(0))/(sqrt((1-v^(2)//c^(2))))=(m_(0))/(sqrt(1-1//4))=(2m_(0))/(sqrt(3))` (ii) momentum `=mc//2=(2m_(0))/(sqrt(3)).(c)/(2)=(m_(0)c)/(sqrt(3))` (iii) total energy `=mc^(2)=(2m_(0))/(sqrt(3))c^(2)` (iv) K.E. `=` total energy `-` rest mass energy `=(2m_(0))/(sqrt(3))c^(2)-m_(0)c^(2)=m_(0)c^(2)[(2)/(sqrt(3))-1]` `=(m_(0)c^(2))/(sqrt(3))(2-sqrt(3))` |
|
| 577. |
In a nuclear reation, the mass defect is 1atomic mass unit. What is the energy released ? |
|
Answer» Here, `Deltam=1am u=1.66xx10^(-27)kg,E=?` `E=(Deltam)c^(2)=1.66xx10^(-27)xx(3xx10^(8))^(2)=1.494xx10^(-10)J` |
|
| 578. |
Estimate the amount of energy released in the nuclear fusion reaction: `_(1)H^(2)+._(1)H^(2)rarr._(2)He^(2)+._(0)n^(1)` Given that `M(._(1)H^(2))=2.0141u, M(._(2)He^(3))=3.0160u` `m_(n)=1.0087u` , where `1u=1.661xx10^(-27)kg` . Express your answer in units of MeV. |
|
Answer» Here, mass of two deutrons `=2xx2.0141=4.0282u` Mass of `_(2)He^(3)` and neutron `=3.0160+1.0087=4.0247u` Loss of mass `= Deltam=4.0282-4.0247` `=0.0035u=0.0035xx1.661xx10^(-27)kg` Energy released `E=(Deltam)c^(2)` `=0.0035xx1.661xx10^(-27)(3xx10^(8))^(2)"joule"` `=52.2xx10^(-14)"joule"` As `1MeV =1.602xx10^(-13)"joule:` ` :. ` Energy released `=(52.2xx10^(-14))/(1.602xx10^(-13))=3.26MeV` |
|
| 579. |
The amount of energy released in burning 1 kg of coal isA. 3 MJB. 30 MJC. 300 MJD. 3000 MJ |
|
Answer» Correct Answer - B When 1 kg of coal is burnt the energy released is 30 MJ `(=3 xx 10^7 J)`. |
|
| 580. |
Energy required to break one bond in DNA is approximatelyA. `10^(-10)` JB. `10^(-18)` JC. `10^(-7)` JD. `10^(-20)` J |
|
Answer» Correct Answer - D The energy required to break one bond in DNA is `10^(-20)` J |
|
| 581. |
What is represented by area under the force displacement curve? |
| Answer» It represents the work done by the force over the given displacement. | |
| 582. |
Force acting on a particle varies with displacement as shown in figure. Find the total work done by the force. |
|
Answer» Work done = Area under F- s curve Hence, W = Triangular Area + Rectangular Arrea `=1/2xx4xx10+(8-4)xx10` `=20+40` `=60J` |
|
| 583. |
A force F acting on an object varies with distance x The work done by the force in moving the object from x=0 to x=8 m isA. zeroB. 80 JC. `-40 J`D. `40 J` |
|
Answer» Correct Answer - A Work done = Area under F-x graph with proper algebraic sign `=1/2xx20 xx4-1/2xx20 xx 4=0J` |
|
| 584. |
The area under force-displacement curve representsA. velocityB. accelerationC. impulseD. work done |
|
Answer» Correct Answer - D The area under force-displacement curve represents work done |
|
| 585. |
The iron `Fe^(27)` nucleus emits a y-axis of energy 14.4KeV. If mass of nucleus is 56.935u, calculate the recoil energy of the nucleus. Take `1u=1.66xx10^(-27)kg.` |
|
Answer» The nuclear decay is represented as Accordinig to be Broglie hypothesis, linear momentum of photon, `p=(E)/(c)=(14.4xx1.6xx10^(-16)J)/(3xx10^(8)m//s)` `=7.68xx10^(-24)kgms^(-1)` From conservation of linear momentum, momentum of daughter nucleus `=` momentum of photon `p=7.68xx10^(-24)ms^(-1)` `:.` Recoil energy of nucleus `K=(p^(2))/(2m)=((7.68xx10^(-24))^(2))/(2xx56.935xx1.66xx10^(-27))` `=0.312xx10^(-21)J` `K=(0.312xx10^(-21))/(1.6xx10^(-16))keV` `=1.95xx10^(-6)keV` |
|
| 586. |
Does a man standing at rest on a moving truck possess KE? |
| Answer» Yes, the amn has KE, as he is moving with the velocity of truck. | |
| 587. |
Two springs have their force constant as `k_(1)` and `k_(2) (k_(1) gt k_(2))`. When they are streched by the same force.A. No work is done in case of both the springsB. Equal work is done in case of both the springsC. More work is done in case of second springD. More work is done in case of first spring |
|
Answer» Correct Answer - C |
|
| 588. |
State the two conditions under which a force does not work. |
|
Answer» 1. Displacement is zero or it is perpendicular to force. 2. Conservative force moves a body over a closed path. |
|
| 589. |
Two bodies stick together after collision. What type of collision is in between these two bodies? . |
|
Answer» Inelastic collision. |
|
| 590. |
Which spring has greater value of spring constant – a hard spring or a delicate spring? |
|
Answer» Hard spring has greater value of spring constant - a hard spring or a delicate spring. |
|
| 591. |
What is work done by centripetal force in moving on body half-cycle on a circular path of radius 30m? |
| Answer» `W=Fscostheta=Fscos90^(@)=Zero` | |
| 592. |
A body of mas 5 kg is placed at origin. A force starts acting on the body given by `vecF=(2+3x)hati` where x is the distance of body from origin in meters. Find the speed acquired by the body as it passes through `x=5m.` |
|
Answer» Here the only force acting is `vecF=(2+3x)hati.` So body starts moving along x-axis. By applying work-energy principle. `W=intvecF.vec(dx)=DeltaK` `impliesunderset(0)overset(5)int[(2+3xx)hati].(dxhati)=1/2xx5v^(2)-0` `impliesunderset(0)overset(5)int(2+3x)dx=5/2v^(2)` `implies[2x+(3x^(2))/(2)]_(0)^(5)=5/2v^(2)` `implies[10+75/2]-[0]=5/2v^(2)` `impliesv=sqrt(19)m//s` |
|
| 593. |
Assertion: Work done by friction over a closed path is not zero and no potential energy can be associated with friction. Reason: Every force encountered in mechanicshave an associated potential energy.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. |
|
Answer» Correct Answer - C Every force encountered in mechanics does not have an associated potential energy. |
|
| 594. |
A force is acting on a body moving alonog x-axis in the direction of motion of the body. If this force produces a potential energy `u=Ax^(4),` where `A=1.2Jm^(-4),` then the what is the force acting on the body when the body is at `x=-0.8m`? |
|
Answer» Here, `u=Ax^(4), A=1.2Jm^(-4), F_(x)=?` `x=-0.8m` As `F_(x)=-(du)/(dx)=-(d)/(dx)(Ax^(4))` `:.F_(x)=-A(4x^(3))=-4Ax^(3)` `=-4xx1.2(-0.8)^(3)=2.46N` |
|
| 595. |
Assertion: A light body and a heavy body have same momentum. Then they also have same kinetic energy. Reason: Kinetic energy does not depand on mass of the body.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. |
|
Answer» Correct Answer - D Kinetic energy of a body of mass `m_1` `K_1=1/2m_1v_1^2=p_1^2/(2m_1)` Again, kinetic energy of a body of mass `m_2`. `K_2=1/2m_2v_2^2=p_2^2/(2m_2)` If `p_1=p_2 , K_1/K_2=m_2/m_1` If `m_2 gt m_1` Then , `K_1 gt K_2` , i.e. the kinetic energy of light body will be more than the kinetic energy of heavy body when both have same momentum. |
|
| 596. |
A force is acting on a body moving along x-axis in the direction of motion of the body. If this force produces a potential energy `u=Ax^(4),` where `A=1.2Jm^(-4),` then the what is the force acting on the body when the body is at `x=-0.8m`? |
|
Answer» Here, `U=Ax^(4)` `F_(x)=-(partialU)/(partialx)=(partial )/(partial x)(Ax^(4))=-4Ax^(3)` When `x=-0.8m, ` then `F_(x)=-4xx1.2(-0.8)^(3)=2.46N` |
|
| 597. |
Work done by the conservative force on a system is equal to :A. The change in kinetic energy of the systemB. The change in potential energy of the systemC. The change in total mechanical energy of the systemD. None of these |
|
Answer» Correct Answer - A::B |
|
| 598. |
Chemical, gravitational and nuclear energies are nothing but potential energies for different types of forces in nature. Explain this statement clearly with examples. |
|
Answer» A system of particles has potential energy when these particles are held certain distance apart agains some force. For example, chemical energy is due to chemical bonding between the atoms. Gravitational energy arises when objects are held at some distance against the gravitational attraction. Similarly, nuclear energy arises on account of nuclear forces. |
|
| 599. |
Assertion:The work done by a conservative force such as gravity depends on the initial and final positions only Reason: The work done by a force can not be calculated if the exact nature of the force is not known.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. |
|
Answer» Correct Answer - C The work done by a force can be calculated sometimes even if the exact nature of the force is not known |
|
| 600. |
A ball of mass m collides with a wall with speed v and rebounds on the same line with the same speed. If the mass of the wall is taken as infinite, then the work done by the ball on the wall isA. `mv^2`B. `1/2mv^2`C. `2mv`D. zero |
|
Answer» Correct Answer - D Since the wall has infinite mass therefore the displacement is zero. |
|