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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.
| 601. |
Thekinetic energyofaparticle movingalong a circle of radius R depends on the distance covered S as K `=alphaS^(2),` where `alpha` is a constant. Find theaorce acting on the particle as a function of S. |
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Answer» In circular motion, thefore acting is tangentialas well as fadial.The two forces are given by the expressions `F_("tangential")=ma_(t)m(dv)/(dt)and F_("radial")=me_(c)=mv^(2)R` Now `K=alphaS^(2)` Also `K=1/2mv^(2)` So, we have `1/2mv^(2)=alphaS^(2)` `v^(2)=(2alphaS^(2))/(m)` Differentiating w.r.t time, we get `2vxx(dv)/(dt)=((2x)/(m))2S(dS)/(dt)` Now `(dS)/(dt)=v` (speed) So we have `(dv)/(dt)=(2alphsS)/(m)` Now `F_("tangential")=(mdv)/(dt)=2alphaS` Also, `F_("radial")=ma_(c)=(mv^(2))/(R)=(2aS^(2))/(R)` Net force `=sqrt(F_("tangential")^(2)+F_("radial")^(2))=2alphaSsqrt(1+(S^(2))/(R^(2)))` |
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| 602. |
Assertion : Time taken by a body to complete a given work has nothing to do with energy of the body. Reason : Because power of a body is the rate of doing work.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - B Both, assertion and reason are true, but th elatter is not a correct explanation of the former. |
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| 603. |
Assertion : Work done by orforce of friction in moving a body through any round trip is zero. Reason : Because friction is a conservative force.A. If both, Assertion and Reason are true and Reason is correct explantion of Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - D Friction is a non conservative force. Work done by or against friction in movting a body over a body round trip is never zero. |
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| 604. |
Two blocks each of mass M are connected to the ends of a light frame as shown in figure. The frame si rotated about the vertical line of symmetry. The rod breaks if the tension in it exceeds `T_0`. Find the maximum frequency with which the frame may be rotted without breaking the rod. A. `1/(2pi)[T_(0)/(Ml)]^(1//2)`B. `1/(2pi)[(Ml)/T_(0))]^(1//2)`C. `1/(2 pi)[(MT_(0))/l)]^(1//2)`D. `1/(2pi)[l/(MT_(0)]^(1//2)` |
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Answer» Correct Answer - A |
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| 605. |
Examine the condition of people living in chawls of Bombay. |
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Answer» The people living in chawls were living in pitiful condition.
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| 606. |
A body of mass `m_(1)` collides elastically with another body of mass `m_(2)` at rest. If the velocity of `m_(1)` after collision is `(2)/(3)` times its initial velocity, the ratio of their masses is `:`A. `1:5`B. `5:1`C. `5:2`D. `2:5` |
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Answer» Correct Answer - B In all elastic collision, `upsilon_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m_(2))/(m_(1)+m_(2)))u_(2)` As second ball is initially at , `u_(2)=0` `:. upsilon_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)` As `upsilon_(1)=(2)/(3)u_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)` `:. 2m_(1)+2m_(2)=3m_(1)-2m_(2)` or `5m_(2)=m_(1)` or`(m_(1))/(m_(2))=(5)/(1)` |
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| 607. |
A heavy truck moving with a velocity of `60 km//h` collides with a light drum at rest. If the collision is elastic, then the velocity of the drum immediately after collision will beA. zeroB. `60km//h`C. `120 km//h`D. `30km//h` |
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Answer» Correct Answer - C In elastic collosion of a very heavy body with a light body at rest, the hearvier body continues to move with its initial velocity, but velocity of lighter body becomes twice the velocity of hearvier body. `:. upsilon_(2)=2u_(1)=2xx60km//h=120km//h` |
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| 608. |
This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible glven as `f(1/2mv^2)` then `f=(m/(M+m))` Statement II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.A. Statement - 1 is ture , statement -2 is true, and statement-2 is correct explanation of statement -1.B. Statement-1 if ture, statement-2 is true, but statement -2 is not a correct explanation of statement-1.C. Statement-1 is true, but statement -2 is false.D. Statement-1 is false, but statement-2 is true. |
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Answer» Correct Answer - D Statement `-2` is true , from the knowledge of theory. Now, maximum energy loss `=(p^(2))/( 2m)-(p^(2))/(2(M+m))` `=(p^(2)M+p^(2)m-p^(2)m)/( 2m(M+m))` `=M(p^(2))/(2m(M+m))` `=((1)/(2) m upsilon^(2))(M)/( (M+m))=f((1)/(2)m upsilon^(2))` where ` f=(M)/(M+m)` `:.` Statement `-1` is false. Choice `(d)` is correct. |
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| 609. |
Prove work energy theorem for a variable force. |
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Answer» Using the relation for kinetic energy, m = Mass of a body u = Initial velocity of the body v = Final velocity of the body a = Acceleration ki = Initial kinetic energy of the body kf = Final kinetic energy of the body The rate of change of kinetic energy is, \(\frac{dk}{dt} = \frac{d}{dt} \Big(\frac{1}{2} mv^2 \Big)\) \(= m \frac{dv}{dt} \times v\) = Fv [\(\because F =m \frac{dv}{dt}\) (Newton's second law)] \(\because \frac{dk}{dt} = F \frac{dx}{dt}\) dk = Fdx On integrating from the initial position (xA) to the final position (xB), we have \(\int \limits ^{k_f}_{k_i} dk = \int \limits ^{x_B}_{x_A} Fdx\) kf - ki = \(\int \limits ^{x_B}_{x_A} Fdx\) = W \(\therefore\)Work done on the body = Increase in K.E. of the body. |
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| 610. |
Write down more activities familiar to you |
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Answer» 1. A man carrying a load 2. Throwing a ball 3. Lifting the bag on to the shoulder 4. Pushing a car into motion |
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| 611. |
Find the work done and power of an engine which can maintain a speed of 50 ms-1 for a train of mass 3 × 106 kg on a rough level track for 5 km. The coefficient of friction is 0.05. (Giving g = 10 ms-2) |
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Answer» Given : m = 3 × 106 kg, v = 50 ms-1 μ = 0.05, g = 10 ms-2 S = 5 km = 5000 m P = ?, W = ? F = force of friction = μmg (Normal reaction, R = mg) ∴ P = Fv = μmgv or P = 0.05 × 3 × 106 × 10 × 50 or P = 75 × 106 watt or P = 75 × 103 kW = 7.5 × 104 kW Work done, W = FS = μmgS or W = 0.05 × 3 × 106 × 10 × 5 × 103 J or W = 75 × 108 J. |
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| 612. |
An engine draws a train up an incline of 1 in 100 at the rate of 36 kmh-1. If the resistance due to friction is 5 kg wt per ton, find out the power of the engine. Mass of train and engine is 100 metric ton. |
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Answer» Given : m = 100 metric ton = 100 × 1000 kg Total force of friction, f1 = 100 × 5 = 500 kg wt = 500 × 9.8 N = 4900 N \(sin\,θ=\frac{1}{100}\) Suppose, f2 = Downward force on the train = component of its weight acting in downward direction parallel to the inclined plane = mg sin θ = 100 × 1000 × 9.8 × \(\frac{1}{100}\) = 9800 N When F be the total force against which engine has to work, then F = f1 + f2 = 4900 + 9800 = 14700 N v = velocity of train = 36 kmh-1 = 36 × \(\frac{5}{18}\) ms-1 = 10 ms-1 ∴ Power of the engine, is given by P = F × v = 14700 × 10 = 147000 watt = 147 kW. |
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| 613. |
Can P.E. of an object be negative? |
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Answer» Yes, it can be negative when forces involved are attractive. |
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| 614. |
A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy? |
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Answer» No, Force of friction on body is due to the tendency of block M to slide down over inclined plane. No, as there is no work done so there no dissipation of energy. |
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| 615. |
Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case? |
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Answer» (i) Power is required to increase the velocity due to free fall. (ii) Because limited or specified power can stop the speed of free falling of passenger along with elevator. |
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| 616. |
Ball A of mass m, after sliding from an inclined plane, strikes elastically another ball B of same mass at rest. Find the minimum heihgt h so that ball B just completes the circular motion of the surface at C. (All surfaces are smooth). A. (a) `h=5/2R`B. (b) `h=2R`C. (c) `h=2/5R`D. (d) `h=3R` |
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Answer» Correct Answer - A `mgh=1/2mv^2+2mgR` `gh=v^2/2+2gR` Required velocity at the top, `v=sqrt(gR)` `gh=(gR)/(2)+2gRimpliesh=5/2R` |
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| 617. |
A uniform board of length L is sliding along a smooth (frictionless) horizontal plane as in figure. The board then slides across the boundary with a rough horizontal surface. The coefficient of kinetic friction between the board and the second surface surface is `mu_k` a. Find the acceleration of the board at the moment its front end has traveled a distance x beyond the boundary. b. The board stops at the moment its back end reaches the boundary, as in figure. Find the initail speed `v` of the board. |
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Answer» a. Let `m` be the mass of the whole board. The portion on the rough surface has mass `mx//L`. The normal force supporting it is `mxg//L` and the frictional force is `mu_kmgx//L=ma`. Then `a=mu_kgx//L` opposite to the motion. b. In an incremental bit of forward motion `dx`, the kinetic energy converted into internal energy is `f_xdx=(mu_kmgx)/(L)dx`. The whole energy converted is `1/2mv^2=underset0oversetLint(mu_kmgx)/(L)dx=(mu_kmg)/(L)x^2/2 :|_0^L=(mu_kmgL)/(2)` `v=sqrt(mu_kgL)` |
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| 618. |
The figure shows the variation of potential energy of a particle as a funcation pf x, the x-coordination of the region. It has been assumed that potential energy depends only on `x`. For all other values `x, U` is zero. i.e. for` x lt - 10` and `x gt 15, U = 0`. If total mechanical energy of the particle is `-40J`, then it can be found in region.A. `xlt10` and `xgt15`B. `-10ltxlt5` and `6ltxlt15`C. `10ltxlt15`D. It is not possible |
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Answer» Correct Answer - D `U=E-K=-40-K` Since `Kge0` `:. U le -40J`. |
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| 619. |
A car of mass 1000 kg moving with a speed `18 km h^(-1)` on a smooth road and colliding with a horizontally mounted spring of spring constant `6.25 xx10^3 N m^(-1)`. The maximum compression of the spring isA. 1 mB. 2 mC. 3 mD. 4 m |
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Answer» Correct Answer - B Here, m=1000 kg , v=18 km `h^(-1) = 18xx 5/18 m s^(-1) =5 m s^(-1)` `k=6.25 xx 10^3 N m^(-1)` At maximum compression `x_m`, the kinetic energy of the car is converted entirely into the potential energy of the spring. `therefore 1/2mv^2=1/2kx_m^2 " " or " " x_m=sqrt(m/k)v` Substituting the given values , we get `x_m=sqrt((1000 kg)/(6.25xx10^3 N m^(-1)))xx 5m s ^(-1) =0.4xx5 m=2 m` |
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| 620. |
A particle moves along X-axis from `x=0` to `x=1` m under the influence of a force given by `F=3x^(2)+2x-10.` Work done in the process isA. `+4J`B. `-4J`C. `+8J`D. `-8J` |
| Answer» Correct Answer - D | |
| 621. |
Calculate work done in raising a stone of mass 5kg and specific gravity 3 lying at the bed of a lake through a height of 4 metre. |
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Answer» Here,`m=5kg`, sp. Gravity `=3,h=4m,W=?` As specify gravity of stone `=("mass of stone in air")/("mass of an equal volume of water")` `:. 3=(5kg)/(m) :. m=(5)/(3)kg` Loss of weight of stone when immersed in water = Weight of water displaced `=(5)/(3)kg` Apparent weight of stone in water `=(5-(5)/(3))kg=(10)/(3)kg` Force required to lift the stone, `F=(10)/(3)xx9.8N` Work done `=Fxxh=(10)/(3)xx9.8xx4=130.7J` |
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| 622. |
System shown in figure is released from rest . Pulley and spring is mass less and friction is absent everywhere. The speed of `5 kg` block when `2 kg` block leaves the constant of with ground is (force constant of spring `k = 40 N//m and g = 10 m//s^(2))` A. `sqrt(2)m//s`B. `2sqrt(2)m//s`C. `2m//s`D. `4sqrt(2)m//s` |
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Answer» Correct Answer - B |
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| 623. |
Consider a uniform chain of mass `m` and length `L`, the chain is released when hanging length is `L_0`. Find the speed with which the chain leaves the table. |
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Answer» `K_1=0`,`U_1=-(1)/(2)((m)/(L)L_0)gL_0` `K_2=(1)/(2)mv^2`,`U_2=-(mgL)/(2)` By mechanical energy conservation `K_1+U_1=K_2+U_2` `0-(mgL_0^2)/(2L)=(1)/(2)mv^2-(mgL)/(2)` `v^2=gL-(gL_0^2)/(L)=(g(L^2-L_0^2))/(L)` `v=sqrt((g(L^2-L_0^2))/(L))` |
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| 624. |
A particle moves with a speed v in a circle of radius R.The x-component of the average velocity of the particle in a half- revolution, as shown in figure`-3.76`,is: A. `-(v)/(pi)`B. `-(v)/(2pi)`C. `-(v)/(4pi)`D. `-(2v)/(pi)` |
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Answer» Correct Answer - A |
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| 625. |
A dparticle of mas m moving with speed u collides elastically with a sphere of radius R and same mass at rest, at an impact parameter d. Find (a) Angle between their final velocities and (b) Magpitude of their final velocities. |
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Answer» As the collision is perfectly elastic and masses of both the particles are same thus their velocity vectors should b e exchanged along the line impact. Thus `v_(2)=u cos theta and v_(3)=0` Now id we focus on the particle along there is no force on it parpendicular to the line of impact. Thus its velocity does not change in this directin and hence. `v_(1)=usin theta` Now, from geometry, `sin theta=(d)/(R)` `cos theta=sqrt(1-(d^(2))/(R^(2)))` `impliesv_(2)=u/Rsqrt(R^(2)-d^(2))` `v_(1)=(ud)/(R)` `v_(3)=0` |
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| 626. |
If the kinetic energy of a body is directly proportional to time t, the magnitude of the force acting on the body isA. directly proportional to `sqrt(t)`B. inversely proportional to `sqrt(t)`C. directly proportional to the speed of the bodyD. inversely proportional to the speed of the body |
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Answer» Correct Answer - B::D |
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| 627. |
Select the correct alternative(s).A. Work done by static friction is always zeroB. Work done by kinetic friction can be positive alsoC. Kinetic energy of a system can not be increased without applying any external force on the sytemD. Work energy theorem is valid in non-inertial frames also |
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Answer» Correct Answer - B::D |
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| 628. |
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 positiveB. Work done by all the forces in region AB is zeroC. Work done by all the forces in region BC is negativeD. Work done by all forces in region OA is negative |
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Answer» Correct Answer - B::C |
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| 629. |
In the figure shown, upper block is given a velocity of `6 m//s` and lower block. `3 m//s`. When relative motion between them is stopped A. Work done by friction on upper block is negativeB. Work done by frictoin on both the blocks is negativeC. Work done by friction on lower block is negativeD. Work done by friction on both the blocks is positive |
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Answer» Correct Answer - A |
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| 630. |
Define spring constant of a spring. |
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Answer» It is the restoring force set up in a string per unit extension. |
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| 631. |
Which spring has greater value of spring constant – a hard spring or a delicate spring ? |
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Answer» Hard spring. |
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| 632. |
Two bodies stick together after collision. What type of collision is in between these two bodies ? |
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Answer» Inelastic collision. |
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| 633. |
Which energy is stored in the spring while winding a clock? In which energy it gets converted when the clock is working? |
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Answer» Potential energy; Kinetic energy. |
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| 634. |
A spring of force constant K is cut into two equal pieces. Calculate force constant of each part. |
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Answer» Force constant of each half becomes twice the force constant of the original spring. |
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| 635. |
A man of mass 60 Kg runs up a flight of 30 steps in 40 seconds. If each step is 20 cm high, calculate the power of the man. |
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Answer» Given, Mass m= 60 kg Number of steps = 30 Height of each step = 20 cm Time taken= 40 s Total height h = 20×30 = 600 cm = 6m Total time (t) = 40×30 = 1200 s We know, Power = \(\frac{Work\,done}{Time\,taken}\) Work done = potential energy stored = m × g × h Let g = 10 ms-2 So, work done will be =60×10×6 =3600 J Substituting value or work in above formula of power: Power = \(\frac{3600\,J}{1200\,s}\) Power of man = 3 J/s = 3 W |
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| 636. |
Write true or false for the following statements: When energy changes from one form to another, the energy that disappears from one form, reappears in exactly equivalents amount in the other form. |
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Answer» True Explanation: Yes, this is true according to law of conservation of energy, which says energy can neither be created nor be destroyed. It can only be transformed from one form to another. Whenever energy changes from one form to another, the total amount of energy remains constant. |
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| 637. |
Write true or false for the following statements: Work done by a force depends upon how fast work is done |
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Answer» False Explanation: Power is defined as the rate of doing work so, how fast work is done depends on the power of a person. If a person has more energy then he will have more power to do the work and faster he can do the work. On other hand, if he does not have enough power, he would not be able to do the work faster. |
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| 638. |
One joule work is said to be done when:A. a force of 1 N displaces a body by 1 cm B. a force of 1 N displaces a body by 1 m C. a force of 1 dyne displaces a body by 1 m D. a force of 1dyne displaces a body by 1 m |
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Answer» SI unit of work is joule which is denoted by J. 1 joule = 1 newton×1 meter 1 J = 1 Nm 1 joule of work is said to be done when a force of 1 Newton moves an object through a distance of 1 meter in direction of force. |
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| 639. |
A 1 kg mass has a kinetic energy of 1 joule when its speed is:A. 0.45 m s–1 B. 1 m s–1 C. 1.4 m s–1 D. 4.4 m s–1 |
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Answer» Given, Mass of body m= 1 kg Kinetic energy = 1 J Velocity v=? Kinetic energy = \(\frac{1}{2}\)mv2 Substituting the given values in the equation: 1 J = \(\frac{1}{2}\) ×1× v2 1×2 = v2 v2 = 2 v= √2 v= 1.4 ms-1 |
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| 640. |
The speed of a particle is doubled, its kinetic energy:A. remains same B. becomes double C. becomes four times D. becomes half |
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Answer» Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula: KE = \(\frac{1}{2}\) mv2 so kinetic energy is directly proportional to the mass of object and to the square of speed of the body. If the speed of particle is doubled then kinetic energy becomes four time. KE1 =\(\frac{1}{2}\) mv12 KE2 = \(\frac{1}{2}\) mv22 where, v2 = 2v1 Substituting the value of v2 in the formula, KE2 = \(\frac{1}{2}\) m (2v1)2 = \(\frac{1}{2}\)(4 mv12 ) KE2 = 4 KE1 So kinetic energy will become 4 times if the speed is doubled. |
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| 641. |
Write true or false for the following statements: When a body falls on the ground and stops, the principle of conservation of energy is violated. |
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Answer» False Explanation: Initially that body contains only potential energy. When that body falls on ground then its potential energy is continuously being transferred into kinetic energy and the total energy which is the sum of kinetic and potential energy remains constant. So, the principle of conservation of energy is not violated. |
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| 642. |
An iron sphere of mass 30 kg has the same diameter as an aluminum sphere whose mass is 10.5 kg. The spheres are dropped simultaneously from a cliff. When they are 10 m from the ground, they have the same:A. acceleration B. momentum C. potential energy D. kinetic energy |
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Answer» Given, Mass of iron m = 30 kg Mass of aluminium M = 10.5 kg Height of iron h = 10m Height of aluminum H = 10m We know, Kinetic energy =\(\frac{1}{2}\)mv2 Potential energy = m × g × h Momentum = mass × velocity So kinetic energy, potential energy, kinetic energy and momentum depends of the mass of the body, they can’t be same for both spheres. But acceleration due to gravity g = 10 ms-2 is same for all the objects at some height, irrespective of their mass and height. |
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| 643. |
When the speed of a moving object is doubled, its:A. acceleration is doubled B. weight is doubled C. kinetic energy is doubled D. kinetic energy increases 4 times |
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Answer» Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula: KE1 = \(\frac{1}{2}\)mv12 If the speed of particle is doubled then KE2 = \(\frac{1}{2}\)mv22 where, v2 = 2v1 Substituting the value of v2 in the formula, KE2 = \(\frac{1}{2}\)m(2v1)2 = \(\frac{1}{2}\) (4mv12 ) KE2 = 4 KE1 So kinetic energy will become 4 times if the speed is doubled. On other hand, acceleration depends on final as well as initial velocity, and here acceleration will be same. Weight is independent of change in speed. |
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| 644. |
When do you apply more power-when you lift a pen or when you lift a book? |
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Answer» Power = \(\frac{Energy\,Consumed}{Time\,taken}\) A book has more mass than a pen, so when a book is lifted more energy will be consumed than that when the pen is lifted. Since power is directly proportional to energy so more power will be applied in lifting a book. |
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| 645. |
An object of mass 1 kg has a potential energy of 1 joule relative to the ground when it is at a height of:A. 0.102 m B. 1 m C. 9.8 m D. 32 m |
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Answer» Given, Mass m= 1 kg Potential energy = 1 J Height h=? Let g = 9.8 \(ms^{-2}\) We know Potential energy = m × g × h Substituting the values in this formula 1 J = 1 kg× 9.8 ms-2 × h h = \(\frac{1\,J}{9.8\,kgms^-2}\) h = 0.102 m |
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| 646. |
An object of mass 5 kg falls from a height of 5 m above the ground. The loss of potential energy of the mass is:A. 250 J B. 25 J C. 2.5 kJ D. 50 J |
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Answer» Given, Mass = 5 kg Height = 5 m Let g = 10 ms-2 Potential energy stored in a body of mass m and height ‘h’ will be given by = m × g × h = 5 kg×10 ms-2×5 m = 250 J When the mass falls completely, same potential energy is lost = 250 J |
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| 647. |
Is any work done by a body moving along a circular path? |
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Answer» To keep a body moving in circular path, a force named ‘centripetal force’ acts towards the center of the path. This force acts along the radius and at right angle to the circular motion of body. So, work done by body in going around the circular path is zero because displacement of body is perpendicular to the direction of centripetal force and by the formula: Work= FS cos θ where θ=90° And cos 90°= 0 So, work done = zero |
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| 648. |
Write true or false for the following statements: Kilowatt-hour is the unit of power. |
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Answer» False Explanation: kilowatt hour is the commercial unit of energy. 1 kilowatt hour is the amount of electrical energy consumed when the appliance have power rating of 1 kilowatt which is used for 1 hour. SI Unit of power is watt and its larger unit is kilowatt. |
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| 649. |
A rowing boat upstream is at rest with respect to the shore. Is he doing work? |
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Answer» Yes, the boat is doing work against the river current but there is no work done by boat against shore because there is no displacement taking place with respect to shore. |
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| 650. |
Differentiate between Work and energy. |
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