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Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 1
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Q.1
If momentum is increased by 20% then kinetic energy increased by..[ AFMC 1997]
0%
a)77%
0%
b) 66%
0%
c)44%
0%
d)55%
Explanation
Formula for kinetic energy in terms of momentum is momentum is increased by 20% thus new momentum p'=1.2pSubstituting of p' in above equation we getThus increase in KE=44%Answer: (c)
Q.2
A bread gives a boy of mass 40kg an energy of 21kJ. If the efficiency is 28% then the height can be climbed by him using this energy is [AFMC 1997]
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a) 22.5 m
0%
b) 15 m
0%
c)10 m
0%
d)5 m
Explanation
28% of energy of bread is available to climb∴ 28% of 21 kJ=5.88kJBy climbing energy form bread is converted to potential energy of boy=mgh∴ 5.88kJ=mghon substituting m=40 kg, g=10 m/s25.88 × 1000=(40)(10) hh=14.7 mAnswer: (b)
Q.3
If a body of mass 200g falls from a height 200m and its total potential energy is conserved into kinetic energy, at the point of contact of the body with the surface, then decrease in potential energy of the body at the contact is .. [ AFMC 1997]
0%
a) 900 J
0%
b) 600 J
0%
c)400 J
0%
d)200 J
Explanation
We consider at the surface potential energy is zero∴ Potential energy lost=Final potential energy - initial potential energyPotential energy lost=0 - mghon substituting values in above equation we get Potential energy lost=(0.2)(200)10)=400J Answer:(c)
Q.4
A body of mass 10 kg and velocity 10m/s collides with a stationary body of mass 5kg. After collision both bodies stick to each other, velocity of the bodies after collision will be. [AFMC 1997]
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a) (3/10) m/s
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b) (18/3) m/s
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c) (9/20) m/s
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d) (20/3) m/s
Explanation
No external force acts on the system of particles so according to law of conservation of momentum momentum before collision=momentum after collision Let m1=10 kg Initial velocity of m1 be u1=10m/sLet m1=5 kgInitial velocity of m2=0 Since both the masses stick together and have common final velocity v ∴ m1 × u1=v( m1 + m2) by substituting the values in above equation we get 10 × 10=v(10 + 5) on solving v=20/3Answer: (d)
Q.5
If a body of mass 3 kg is dropped from the top of a tower of height 25m. Then its kinetic energy after 3 sec is ...[ AFMC 1998]
0%
a)557 J
0%
b) 735 J
0%
c)1050 J
0%
d)1296 J
Explanation
Body is dropped from the top of tower it means it is free fall, initial velocity is zero,now final velocity after 3 sec can be calculated by following formulaDistance travelled by the object in 3 sec can be calculated by using formula s=ut + ½at2on substituting and solving the equation we get s=44.1 mbut our tower is of height 25m there fore body hit the surface before end of three secondTo calculate the final velocity we will use formula v2=u2 + 2gssubstituting the values we getv2=0 + 2 × 9.8 × 25v2=490 m/sFormula for kinetic energy=½ m v2Substituting values in above equation we get K. E.=735 J Answer: (b)
Q.6
A body of mass 100 gm is rotating in a circular path of radius r with constant velocity. The work done in one complete revolution is [ AFMC 1998]
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a) (r/100) J
0%
b) (100/r) J
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c)zero
0%
d)100 rJ
Explanation
For a circular motion force and displacement are mutually perpendicular. Thus work done is zeroAnswer: (c)
Q.7
The work done in pulling up a block of wood weighing 2kN for a length of 10m on a smooth plane inclined at an angle of 15° with the horizontal is .. [ AFMC 1999]
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a) 9.82 kJ
0%
b) 8.91 kJ
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c)5.17 kJ
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d)4.36 kJ
Explanation
Here friction is not considered. Gravitational field is conservative field hence work done is independent of path. Object is moved to the height h=s sin θ , s is the length of inclined plane Thus work done=change in potential energyWork=mghWork=mg (s sin θ)on substituting the values in above equationWork=2 × 1000 × sin15° × 10=5.17 kJ Answer:(c)
Q.8
A 30 gm bullet initially traveling at 120 m/s penetrates 12 cm into wooden block. The average resistance exerted by the wooden block is ..... [AFMC 1999]
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a) 1800 N
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b) 2000 N
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c) 2200 N
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d) 2850 N
Explanation
Bullet stopped after penetrating the block hence work is down by the block Work done=Change in kinetic energy Work done=Final Kinetic energy - Initial kinetic energy Work done=0 - ½(m)(v 2) substituting the values in above equation we get Work done=0 - ½(0.03)(120)2 Work done=216 J Now if F is the average resistance force then work done by resistive force=F × (s) Here s=12cm=0.12m F(0.12)=216 F=216/ (0.12)=1800N Answer: (a)
Q.9
The mass of two substances are 25g and 81g respectively. If their kinetic energies are same then the ratio of their momentum is ... [ AFMC 2000]
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a)9:5
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b) 1:4
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c)5:9
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d)2:1
Explanation
Form the equation if kinetic energy is constant then p ∝(m)1/2Answer: (c)
Q.10
Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with the same force. They will have the potential energies in the ratio of ... [ AFMC 2000]
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a) 1 : 2
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b) 1 : 4
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c)4 : 1
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d)2 : 1
Explanation
Formula for potential energy of spring is U=½ k x2here is spring constant an x is extensionAlso F=kx or x=F/ksubstituting value of x in above equation we getU=½ k ( F/k)2Force is constant thus potential energy ∝ (1/k)Let k1=1500 Nk2=3000NNow for first spring U1 ∝( 1/ k1) for second spring U1 ∝( 1/ k2) taking the ratio of above two we get U1:U2=(k2)/ (k1)U1:U2=2 : 1Answer: (d)
Q.11
A object of mass 40kg having velocity 4 m/s collides with another object m=60 kg having velocity 2m/s. The collision is perfectly inelastic. The loss in energy is ....[ AFMC 2001]
0%
a) 110 J
0%
b) 48 J
0%
c)392 J
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d)440 J
Explanation
Since the collision is perfectly inelastic both object will stick together have common velocity No external force is acting on the system according to law of conservation of momentumInitial momentum=Final momentumLet m1=40 kgInitial velocity of m1 be u1=4 am/sLet m2=60 kgInitial velocity of m2 be u2=2m/s Let common velocity after collision be=vthus m1 × u1 + m2 × u2=(m1 + m2) vOn substituting the values we get(40)(4) + (60) (2)=(40 + 60) v on simplification we get v=2.8 m/s formula for kinetic energy=½m v2 Final kinetic energy :- m=(40 + 60)=100 kg v=2.8 m/s Final kinetic energy=½(100) (2.8)2 Final kinetic energy=392 J Initial kinetic energy before collision:- Initial kinetic energy=½m1u12 + ½m2u22 Initial Kinetic Energy=½(40)(4)2 + ½(60)(2)2 Initial Kinetic Energy=440 J Loss in kinetic energy=Initial kinetic energy - Final kinetic energy Loss in Kinetic Energy=440 - 392=48 J Answer:(b)
Q.12
A block of mass m1 rest on horizontal table. A string tied to this block is passed over a frictional pulley fixed at one end of the table and another block of mass m2 is hung to the other end of string. The acceleration (a) of the system is .. [AFMC 2001]
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a)
0%
b)
0%
c)
0%
d) g
Explanation
both the masses form a system hence will have common acceleration let it be 'a' and mass m2 moves downas shown in figureHere we are considering frictional force is not actingMass m1 is pulled by tensionMass m2 is pulled down by gravityMass m2 is pulled up by tension∴ For mass m1T=m1(a) --eq(1) For m2m2=mg - T --eq(2)eliminating T by adding eq(1) and eq(2) we get Answer: (b)
Q.13
A force (3i + 4j) newton acts on a body and displaced it by (3i + 4j) meter. The work done by this force is ...[ AIIMS 2001]
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a)5 J
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b) 25 J
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c)10 J
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d)30 J
Explanation
Work done=F. SWork done=(3i + 4j) . (3i + 4j)Work done=9 + 16=25 JouleAnswer: (b)
Q.14
If a water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbines is ( take g=9.8 m/s2) [ AIIMS 2007]
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a) 9.8 m/s
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b) 19.6 m/s
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c)39.2 m/s
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d)98.0 m/s
Explanation
According to law of conservation of energy Potential energy=Kinetic energymgh=½(mv2)∴ v=(2gh)1/2v=19.6 m/sAnswer: (b)
Q.15
When a ball is thrown up vertically with velocity vo it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity.. [ AIIMS 2005]
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a)vo√3
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b) 3vo
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c)9vo
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d)(3/2)vo
Explanation
According to Law of conservation of Energy½(m vo2 )=mghThus vo2=2gh --(1)To triple the height let velocity be v' v'2=2g(3h) --(2) Take the ratio of eq(2) and eq(1) we get Answer:(a)
Q.16
A body of mass 5kg has momentum of 10kg m/s. When a force of 0.2N is applied on it for 10s, the change in its kinetic energy is .. [ AIIMS 2000]
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a) 4.4 J
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b) 3.3 J
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c)5.5 J
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d)1.1 J
Explanation
Initial Velocity : momentum p=mv v=p/m=10/5=2 m/s Initial Kinetic energy : Ei=½(mv2) Ei=½ ( 5 × 22)=10 J Change in momentum on application of force: Δp=F.t=0.2 × 10=2 kg m/sFinal momentum p'=Initial momentum + change in momentumFinal momentum p'=10 + 2=12 kg m/sFinal velocity:v'=p'/ m=12/5 m/sFinal kinetic energy Ef=½( mv'2)Change in kinetic energy=Final K.E. - Initial=14.4 - 10=4.4 J Answer:(a)
Q.17
Two bodies of masses 'm' and '4m' are moving with equal kinetic energy. Then the ratio of their linear momentum will be... [ AIIMS 1999]
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a) 1:1
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b) 2:1
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c) 4:1
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d) 1:2
Explanation
relation between Kinetic energy and momentum is given by equation Let P1 be the momentum of first object P2 be the momentum of first object now both have same kinetic energy thereforeAnswer: (d)
Q.18
A spring 40 mm long is stretched by applying a force. If 10N force is required to stretch the spring through one mm, then work done in stretching the spring through 40mm is [ AIIMS 1998]
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a)24 J
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b) 8 J
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c)56 J
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d)54 J
Explanation
1 mm=0.001 m; 40 mm=4 × 10-2mForce constant k=F / x=10 /0.001=104 N /m Work done=Potential energy of springWork done=½(k x2)Work done=½ ( 104 × 16 × 10-4)Work done=8 JouleAnswer: (b)
Q.19
A ball loses 15.0% of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of 12.4 m to have it bounce back to the same height ( ignore resistance)? [ AIIMS 2010]
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a) 6.55 m/s
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b) 12.0 m/s
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c)8.6 m /s
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d)4.55 m/s
Explanation
It is given in the problem 15% of K.E is lost during collision with concrete, which is equalto loss of initial potential energy + Potential energy Now we want to ball to come back to the same height hence Initial and final potential energy is same and final kinetic energy zero According to law of conservation of energy Initial K.E. + Initial Potential energy = Final K.E. + Final Potential Energy+ 15% loss in Initial K.E As Ball reached same height Initial Potential energy = Final Potential energy ∴ Initial k.E. = Final KE + 15% loss in Initial K.E Now Final K.E. = 0 as ball reached its maximum height ∴Initial k.E. = 15% loss in Initial K.E ...(1) Now while ball bounce back, it losses 15%of energy thus could not reached initial height ∴ 15 % of loss in initial K.E = (15% loss in initial KE + 15% loss initial PE) (As energy is lost when it bounce back) ...(2) From (1) and (2) we get ∴Initial k.E. = (0.15 initial k.E. energy + 0.15 initial PE) 0.85 Initial k.E. = 0.15 initial PE ∴Initial K.E. = (15/85) initial PE. ½ ( m vi2)=(15/85) mghvi 2=(15/85) gh vi=[2 (15/85) g h]1/2 vi=[2 (15/85) (9.8)( 12.4)]1/2 vi=6.55 m/s Answer: (a)
Q.20
The potential energy of a certain particle is given by The force on it is ... [ AIIMS 2011]
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a) -xi + zk
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b) xi + z k
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c)½ ( xi + zk )
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d)½ ( xi - zk)
Explanation
We know that for conservative force F(r)=-dU/drForce along x-axis is given bySimilarly force along z-axis is ∴ F=-xi + zk Answer:(a)
Q.21
A mass of 1kg is hanging from a spring of spring constant 1N/m. If Saroj pulls the mass down by 2m. The work done by Saroj is .. [ AIIMS 2009]
0%
a) 1 J
0%
b) 2 J
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c) 3 J
0%
d) 4 J
Explanation
since restoring spring is position dependant hence we should integrate small work done dw Answer: (b)
Q.22
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work done against the gravitational force between them to take the particle far away from the sphere ( G=6.67 × 10-11 Nm2 / kg2 ) [ AIIMS 2008]
0%
a)3.33 × 10-10 J
0%
b) 13.34 × 10-10 J
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c)6.67 × 10-10 J
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d)6.67 × 10-9 J
Explanation
W=GMm /R Answer: (c)
Q.23
Which of the following is true? [ AIIMS 2000]
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a) momentum is conserved in all collisions but kinetic energy is conserved in elastic collisions.
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b) momentum is conserved in all collisions but not kinetic energy
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c)both momentum and kinetic energy are conserved in all collisions.
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d)neither momentum nor kinetic energy is conserved in elastic collisions.
Explanation
Answer: (a)
Q.24
A ball of mass 10 kg is moving with a velocity of 10 m/s. It strikes another ball of mass 5 kg, which is moving in the same direction with a velocity of 4m/s. If the collision is elastic their velocities after collision will be respectively.. [ AIIMS 2000]
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a) 12m/s, 6m/s
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b) 12 m/s, 25 m/s
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c)6m/s, 12m/s
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d)8m/s, 20m/s
Explanation
Let v1 and v2 are the final velocities after collision Relative velocity of approach=relative velocity of separation10 - 4=v2 - v16=v2 - v1v1=v2 - 6 --eq(1) Applying conservation of momentum 10 × 10 + 5 × 4=10v2 + 5v1substituting value of v1 from equation (1) in above equation we get120=10(v2 - 6 ) + 5 v2 15v2=180 v2=12 cm/secand v1=6 cm/sec Answer:(c)
Q.25
Two bodies of masses 0.1kg and 0.4kg move towards each other with velocities 1 m/s and 0.1m/s respectively. After collision they stick together. In 10sec the combined mass travels.. [ AIIMS 2010]
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a) 120 m
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b) 0.12 m
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c) 12 m
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d) 1.2 m
Explanation
let combine velocity be v now according to law of conservation of momentum 0.1 × 1 + 0.4 × (-0.1)=v(0.1 + 0.4) here velocity of second object is taken negative as both the objects are approaching each other on solving above equation we get common velocity v=0.12 m/s Now distance traveled in 10 sec=velocity × time distance=0.12 × 10=1.2m Answer: (d)
Q.26
A shell of mass 'm' moving with velocity v suddenly break into two pieces the part having mass m/3 remains stationary. The velocity of other part will be..[ AIIMS 2009]
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a)(2/3) v
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b) (7/5) v
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c)(3/2)v
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d)none of these
Explanation
According to law of conservation of momentummv=(m/3) × 0 + (2m/3)v'mv=2mv' / 3 v=(3/2) v'Answer: (c)
Q.27
If the force applied is F and the velocity gained is 'v', then the power developed is... [ AIIMS 1998]
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a) v/F
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b) Fv
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c)Fv2
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d)F / v
Explanation
Power=Work / time=(F.S) /t=F .(S/t)=F.vAnswer: (b)
Q.28
A particle of mass 'm' moving with velocity 'v' collides with stationary particle of mass 2m. Then the speed of the system after collision is .. [ AIIMS 1999]
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a) 2v
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b) v/2
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c)3v
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d)v/3
Explanation
Applying law of conservation of momentummv + 0=(2m + m)v'=3mv'v'=mv/ 3m=v/3 Answer:(d)
Q.29
A bullet is fired from a riffle. If the riffle recoils freely, then the kinetic energy of the rifle will be..[ AIIMS 1998]
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a) same as that of bullet
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b) more than that of bullet
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c) less than that of bullet
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d) none of these
Explanation
For recoil of riffle, momentum will be conserved.. MV=mv V /v=m/M by substituting Value of V/v in the above equation we get as m < M, kinetic energy of riffle < kinetic energy of bullet Answer: (c)
Q.30
A metal ball of mass 2 kg moving with speed of 36km/h is having a collision with a stationary ball of mass 3kg. If after collision, both the balls move together, the loss in kinetic energy due to collision is .. [ AIIMS 2001]
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a)80 J
0%
b) 40 J
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c)60 J
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d)160 J
Explanation
Let 'v' be the common velocity36km/h=10 m/secApplying conservation of momentum2 × 10 + 3 × 0=(2 + 3)v v=4 m/secInitial energy=½ × 2 × 102 + 0=100 JFinal Energy=½ × 5 × 42=40 J Loss of energy=100 - 40=60 JAnswer: (c)
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