MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
When a ling spring is stretched by 2cm its potential energy is U. If the spring is stretched by 10cm, the potential energy stored in it will be [ CBSE-PMT 2003]
0%
a) 25U
0%
b) U/5
0%
c) 5U
0%
d) 10U
Explanation
Potential energy of spring U=½ k × x2 Thus for 2cm stretching U=2k For 10 cm stretching U'=50 k Thus U'=25U Answer: (a)
Q.2
A stationary particle explodes into two particles of mass m1 and m2 which moves with velocities m1 and vThe ratio of there kinetic energies E1 / E2 is [ CBSE-PMT 2003]
0%
a)m1v2 /m2v1
0%
b) m1 / m2
0%
c) m2 / m1
0%
d)1:1
Explanation
Applying law of conservation of momentumm1v1 + m2v2=0 m1v1=-m2v2 Thus magnitude of momentum both particle is same Answer: (c)
Q.3
A vertical spring with force constant k is fixed on a table. A ball of mass m at height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance 'd'. The net work done in the process is [ CBSE-PMT 2007]
0%
a) mg(h + d) - ½ k d2
0%
b) mg(h - d) - ½ k d2
0%
c)mg(h - d) + ½ k d2
0%
d)mg(h + d) + ½ k d2
Explanation
Object falls on the spring form height and compressed by distance d Thus total height h'=h + dPotential energy of object=mgh'U=mg(h + d) After impact object velocity comes to zero and its energy is stored as elastic potential energy Elastic potential energy U'=½ k d2 According to law of conservation of energy U=U'mg(h + d)=½ k d2 Thus net work=mg(h + d) - ½ k d2=0 Answer: (a)
Q.4
Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of energy. How much power is generated by the turbine? (g=10 m/s2)
0%
a)8.1 kW
0%
b) 10.2 kW
0%
c)12.3kW
0%
d)7.0kW
Explanation
Given h=60 m, g=10 ms2Rate of flow of water=15 kg/s∴ Power of the falling water=Change in potential energy per secondPower of the falling water=Rate of flow of water × g × h Rate of flow of water=15 × 10 × 60=9000 watt Energy loss 10%=900 watt ∴ Power generated by the turbine=9000 - 900=8100 watt=8.1 kW Answer:(a)
Q.5
A shell of mass 200gm is ejected from a gun of mass 4kg by an explosion that generates 1.05kJ of energy. The initial velocity of the shell is .. [ CBSE-PMT 2008]
0%
a) 100 m/s
0%
b) 80 m/s
0%
c) 40 m/s
0%
d) 120 m/s
Explanation
Due to explosion gun and shell will have kinetic energy According to law of conservation of momentum mv=Mv' here m=mass of shell, M=mass of gun, v=velocity of shell, v'=velocity of gun Thus v'=mv/M Energy of explosion=energy of shell + energy of gun Answer: (a)
Q.6
A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attending a height of 18m . How much energy is lot due to air friction( g=10m/s2 [CBSE-PMT 2009]
0%
a)30J
0%
b) 40J
0%
c)10J
0%
d)20J
Explanation
Kinetic energy=potential + Loss in energy ½ m v2=mgh + loss in energy ½ × 1 × (20)2=1 × 10 × 18 + Loss of energy200=180 + loss of energy Loss of energy=20 JAnswer: (d)
Q.7
A ball moving with velocity 2m/s collides head on with another stationary ball of double the mass. If coefficient of restitution is 0.5, then their velocities after collision will be [ CBSE-PMT 2010]
0%
c)1, 0.5
0%
a) 0, 1
0%
b) 1, 1
0%
d)0, 2
Explanation
Formula for restitutionHere u1=2 m/s , u2=0m1=m and m2=2mAccording to formula for v1 on substituting valuesAccording to formula for v2 on substituting valuesAnswer: (a)
Q.8
A body projected vertically from the earth reaches a height equal to radius of earth before returning to earth. The power exerted by gravitational force is greatest [ CBSE-PMT 2011]
0%
a)at the highest position of the body
0%
b) at the instant just before the body hits the earth
0%
c)it remains constant all through
0%
d)at the instant just after the body is projected
Explanation
Power exerted by the force is given by P=F × v When body is just above the earth's surface, its velocity is greatest. At this instant, gravitational force is also maximum. Hence, the power by gravitational force is greatest at the instant just before the body hits the earth Answer:(b)
Q.9
A mass m moving horizontally ( along x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along y-axis) with velocity 2v. The final velocity of the combination is .. [ CBSE-PMT 2011]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
As the two masses stick together collision is inelastic collision thus only momentum is conserved mvi + 3m(2v)j=(4m)v Thus v=Answer: (a)
Q.10
A particle of mass m is moving in circular path of constant radius 'r' such that its centripetal acceleration ac is varying with time t as ac=k2rt2 where k is a constant. The power delivered to the particles by the force acting on it is [ IIT 1994]
0%
a)2π mk2r2t
0%
b) mk2r2t
0%
c)(mk4r2t5 )/3
0%
d)zero
Explanation
The centripetal acceleration ac=k2rt2 v2 / r=k2rt2 ∴ v=krt So, tangential acceleration a=dv/dt=krWork done by tangential forcePower=Fvcos0 Power=(ma)(krt) Power=mk2r2tAnswer: (b)
Q.11
A wind-power generator converts wind energy into electrical energy. Assume that the generator converts fixed fraction of the wind energy. For wind speed v, the electrical power output will be proportional to. [ IIT 2005]
0%
a) v
0%
b) v2
0%
c)v3
0%
d)v4
Explanation
Power=Force × velocity Force due to air dm/dt = V ρ dm/dt = Av ρ F = Av2 ρ Here A is the cross section of blade and v is the velocity of wind Power= (Av2 ρ)×v Power ∝ v3Answer: (c)
Q.12
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unscratched. The maximum extension in the spring is [ IIT 2002]
0%
a) 4Mg/k
0%
b) 2Mg/k
0%
c)Mg/k
0%
d)Mg/2k
Explanation
Let the maximum extension of the spring be x. Decrease in gravitational potential energy=gain in elastic potential energy Thus Mgx=½ kx2 x=2Mg / k Answer:(b)
Q.13
A block (B) is attached to two unscratched springs S1 and S2 with spring constant k and 4k, respectively as shown in figure. The other ends are attached to identical supports M1 and M2 not attached to walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards all 1 by a small distance x and released. The block returns and move a maximum distance y towards wallDisplacement x and y are measured with respect to the equilibrium position of the block b. The ratio y/x is .. [ IIT 2008]
0%
a) 4
0%
b) 2
0%
c) 1/2
0%
d) 1/4
Explanation
When the block B is displaced towards wall 1, only spring S1 is compressed and S2 is in natural state. This happens because the other end of S2 is not attached to the wall but is free. Therefore the energy stored in the system=½ k1x2. When the block is released, it will come back to the equilibrium position, gain momentum, overshoot to equilibrium position and move towards wall 2. As this happens, the springs S1 comes to its natural length and S2 gets compressed. As there are no frictional force involved, the potential Energy of spring S1 gets stored as the Potential energy of spring S2 by y ∴ ½ k1x 2=½ k2y2∴ ½ k x 2=½ 4ky2∴ y /x=1/2 Answer: (c)
Q.14
A piece of wire is bent in the shape of a parabola y=kx2 ( y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead where the bead can stay at rest with respect to the wire from y-axis is [ IIT 2009]
0%
a)a/ gk
0%
b) a/ 2gk
0%
c)2a/ gk
0%
d)a/ 4gk
Explanation
The force acting on the bead as shown in figure here ma is pseudo force Let θ be the angle which the tangent at P makes with the X-axis. As the bead is in equilibrium with respect to the wire, thereforeNsinθ=ma Ncosθ=mg ∴ tanθ=a/g --(1) But y=kx2, therefore tanθ=2kx=tanθ --(2) from above two equations 2kx=a/g x=a/2kg Answer: (b)
Q.15
If a machine is lubricated with oil [ IIT 1980]
0%
a) the mechanical advantage of the machine increases
0%
b) the mechanical efficiency of the machine increases
0%
c)both its mechanical advantage and efficiency increases
0%
d)its efficiency increases, but its mechanical advantage decreases
Explanation
Mechanical efficiency=Output work / Input energy The output work will increase because the friction becomes less. Thus the mechanical efficiency increasesAnswer: (b)
Q.16
Consider the following two statements [ AIEEE 2003] A. Linear momentum of a system of particle is zero B. Kinetic energy of a system of particle is zero then
0%
a) A does not imply B and B does not imply A
0%
b) A implies B but B does not imply A
0%
c)A does not imply B but B implies A
0%
d)A implies B and B implies A
Explanation
Kinetic energy of system particle is zero only when the speed of each particle is zero. And if the speed of each particle is zero, the linear momentum of the system of particle has to be zeroAlso the linear momentum of the system may be zero even when the particle are moving. This is because linear momentum is a vector quantity. In this case the kinetic energy of the system of particles will not be zero. Therefore A does not imply B but B implies A Answer:(c)
Q.17
A wire suspended vertically from one of its ends is stretched by attaching a weight of 200N to the lower end. The weight stretched the wire by 1mm. Then the elastic energy stored in the wire is [ AIEEE 2003]
0%
a) 0.2 J
0%
b) 10 J
0%
c) 20 J
0%
d) 0.1 J
Explanation
The elastic potential energy=½ × Force × extension Elastic potential energy=½ × 200 × 0.001=0.1J Answer: (d)
Q.18
A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unscratched position. Then the work required to stretch it further by another 5 cm is [ AIEEE 2003]
0%
a)12.5 N-m
0%
b) 18.75 N-m
0%
c)25.00 N-m
0%
d)6.25 N-m
Explanation
k=5×103N/m According to work energy theorem work=change in potential energy W=½ k ( x22 - x12 ) W=½ × 5 × 103 [ (0.1)2 - (0.05)2] W=18.75 N-mAnswer: (b)
Q.19
A body moves along a straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to [ AIEEE 2003]
0%
a) t 3/4
0%
b) t 3/2
0%
c)t1/4
0%
d)t1/2
Explanation
Power P=Fv Dimensional formula of Power=ML2T-3=given P is constant and m is constant Thus L2 ∝ T 3 L ∝ T3/2 Thus displacement x ∝ t3/2 Answer: (b)
Q.20
A particle moves in a straight line with retardation proportional to its displacement, Its loss of kinetic energy for any displacement x is proportional to [ AIEEE 2004]
0%
a) x
0%
b) ex
0%
c)x2
0%
d)log x
Explanation
Given : retardation ∝ displacement Kinetic energy dimension formula=ML2T-2 Now M and T is constant thus change in kinetic energy ∝ L2 Change in kinetic energy ∝ x2 Answer:(c)
Q.21
A uniform chain of length 2m is kept on a table such that a length of 60cm hangs freely from the edge of the table. The total mass of the chain is 4kg. What is the work done in pulling the entire chain on the table? [ AIEEE 2004]
0%
a) 12 J
0%
b) 3.6J
0%
c) 7.2 J
0%
d) 1200 J
Explanation
Mass per unit length of chain=λ=4/2=2 kg/meter mass of hanging chain m'=λ (0.6)=1.2 kg Let Potential energy of chain at the surface=0 Centre of mass of hanging chain is=0.3m below the table Potential energy of hanging chain U=-mgx=-1.2 × 10 × 0.3 J U=-3.6 J Thus ΔU=3.6J=work done in putting entire chain on the table Answer: (b)
Q.22
A body of mass 'm' accelerated uniformly from rest to 'v1' in time tThe instantaneous power delivered to the body as a function of time 't' is [ AIEEE 2004]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
v1=u + at1 u=0 thus a=v1/t1∴ v=at=(v1/t1)tpower P=Fv=mav Answer: (b)
Q.23
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles take place in the plane. It follows that [ AIEEE 204]
0%
a) its kinetic energy is constant
0%
b) its acceleration is constant
0%
c)its velocity is constant
0%
d)it moves in a straight line
Explanation
Work done by such force is always zero thus kinetic energy remains constantAnswer: (a)
Q.24
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collision is [ AIEEE 2005]
0%
a) kL2 / 2M
0%
b) L√(Mk)
0%
c)ML2 / k
0%
d)zero
Explanation
Block stops after collision thus Kinetic energy converted in spring potential energy Thus ½ M v2=½ kL2 Thus v=L√( k/M)Momentum=M × v Momentum=M × L√( k/M)Momentum=L√(kM) Answer:(b)
Q.25
A spherical ball of mass 20kg is stationary at the top of a hill of height 100m. It rolls down a smooth surface to the ground, then climb up another hill of height 30m and finally rolls down to a horizontal base at height of 20m above the ground. The velocity attended by the ball is [ AIEEE 2005]
0%
a) 20 m/s
0%
b) 40 m/s
0%
c) 10√30 m/s
0%
d) 10 m/s
Explanation
When ball is at height of 100 meters it has only potential and when ball is at 20 meters it will have potential and kinetic energy Thus mgh=½ m v2 + mgh' gh=½v2 + gh' 10 × 100=½ v2 + 10 × 20 v=40 m/s Answer: (b)
Q.26
A body of mass m is accelerated uniformly from rest to a speed v in time T. The instantaneous power delivered to the body as a function of time is given by [ AIEEE 2005]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
u=0, v=u + aT v=aT Power=Fv Power=ma × at=ma2t Power=mv2t / T2Answer: (b)
Q.27
A particle of mass 100g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is [ AIEEE 2006]
0%
a) -0.5 J
0%
b) -1.25 J
0%
c)1.25 J
0%
d)0.5 J
Explanation
According to law of conservation of energy Change in K.E=-Change in P.E K.E=½ mv2 KE=½ × 0.1 × 52 K.E=1.25 J Thus Potential energy=-1.25 J Work done=Potential energy W=-1.25 JAnswer: (b)
Q.28
The potential energy of a 1kg particle free to move along the x-axis is given by The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s ) is [ AIEEE 2006]
0%
a)3/ √2
0%
b) √2
0%
c)1/√2
0%
d)2
Explanation
When Potential energy is minimum Kinetic energy is maximum In given equation potential energy is function of position thus for minimum energy dV/dx=0 x3 - x=0∴ x=±1 on substituting x=1 in given equation for potential energy Minimum potential energy=1/4 - 1/ 2=(-1/4)J Given K.E + P.E=2 K.E=2 + (1/4)=9/4 Now K.E=½ m v2 9/4=½ × 1 × v2 v max=3/ √2 Answer:(a)
Q.29
A 2kg block slides on a horizontal floor with a speed of 4m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction is 15N and spring constant is 10,000 N/m. The spring compresses by [ AIEEE 2007]
0%
a) 8.5 cm
0%
b) 5.5 cm
0%
c) 2.5 cm
0%
d) 11.0 cm
Explanation
Let block compress spring by x before stopping Kinetic energy of block=P.E of compressed spring + work done against friction ½ × m v2=½ × k × x2 + f × x ½ × 2 × 42=½ × 10000 × x2 + 15x 10000x2 + 15x - 16=0 ∴ x=0.055 m=5.5 cm Answer: (b)
Q.30
An athlete in the olympic games covers a distance of 100m in 10s. His kinetic energy can be estimated to be in the range [ AIEEE 2008]
0%
a)200J - 500J
0%
b) 2×105J - 3×105 J
0%
c)20,000J - 50,000J
0%
d)2,000J - 5,000J
Explanation
Speed of athlete v=100 / 10=10 m/s Kinetic energy=½ m v2 If athlete is 40 kg then:Kinetic energy=½ 40 × 102=2000J If athlete is 100 kg the : Kinetic energy ½ 100 × 102=5000JAnswer: (d)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page