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Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 6
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Q.1
A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two sphere is [ CET 1994]
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a) 90°
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b) 0°
0%
c) 45°
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d) different from 90°
Explanation
collison is inelastic thus spheres will stick together Answer: (b)
Q.2
A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest. After collision, their final velocities are V and v, respectively. The value of v is [ MPPMT 1995]
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a)
0%
b)
0%
c)
0%
d)
Explanation
Mu=MV + mv --eq(1) given collision is elastic e=1 putting in eq(1) Mu=M( v - u) + mv Answer: (c)
Q.3
The average power required to lift a 100kg mass through a height of 50 metres in approximately 50 seconds would be ( in J/s)
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a) 50
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b) 5000
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c)100
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d)980
Explanation
power=work/times=(100 × 9.8 × 50)/50=980Answer: (d)
Q.4
A single conservative force F(x) acts on a 2.5 kg particle that moves along the x-axis. The potential energy U(x) is given by U(x)=(10 + (x - 4)2) where x is in meter. At x=6.0m the particle has kinetic energy of 20J. what is the mechanical energy of the system ?
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a)34 J
0%
b)45 J
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c) 48 J
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d)49 J
Explanation
P.E. at X=6m is U=10 + (6 - 4)2=14 JMechanical energy=K.E. + P.E.=20 + 14=34 J Answer:(a)
Q.5
How much is 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 30° with the horizontal ?
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a) 1.732 KJ
0%
b) 17.32 KJ
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c) 10 KJ
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d) 100 KJ
Explanation
W=2sin30 × 10 = 10KJ Answer: (c)
Q.6
A force of 7N, making an angle θ with the horizontal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains K.E. of 2J, what is the horizontal component of the force ?
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a)2N
0%
b) 4 N
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c)1 N
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d)14 N
Explanation
Gain in K.E=Work done 2=Fx × 0.5 Fx=4 NAnswer: (b)
Q.7
A ball of mass 5 kg is sliding on a plane with initial velocity of 10 m/s. If coefficient of friction between surface and ball is 1/2 , then before stopping it will describe...... (g=10 m/s2)
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a)12.5 m
0%
b)5 m
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c)7.5 m
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d)10 m
Explanation
Frictional force f=µN=(1/2) (5 × 10)=25 N and Change in Kinetic energy=work ½ (5) (10)2=25 × d d=10 m Answer: (d)
Q.8
The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing abody from x=0 cm to x=5 cm
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a)30 ergs
0%
b) 70 ergs
0%
c)20 ergs
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d)60 ergs
Explanation
Area under the graph gives work done Answer:(a)
Q.9
A ball is released from the top of a tower. what is the ratio of work done by forceof gravity in first, second and third second of the motion of the ball ?
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a)1 : 2 : 3
0%
b) 1 : 4 : 9
0%
c) 1 : 3 : 5
0%
d)1: 5 : 3
Explanation
Force is constant for all cases Thus work ∝ dispalcement From the formula for displacement in nth second we get dispalcement in first second=g/2 Displacement in second second=3g/2 Displacement in third second=5g/2Thus work done ratio is 1:3:5Answer: (c)
Q.10
A spring of spring constant 103 N/m is stretched initially 4cm from the unstretched position. How much the work required to stretched it further by another 5 cm ?
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a) 6.5 NM
0%
b) 2.5 NM
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c)3.25 NM
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d)6.75 NM
Explanation
Work done=change in Potential energy of spring Answer: (c)
Q.11
A 8 kg mass moves along x - axis. Its accelerations as a function of its position is shown in the figure. What is the total work done on the mass by theforce as the mass moves from x=0 tox=6cm ?
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a) 48 × 10-3 J
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b) 98 × 10-3 J
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c)4.8 × 10-3 J
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d)9.8 × 10-3
Explanation
E=F ×d W=mad ad=Area under the curveW=m ( Area under the curve) Answer:(a)
Q.12
The work done by a force acting on a body is as shown in the graph. what is the total work done in covering an initial distance of 15m ?
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a) 50 J
0%
b) 75 J
0%
c) 100 J
0%
d) 25 J
Explanation
Work=Area under the curve Answer: (b)
Q.13
A spring gun of spring constant 90 × 102 N/M is compressed 4cm by a ball of mass16g. If the trigger is pulled, calculate the velocity of the ball.
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a)60 m/s
0%
b) 3 m/s
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c)90 m/s
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d)30 m/s
Explanation
Loss of potential energy of spring=gain in kinetic energy of ballAnswer: (d)
Q.14
A uniform chain of length 2m is kept on a table such that a length of 50cm hangs freely from the edge of the table. The total mass of the chain is 5kg. What is the work done in pulling the entire chain on the table. (g=10 m\\s2)
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a) 7.2 J
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b)3 J
0%
c)4.6 J
0%
d)120 J
Explanation
Mass of a chain is uniform . Thus centre of mass of hanging chain is blow 0.25 m of the table. Mass per unit length of the chain=5/2=2.5kgs. mass of hanging chain of 0.5 m=1.25 kg Now work done=change in potential energy of hanging chain W=(1.25) (0.2)(10)=3.125 ≈ 3.0 JAnswer: (b)
Q.15
A uniform chain of length L ans mass M is lying on a smooth table and one third of its is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is
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a)MgL
0%
b) MgL / 3
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c)MgL/9
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d)MgL / 18
Explanation
Chain is uniform and one third is hanging thus centre of mass of chain from table=L/6Mass of hanging chain=M/3Thus work done=Change in potential energy W=(L/6)(M/3) g=MgL/18 Answer:(d)
Q.16
A cord is used to lower vertically a block of mass M by a distance d with constantdown-word acceleration g/2 work done by the cord on the block is
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a) - Mgd/2
0%
b) Mgd/4
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c) -3Mgd/4
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d) Mgd
Explanation
Tension in the string=M( g - g/2)=-Mg/2 Work=Tension × displacement=-Mgd/2 Answer: (a)
Q.17
A block of mass 5 kg is resting on a smooth surface. At what angle a force of 20N be acted on the body so that it will acquired a kinetic energy of 40J after moving 4m
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a) 30°
0%
b) 45°
0%
c) 60°
0%
d) 120°
Explanation
Work = Chagne in kinetic energy 40 = 20 × 4 × cosθ cosθ = 1/2 θ = 60° Answer: (c)
Q.18
The potential energy of a body is given by U = A - Bx2 (where x is displacement). The magnitude of force acting on the particle is
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a)constant
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b)proportional to x
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c) proportional to x2
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d) Inversely proportional to x
Explanation
For conservative field F = -dU/dx F = -2Bx Answer: (b)
Q.19
An open knife edge of mass m is dropped from a height h on a wooden floor. If the blade penetrates upto the depth d into the wood, the average resistance offered by the wood to the knife edge is
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a) mg
0%
b)
0%
c)
0%
d)
Explanation
Work done by resistive force = Loss in potential energy of knife Fd = mg(h + d) F = mg [1 + (h/d)] Answer:(b)
Q.20
A toy car of mass 3.2 kg moves up a ramp under the influence of force F plotted against displacement x. The maximum height attained is given by
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a) ymax = 5 m
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b) ymax = 10 m
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c) ymax = 15 m
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d) ymax = 20 m
Explanation
Work done = Area under the graph Work done = change in Potential energy Area under graph = Change in potential energy 320 = (3.2)(10) y y = 10 cm Answer: (b)
Q.21
A particle of mass 0.5kg travels in a straight line with velocity v=a x3/2 . Where a=5 m-1/2 s-1 The work done by the net force during its displacement from x=0 to x=2m is
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a)50 J
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b)45 J
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c)25 J
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d)None of these
Explanation
Initial velocity at x=0 v1=0Final velocity v2=5 × 23/2 Work=Change in kinetic energy Answer: (a)
Q.22
Velocity time graph of a particle of mass 2kg moving in a straight line is as shown in figure. Work done by all forces on the particle is
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a) 400 J
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b) -400 J
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c)-200 J
0%
d)200 J
Explanation
Intial velocity of particle v1=20 m/sFinal velocity of particle v2=0Work dine=change in kinetic energy Answer: (b)
Q.23
A mass of M kg is suspended by a weight-less string, the horizontal force that is required to displace it until the string makes an angle of 60° with the initial vertical direction is
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a) Mg /√3
0%
b) Mg √2
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c)Mg / √2
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d)Mg √ 3
Explanation
Work done=change in potential energy F( lsin60 )=mgl(1 - cos60) Fsin60=mg( 1 - cos60) F(√3/2)=mg[ 1 - (1/2)] F=Mg /√3 Answer:(a)
Q.24
A spring with spring constant K when stretched through 2cm the potential energyis U. If it is stretched by 6cm. The potential energy will be......
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a) 6U
0%
b) 3U
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c) 9U
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d) 18U
Explanation
Potential energy of spring U ∝ X2 Thus U'/U=(6/2)2 U'/U=32 U'=9U Answer: (c)
Q.25
If linear momentum of body is increased by 1.5%, its kinetic energy increasesby.......%
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a)0%
0%
b)10%
0%
c)2.25%
0%
d)3%
Explanation
We know that E ∝ P2 Thus dE ∝ 2PdP dE/E=2PdP/P2 dE/E=2(dP/P) dE/E=2 × 1.5=3%Answer: (d)
Q.26
A body having a mass of 0.5 kg slips along the wall of a semi spherical smooth surface of radius 20 cm shown in figure.What is the velocity of body at the bottom of the surface ? (g=10 m/s2)
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a) √2 m/s
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b) 2 m/s
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c)2√2m/s
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d)4 m/s
Explanation
Loss of PE=kinetic energyAnswer: (b)
Q.27
An engine pump is used to pump a liquid of density ρ continuosly through apipe of cross-sectional area A. If the speed of flow of the liquid in the pipeis v, then the rate at which kinetic energy is being imparted to the liquid is
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a) ½ A ρ v3
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b) ½ A ρ v2
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c) ½ A ρ v
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d)A ρ v
Explanation
mass of liquid=density volume m=ρ ( Av) Kinetic energy=mv2 Kinetic energy=½ A ρ v3 Answer:(a)
Q.28
when 2kg mass hangs to a spring of length 50 cm, the spring stretches by 2 cm.The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if g=10 m/s2
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a) 10 J
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b) 2 J
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c) 2.5 J
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d) 5 J
Explanation
Use equation F=Kx then U=½ kx2 Answer: (d)
Q.29
Two masses 10kg and 20 kg are connected by a massless spring. A force of 200N acts on 20kg mass. At the instant when the 10 kg mass has an acceleration 12 m/s2 the energy stored in the spring ( k=2400 N/m) will be
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a)30 J
0%
b) √3 J
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c)3 J
0%
d)8 J
Explanation
Force on 10 kg mass F=10 × 12=120 N which is exerted by spring But for spring F=Kx Thus F=2400x From above 2400x=120 x=1/20 is the elongation of spring Energy stored in spring E=½ Kx2 E=½ × 2400 × (1/400)=3 JAnswer: (c)
Q.30
A block of mass 1 kg is released from rest at t=0 to fall freely under gravity. Power of gravitational force acting on it at t=2s is ( g=10 m/s2)
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a)200 watt
0%
b) 10 watt
0%
c)100 watt
0%
d)zero
Explanation
After 2 second velocity is 2g power P=F . v=mg(2g)=200 watt Answer:(a)
0 h : 0 m : 1 s
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