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Quiz 2
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Q.1
For inelastic collision between two spherical rigid bodies [ AIIMS 2006]
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c)the linear momentum is not conserved
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a) the total kinetic energy is conserved
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b) the total potential energy is conserved
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d)the linear momentum is conserved.
Explanation
In inelastic collision, linear momentum is conserved.Answer: (d)
Q.2
The force constant of a wire is k and that of the another wire is 3k when both thewires are stretched through same distance, if work done are W1 and W2, then...
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a) W1=3W2
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b) W2=0.33W1
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c) W2=W1
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d)W2=3W1
Explanation
W1=kx × x=kx2 W2=3kx × x=3kx2 Thus W2=3W1 Answer: (d)
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... [ AIIMS 2008]
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a) mg(h + d) - (1/2)kd2
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b) mg(h - d) - (1/2)kd2
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c) mg(h - d) + (1/2)kd2
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d) mg(h + d) (1/2)kd2
Explanation
Let spring compressed by distance 'd'. Thus total height=h+d Now according to law of conservation of energy Gravitational potential energy=potential energy of spring mg(h + d)=(1/2) kd2 Net work done=mg(h + d) - (1/2)kd2=0 Answer: (a)
Q.4
A block of mass 10kg is moving in x-direction with a constant speed of 10 m/s . It is subjected to a retarding force F=-0.1 x joule / meter during its travel from x=20 meter to x=30 meter. Its final kinetic energy will be.. [AIIMS 2005]
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a)475 J
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b) 450 J
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c)275 J
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d)250 J
Explanation
We know that, Change in kinetic energy=work done on the object by forceHere, work done=∫ Fdx Now initial kinetic energy=½(10×100)=500 JFinal kinetic energy=500 - 25=475 jouleDirection of force is opposite to direction of motion Answer: (a)
Q.5
If spring extends by x on loading, then energy stored by the spring is ( if T is the tension in the spring and k is spring constant) [ AIIMS 1997]
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a)
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b)
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c)
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d)
Explanation
Energy of spring=(1/2)kx2T=kx ∴ x=T/kE=(1/2)k(T/k)2Answer: (c)
Q.6
The co-efficient of restitution e for a perfectly elastic collision is [1988]
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a)1
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b) 0
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c)∞
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d) -1
Explanation
Answer:(a)
Q.7
A body of mass m moving with velocity 3 km/h collides with a body of mass 2 m at rest. Now the coalesced mass starts to move with a velocity [1996]
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a) 1 km/h
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b) 2 km/h
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c) 3 km/h
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d) 4 km/h
Explanation
Since two objects stick together will have common velocity v Applying law of conservation of momentumm1u=(m1+m2)vm × 3=(m + 2m)vv=1 km/h Answer: (a)
Q.8
Two equal masses m1, and m2, moving along the same straight line with velocities + 3 m/s and - 5m/s respectively, collide elastically. Their velocities after the collision will be respectively. [ CBSE-PMT 1998]
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a)-3 m/s and +5 m/s
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b) + 4 m/s for both
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c)-4 m/s and -4 m/s
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d)-5 m/s and +3 m/s
Explanation
In elastic collision, the velocities get inter changed if the colliding objects have equal massesAnswer: (d)
Q.9
A position dependent force, F=(7 - 2x + 3x2) N acts on a small body of mass 2 kg and displaces it from x=0 to x=5 m. Work done in joule is [ CBSE-PMT 1992]
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a) 35
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b) 70
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c)135
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d) 270
Explanation
Here force is not constant it is position dependant. To obtain work we have to integrate the functiondw=FdxAnswer: (c)
Q.10
A force acts on a 30 gm particle in such a way that the position of the particle as a function of time is given by x=3t- 4t2 + t3, where x is in meters and t is in seconds. The work done during the first 4 seconds is [1998]
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a)5.28J
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b) 450mJ
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c)5760mJ
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d)530mJ
Explanation
Velocity v=dx/dt=3 - 8t + 3t2 acceleration a=dv/dt=-8 + 6t Note that acceleration is not constant, so it is varible force problem, can be solved by using calculus method, but esasy way is We know that work done = Change in kinetic energy Final velocity at time t = 4sec v = [3- 8(4) + 3(4)2] = 3 - 32 + 48 = 19m/s Initial velocity at time t = 0 s v = [3- 8(0) + 3(0)2] = 3m/s Answer:(a)
Q.11
A bomb of mass 1 kg is thrown vertically upwards with a speed of 100 m/s. After 5 seconds it explodes into two fragments. One fragment of mass 400 gm is found to go down with a speed of 25 m/s. What will happen to the second fragment just after the explosion? (g=10 m/s2)[CBSE-PMT 2000]
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a) It will go upward with speed 40 m/s
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b) It will go upward with speed 100 m/s
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c) It will go upward with speed 60 m/s
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d) It wi11 also go downward with speed 40m/s
Explanation
Speed of bomb after 5 second v=u - gt v=100 - 10(5) v=50 m/s Now according to law of conservation of momentum (50 × 1) j=0.4 × 25(-j) + 0.6v v=100 m/s Upward Answer: (b)
Q.12
A force of 250 N is required to lift a 75 kg mass through a pulley system. In order to lift the mass through 3 m, the rope has to be pulled through 12m. The efficiency of system is [ CBSE-PMT 2001]
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a)50%
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b) 75%
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c)33%
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d)90%
Explanation
Efficiency=output work / input workoutput work=mgh=75 × 10 × 3=2250 J Input work=Force × displacement=250 ×12=3000J ∴ Efficiency=2250 / 3000=0.75=75%Answer: (b)
Q.13
The potential energy of a system increases if work is done [2011]
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a) upon the system by a non conservative force
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b) by the system against a conservative force
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c) by the system against a non conservative force
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d)upon the system by a conservative force
Explanation
When a work is done upon a system by a conservative force then its potential energy increasesAnswer: (d)
Q.14
A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/s at an angle of 60°. It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for 0.20s, what is the average force exerted on the ball by the wall [ CBSE-PMT 2000]
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a)150N
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b)zero
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c)150√3N
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d) 300N
Explanation
As shown in figure sine component are in same direction hence no change. However cos component changes direction ∴ Change in momentum of ball=mvcos30 -(-mvcos30 )=2mvcos30Change in momentum=2 × 3 × 10 × √(3/4)]=30√3Now Force on the ball=Change in momentum of the ball / time of contactForce on ball=30√3 / 0.2=150√3 N Answer:(c)
Q.15
The kinetic energy acquired by a mass (m) in traveling distance (s) starting from rest under the action of a constant force is directly proportional to [ CBSE-PMT 1994]
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a) 1/√m
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b) 1/m
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c) √m
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d) m0
Explanation
Initial velocity u=0 F=ma a=F/m Now v2=u2 + 2as v2=2(F/m)s K.E=½(m)v2 K.E.=½( m) × 2(F/m) d=Fd K.E=constant, and is independent of mass Answer: (d)
Q.16
A rubber ball is dropped from a height of 5m on a plane, where the acceleration due to gravity is not shown. On bouncing it rises to l.8 m. The ball loses its velocity on bouncing by a factor of [ CBSE-PMT 1998]
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a)16/25
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b) 2/5
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c)3/5
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d)9/25
Explanation
According to law of conservation of energyLoss of potential energy=Gain in kinetic energy∴ mgh=½(mv2)If h1 and h2 are initial and final heights, then ∴ fraction of velocity=Answer: (b)
Q.17
A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a nearly weightless spring of force constant k=50 N/m. The maximum compression of the spring would be [ CBSE-PMT 2004]
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a) 0.5 m
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b) 0.15 m
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c)0.12 m
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d)1.5 m
Explanation
When mass colloids with spring, spring gets compress so that kinetic energy of mass equal the springs potential energy let x be the compression ½mv2=½kx2 here mass m=0.5kg velocity v=1.5 m/s restoring force constant k=50N/m. substituting values in above equation and solving equation for x we get ½(0.5)(1.5)2=½(50) x2 x=0.15mAnswer: (b)
Q.18
A particle of mass m1 is moving with a velocity v1 and another particle of mass m2 is moving with a velocity vBoth of them have the same momentum but their different kinetic energies are E1 and E2 respectively. If m1 > m2 then [ CBSE-PMT 2004]
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a)E1=E2
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b)E1 < E2
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c) E1 / E2=m2 / m1
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d)E1 > E2
Explanation
Form the law of conservation of momentum, before collision and after collision linear momentum (p) will be same. That is Initial momentum=Final momentum∴ 0=m1v1 - m2v2∴ p1=p2Now E=p2 / 2msince p1=p2E1 / E2=m2 / m1 Answer:(c)
Q.19
A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of [ CBSE- PMT 2004]
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a) 1 : √2
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b) √2 : 1
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c) 1:4
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d) 1:2
Explanation
since height is same for both balls, their velocities on reaching the ground will be same thus K.E ∝ mass K.E.1 / K.E. 2=m1 / m2 K.E.1 / K.E. 2=1/2 Answer: (d)
Q.20
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms-1 The kinetic energy of the other mass is [CBSE-PMT 2005]
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a) 324 J
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b) 486 J
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c)256 J
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d)524 J
Explanation
From the law of conservation of linear momentumm1v1 + m2v2=0Now K.E.=½(mv2) on substituting value of v in above equation and on solving we getK.E=486 JAnswer: (b)
Q.21
An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water? [ CBSE- PMT 2009]
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a) mv2
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b) ½(m2v2)
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c) ½(m v2)
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d) ½(m v3)
Explanation
m : Mass per unit length ∴ rate of mass leaving the hose per sec = mx/ t = mv Rate of K.E = ½(mv)v2 Rate of K.E. = ½m v3 Answer: (d)
Q.22
An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/in. What is the power of the engine? [ CBSE-PMT 2010]
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a) 400 W
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b) 200 W
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c)100W
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d)800W
Explanation
Amount of water flowing per second from the pipe=m /timepower=K.E. of water flowing per second Answer:(c)
Q.23
A body of mass 3 kg is under a constant force which causes a displacement s in meters in it, given by the relation s=(1/3) t2,where t is in seconds. Work done by the force in 2 seconds is [ CBSE-PMT 2006]
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a) 3/8 J
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b) 8/3 J
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c) 19/5 J
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d) 5/19 J
Explanation
Acceleration= Force acting on the body=mass × acceleration F=3 × (2/3)=2 Newton Displacement in 2secs=(1/3) × 2 × 2=4/3 m Work done=2 × 4/3=8/3 JAnswer: (b)
Q.24
A bullet of mass 10g leaves a rifle at an initial velocity of 1000m/s and strikes the earth at the same level with a velocity of 500 m/s. the work done in joules overcoming the resistance of air will be [ CBSE_PMT 1989]
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a)375
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b) 3750
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c)5000
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d)500
Explanation
According to work energy theorem W=ΔEW=½ m ( v12 - v22)W=½ × 0.01 [ (1000)2 - (500)2]W=3750JAnswer: (b)
Q.25
How much water, a pump of 2kW can rise in one minute to height of 10 m, taking g=10 m/s2 [ CBSE-PMT 1990]
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a) 1000 liters
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b) 1200 liters
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c)100 liters
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d)2000 liters
Explanation
Power P=work/timeWork W=mgh=m × 10 × 10=100m P=2000W and t=60s2000=100m/60m=1200 kgsince density of water is 1 gm/cc Volume of water is 1200 literAnswer: (b)
Q.26
The kinetic energy acquired by a mass (m) in traveling distance starting from rest under the action of constant force is directly proportional to [ CBSE-PMT 1996]
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a) 1 / √m
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b) 1/m
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c)√m
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d)m0
Explanation
K.E acquired=work done K.E acquired=F × d since F is constant , thus K.E is independent of mass Answer:(d)
Q.27
Two body of masses m and 4m are moving with equal kinetic energy. The ratio of their linear momentum is equal to .. [ CBSE-PMT 1997]
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a) 4:1
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b) 1:1
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c) 1:2
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d) 1:4
Explanation
Given Kinetic nergy of mass m E1=kinetic energy of mass 4m E2 Thus ½ mv12=½ 4mv22 Thus v12 /v22=4 Thus v1=2v2 momentum of mass m is p1=mv1 momentum of mass 4m is p2=4mv2 Thus p1 / p2=m × 2v2 / 4 m × v2 p1 / p2=1/2 Answer: (c)
Q.28
A metal ball of mass 2kg moving with velocity of 36km/h has a head on collision with a stationary ball of mass 3kg. If after the collision the two balls moves together, the loss in kinetic energy due to collision is [ CBSE-PMT 1997]
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a)140J
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b) 100J
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c)60J
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d)40 J
Explanation
Applying law of conservation of momentumm1v1=(m1 + m2) v Here v1=36 km/h=10 m/smass m1=2kg , m2=3kg Thus v=10 × 2 / 5=4m/sInitial kinetic energy=½ × 2 × (10)2=100 J Final kinetic energy=½ × ( 3 + 2) × (4)2=40JLoss in K.E=100 - 40=60 JAnswer: (c)
Q.29
Two bodies with kinetic energies in the ratio 4:1 are moving with equal linear momentum. The ratio of their masses is .. [ CBSE-PMT 1999]
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a) 1:2
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b) 1:1
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c)4:1
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d)1:4
Explanation
Kinetic energy is same kinetic energy is given by formula E=p2 / (2m) It is given p1=p2 Answer: (d)
Q.30
in a simple pendulum of length l the bob is pulled aside from its equilibrium position through an angle θ and then released. the bob passes through the equilibrium position with speed [CBSE-PMT 2000]
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a)
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b)
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c)
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d)
Explanation
l be the length of pendulum makes angle of θ at extreme position as shown in figurefrom figure l=d + h h=l-dand d=lcosθ h=l (1 - cosθ)Now as per law of conservation of energy ΔPE=ΔK.Emgh=½ m v2 v2=2gh Answer:(d)
0 h : 0 m : 1 s
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