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Work, Energy And Power Mcq
Quiz 4
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Q.1
A block of mass 0.5kg is moving with a speed of 2.0m/s on smooth surface. It strikes another mass of 1.0kg and then they move together as a single body. The energy loss during collision is [AIEEE 2008]
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a) 0.16 J
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b) 1.0J
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c)0.67J
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d)0.34J
Explanation
Initial kinetic energy of the system Ei=½ mu2 + ½ M(0)2 Ei=½ × 0.5 × × 22 + 0 br/>Ei=1 J Applying law of conservation of momentum m × u=(m + M) × v 0.5 × 2=(0.5 + 1) × v v=(2/3) m/sFinal kinetic energy of the system is Ef=½ (m + M)v2 Ef=½ (0.5 + 1.0)(2/3)2 Ef=(1/3)J Energy loss in collision=Ef - EiEnergy loss in collision=(1 - 1/3)=0.67JAnswer: (c)
Q.2
A body of mass 2kg is thrown up vertically with a kinetic energy 490J. If the gravitational acceleration due to gravity is 9.8 m/s2, the height at which the kinetic energy of the body becomes half of the original value is [ EAMCET 1986]
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a) 50 m
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b) 25 m
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c)12.5 m
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d)10 m
Explanation
We know that ΔK.E=-ΔP.E Potential energy at maximum height=mgh=490J Thus at half of the height P.E will be half Thus 490/ 2=mgh' 245=2 × 9.8 × h' h'=12.5 m Answer:(c)
Q.3
A 238U nucleus decays by emitting an alpha particle of speed v m/s. The recoil speed of the residual nucleus is ( m/s) [ CBSE 1995]
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a) -4v / 234
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b) v/4
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c) -4v/234
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d) 4v/238
Explanation
mass of alpha particle is 4units since no external force is applied momentum is conserved 0=234V + 4v V=-4v/ 234 Answer: (a)
Q.4
You lift a heavy book from the floor of the room an put it in the book self having a height 2m. In this process you take 5sec. The work done by you will depend upon [ MPPET 1993]
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a)mass of the book and time taken
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b) weight of the book and the height of the book shelf
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c)height of the book shelf and the time taken
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d)mass of the book, height of the book shelf and the time taken
Explanation
Change in Potential energy=Work done Potential energy=mgh Thus option (b) is correctAnswer: (b)
Q.5
A particle moves under the effect of force F=cx, from x=0 to x=x1 the work done in the process is
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a) (cx12) /2
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b) cx12
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c)cx13
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d)zero
Explanation
force is variable, depends on positionwork done=∫fdxAnswer: (a)
Q.6
Which of the following statement is true?
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a) momentum is conserved in elastic collisions but not in inelastic collisions
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b) total kinetic energy is conserved in elastic collisions but momentum is not conserved in elastic collisions
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c)total kinetic energy is not conserved but momentum is conserved in inelastic collisions
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d)kinetic energy and momentum are conserved in all types of collisions.
Explanation
Answer:(c)
Q.7
The principle of conservation of momentum can be strictly applied during a collision between two particles provided the time of impact is
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a) extremely small
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b) moderately small
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c) extremely large
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d) depends on a particular case
Explanation
Answer: (a)
Q.8
Work is done on the body when
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a)force acts on it
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b) it moves through a certain distance
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c)it experiences an increase in energy through a mechanical influence
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d)none of the above
Explanation
When work is done on the object at rest its kinetic energy increasesAnswer: (c)
Q.9
Which one of the following statements does not hold good when two balls of mass m1 and m2 undergo elastic collision?
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a) when m1 < m2 and m2 are at rest, there will be maximum transfer of momentum
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b) when m1 > m2 and m2 are at rest, after collision the ball of m2 moves with four times the velocity of m1
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c)When m1 and m2 and m2 are at rest, there will be maximum transfer of K.E.
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d)when collision in oblique and m2 at rest with m1=m2, after collision the ball moves in same direction
Explanation
Velocity of ball of mass m1 is given by : since collision is elastic e=1 and m2 is at rest the u2=0 from above equation it is clear that velocity of m2 can not be four times the velocity of m1Answer: (b)
Q.10
A block of weight W is pulled a distance l along a horizontal table. The work done by the weight is
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a) Wl
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b) 0
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c)Wgl
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d)Wl/g
Explanation
Work done=F Scosθ Here Force ( weight ) is perpendicular to displacement thus θ=90°Answer: (b)
Q.11
A particle of mass m moving with a velocity v makes a head-on elastic collision with another particle of mass m initially at rest. The velocity of the first particle after collision will be [ MPPMT 1997]
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a) v
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b) - v
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c)-2v
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d)zero
Explanation
Form the formula given e=1 and u2=0 , both the objects have same massThus velocity=0 Answer:(d)
Q.12
A car of mass 'm' is driven with acceleration 'a' along a straight road against a constant external resistance 'R'. When the velocity is 'v', the power of the engine is : [ MPPMT 1998]
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a) Rv
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b) mav
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c) (R+ma)v
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d) (R-ma)v
Explanation
Total force=(R+ma) Power=(R+ma)v Answer: (c)
Q.13
A uniform chain of mass M and length L lies on a frictionless table with length l0 hanging over the edge. The chain begins to slide down. Then the speed with which the end slides away from the edge is given by :
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a)
0%
b)
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c)
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d)
Explanation
Let the mass per unit length of the chain be m mass of hanging chain is ml0 Center of mass of hanging chain is l0/2 from top and at H - l0 /2 from the ground Potential energy of hanging chain is U Mass of rest of the of the chain is m(L - l0 ) Potential energy of remaining chain on the table U'=m(L - l0 )gHThus total potential energy of chain=U + U' When chain of length L falls of the table center of mass of chain is at distance L/2 from edge and at H - L/2 from ground Thus potential energy of chain when slide off U"=mLg( H - L/2)Let the velocity of center of mass be v Then Kinetic energy=½ × mLV2Total energy after sliding Now according to law of conservation of energy Energy before sliding=Energy after slidingAnswer: (d)
Q.14
A car is moving along a straight horizontal road with a speed v0 If the coefficient of friction between the tyres and the road is µ, the shortest distance in which the car can be stopped is .. [ MPPMT 1985]
0%
a)
0%
b)
0%
c)
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d)
Explanation
Work done by friction=F s=µmg sNow work done by friction=Change in kinetic energy=½ m v2 Thus ½ m v2=µmg ss=v02/2µ g Answer: (a)
Q.15
A spring of force constant 10 N per meter has an initial stretch 0.20 m. In changing the stretch to 0.25m, the increase in potential energy is about
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a) 0.1 joule
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b) 0.2 joule
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c)0.3 joule
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d)0.5 joule
Explanation
Elastic potential energy of spring=½ kx2 Change in energy=½10 ( 0.252 - 0.202) Change in energy=0.1125 J Answer:(a)
Q.16
A particle moves in the x-y plane under the action of a force F such that the value of its linear momentum P at any time t is Px=2cost , Py=2sint the angle θ between F and P at any given time t will be [ MNR 1991]
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a) 90°
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b) 0°
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c) 180°
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d) 30°
Explanation
Px=2cos t ,Py=2sin t p is independent of time and is constant. This is possible only if angle between force and momentum is 90° Answer: (a)
Q.17
An engine pumps a liquid of density 'd' constantly through a pipe of area of cros section A. If the speed with which the liquid passes through a pipe is 'v', then the rate at which the kinetic energy is being imparted to the liquid is
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a)Adv3 / 2
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b) Ad v /2
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c)Adv2 / 2
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d)Adv2
Explanation
Volume of liquid per sec=Av Mass of liquid pumped per sec=AvdKinetic energy=½ m v2Kinetic energy imparted per second (power)=½ Avd v2 Kinetic energy imparted per second (power)=½ Ad v3 Answer: (a)
Q.18
A spring is compressed between two toy car of masses m and M. When the toy-cars are released, the spring exerts on each toy-car equal and opposite force for the same time t. If the coefficient of friction 'µ' between the ground and the toy-car are equal, then the displacement of the two toy-cars:
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a)
0%
b)
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c)
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d)
Explanation
Applying law of conservation of momentum mv + MV=0 v/V=-M/m --eq(1)Work done by friction=Change in KEµmg S1=½ m v2 µMg S2=½ m V2 Taking ratio From equation (1) and S1 and S2 are opposite Answer: (c)
Q.19
If V, P and E denote the velocity, momentum and kinetic energy of the particle then
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a) P=dE/dV
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b) P=dE/dt
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c)P=dV/dt
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d)P=dE/dV × dE/dt
Explanation
Kinetic energy E=½ mV2 taking derivative with respect to V we get dE/dV=mV dE/dV=P Answer:(a)
Q.20
A weight hangs freely from the end of a spring. A boy then slowly pushes the weight upwards until the spring becomes slack. The gain in gravitational potential energy of the weight during this process is equal to
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a) the loss of stored energy by the spring minus the work done by the tension in the spring
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b) the work done on the weight by the boy plus the stored energy lost by the spring
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c) the work done on the weight by the body minus the work done by the tension in the spring plus the stored energy lost by the spring
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d) the work done on the weight by the boy minus the work done by the tension in the spring
Explanation
Answer: (d)
Q.21
A bullet is fired freely. If the rifle recoils freely, then
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a) the kinetic energy of the rifle is less than that of the bullet
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b) the kinetic energy of the rifle is greater than that of the bullet
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c) the kinetic energy of the rifle is equal to that of bullet
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d) the kinetic energy of rifle is zero
Explanation
mass of bullet m , velocity v mass of rifle M , velocity V Applying law of conservation of mass mv + MV=0 mv=-MVp=-P Kinetic energy of bullet Eb=p2 / 2m Kinetic energy of rifle Er=p2 / 2M Taking ratio Eb / Er=M /m ∴ Eb / Er > 1 Thus Kinetic energy of bullet is more than rifle Answer: (a)
Q.22
A man pushes a wall but fails to displace it, it does : [ CPMT 1992]
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a) negative work
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b) positive work
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c)no work at all
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d)maximum positive work
Explanation
There no displacement thus work is zeroAnswer: (c)
Q.23
A lighter and a heavy body has equal kinetic energy which has greater momentum [ CPMT 1985]
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a)the heavy body
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b) the lighter body
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c)both have equal momentum
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d)none
Explanation
Let mass of lighter body be m and its momentum p Let mass of heavy body be M and momentum P Both the object have same kinetic energy hence Thus heavy body have higher momentum Answer:(a)
Q.24
A bullet is fired freely. If the rifle recoils freely, then
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a) the kinetic energy of the rifle is less than that of the bullet
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b) the kinetic energy of the rifle is greater than that of the bullet
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c) the kinetic energy of the rifle is equal to that of bullet
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d) the kinetic energy of rifle is zero
Explanation
mass of bullet m , velocity v mass of rifle M , velocity V Applying law of conservation of mass mv + MV=0 mv=-MVp=-P Kinetic energy of bullet Eb=p2 / 2m Kinetic energy of rifle Er=p2 / 2M Taking ratio Eb / Er=M /m ∴ Eb / Er > 1 Thus Kinetic energy of bullet is more than rifle Answer: (a)
Q.25
A particle is moving with a momentum of 10 kgm/s. A force of 0.2N acts on it in the direction of motion of the particle for 10 s. If the mass of the particle is 5kg, the increase in kinetic energy is [ MPPET 1999]
0%
a)4.4J
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b) 3.8J
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c)3.2J
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d)2.8J
Explanation
Force × time=change in momentum Change in momentum=0.2 × 10=2 Initial momentum is pi=10 kgm/s thus final momentum is pf=12 kgm/s Change in kinetic energy=1/2m ( pf2 - pi2 ) Change in kinetic energy=(1/10) ( 144 - 100)=4.4 J Answer: (a)
Q.26
A block of mass M slides along the sides of a flat bottomed bowl. The sides of the bowl are frictionless and the base has a coefficient of friction 0.If the block is released from the top of the side which is 1.5m height, where will the block come to rest. Given, the length of the base is 15m
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a) 1 m from P
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b) mid point of flat part PQ
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c)2m from P
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d)at Q
Explanation
K.E of the block at P is 1.5mg. This is wasted away doing work on the rough flat surface Work done by friction=µmgx 1.5mg=µmgx x=1.5 / µ=7.5Thus block will come to rest at mid point of PQAnswer: (b)
Q.27
If a man increases his speed by 2m/s, his Kinetic energy is doubled. The original speed of the man is
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a) (2 + √2) m/s
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b) (2+2√2) m/s
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c)4 m/s
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d)( 1 +2√2) m/s
Explanation
mass is constant thus E ∝ v2 Increase in velocity by 2 m/s, kinetic energy becomes doubled Thus 2E/E=(v + 2) 2 / v2 √2 v=v + 2∴ v=2 / (√2 - 1)=2(√2 + 1)=(2 + 2√2) Answer:(b)
Q.28
Work done in raising a box depends on
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a) How fast it is raised
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b) The height by which it is raised
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c)The strength of the man
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d)None of the above
Explanation
Work in raising a box=(weight of the box) × (height by which it is raised)Answer: (b)
Q.29
The potential energy between to atoms in a molecule is given by where a and b are positive constants and x is the distance between the atoms. The atoms is in stable equilibrium when:
0%
a)
0%
b)
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c) zero
0%
d)
Explanation
we know that for conservative force field F=-dU/dx For stable equilibrium F=0 Thus - dU/dx=0∴ -12ax-13 + 6bx-7=0 Thus x6=2a/b Answer: (d)
Q.30
A smooth sphere of M moving with velocity u directly collides elastically with another sphere of mass m at rest. After collision, their final velocity are V and v, respectively. The value of v is : [ MPPMT 1995]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
applying law of conservation of momentum Mu=MV + mv given elastic collision e=1 1=(v - V) / (u - 0) v - V=u V=v - u Putting in equation (1) Mu=M(v - u) + mv v=2Mu / ( M + m) Answer: (c)
0 h : 0 m : 1 s
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