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Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 5
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Q.1
The position vector of a particle is r=(cosωt)i + (a sinωt)jThe velocity of the particle is: [ CBSE 1995]
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a) parallel to the position vector
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b) perpendicular to the position vector
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c)perpendicular to the origin
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d)directed away from the origin
Explanation
By taking derivative of given equation we get velocity vector v=-ω sinωt i + ω cosωt j Now r ⋅ v=-ω sinωt cosωt + ω sinωt cosωt since r ⋅ v=0 r and v are perpendicular Answer: (b)
Q.2
The kinetic energy acquired by a mass m in traveling a certain distance d, starting from rest, under the action of a constant force is directly proportional to [ CBSE 1994]
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a)√m
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b) independent of m
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c)1 / √m
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d)m
Explanation
We know that work done=change in kinetic energy Work done by the force=FdThus kinetic energy acquired by the object is independent of mass Answer:(b)
Q.3
A sphere colloids 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°
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c) 45°
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d) different from 90°
Explanation
Answer: (b)
Q.4
The work performed on an object does not depend upon
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a)force applied
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b) angle at which force is inclined to the displacement
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c)initial velocity of the object
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d)displacement
Explanation
Work W=FScosθ Thus does not depends on initial velocityAnswer: (c)
Q.5
Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance 'l' on a rough surface. The following quantities will be same as observed by the two observer
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a) work done by the friction
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b) acceleration of the block
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c)kinetic energy of the block at time t
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d)total work done
Explanation
When two observers are moving with respect to each other at a speed of v,along a straight line . Acceleration of block if any will be same. Distance moved may be different. Therefore, work done will ,K.E of the block may appear differentAnswer: (b)
Q.6
In which case potential energy decreases [MPPMT 1996]
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a) on compressing the spring
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b) on stretching a spring
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c)on moving a body against gravitational pull
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d)on the rising of an air bubble in water
Explanation
Potential energy decreases when air bubble rises in water because work is done by upthrust Answer:(d)
Q.7
The first ball of mass m moving with velocity v collides head on with the second ball of mass m at rest. If the coefficient of restitution is e, then the ratio of the velocities of the first and the second ball after collision is
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a) (1 - e) / (1 + e)
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b) (1 + e) / (1 - e)
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c) (1 + e) / 2
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d) (1 - e) / 2
Explanation
Formula for restitution is since u2=0 thus v2 - v1=eu1 --eq(1) Applying law of conservation of momentum mu1=mv1 + mv2 v1 + v2=u1 --eq(2) Add eq(1) and eq(2) we get Subtract eq(2) from eq(1) we get Taking ratio Answer: (a)
Q.8
A ball is dropped on the ground from the height of 1m. The coefficient of restitution is 0.The height to which the ball will rebound is
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a)0.6 m
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b) 0.4m
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c)0.36m
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d)0.16m
Explanation
From the formula of restitutionHere u2=and v2=0 Thus e=v1 / u1e2=(v1 / u1 ) 2 --(1)Now v2 - u2=2ah When ball dropped from height h1 u=0, let final velocity is u1u12=2ah1 When ball bounce back final velocity=0 , Initial velocity u2, let height be h2v12=2ah2By taking ratio u12 / v12=h1 / h2 --(2)from equation 1 and 2 e2=h2 / h1 0.36=h2 / 1 h2=0.36 m Answer: (c)
Q.9
A body is initially at rest. It undergoes one dimensional motion with constant acceleration. The power delivered to it at a time t is proportional to [ kerala PMT 2000]
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a) t1/2
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b) t
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c)t 3/2
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d)t2
Explanation
v=u + at v=at as u=0 P=Fv=(ma) (at) P=ma2 t as m and a constant thus P ∝ t Answer: (b)
Q.10
A rod elongates by l when a body of mass M is suspended from it. Work done is [ CPMT 1999]
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a) Mgl
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b) Mgl/2
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c)2Mgl
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d)zero
Explanation
Let length of the rod is L . Then CM is at L/2 Rod is elongated by l thus new CM will be at (L+l) /2 Change in position of CM=h=l/2 Work done=Force × displacement Work done=Mg×(l/2) Answer:(b)
Q.11
A block of mass 4 kg is placed on a rough horizontal plane. A time dependent force F=kt2 acts on the block, where k=2N/sForce of friction between block and the plane at t=2 sec is [ CPMT 2000]
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a) 8N
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b) 4N
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c) 2N
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d) zero
Explanation
Unless block start to move Frictional force=Applied force Frictional force=kt2 k=2 N/s2 and t=2 sec Frictional force=2 × 22 Frictional force=8N Answer: (a)
Q.12
A ball falls vertically on the floor, with momentum p, and then bounces repeatedly, the coefficient of restitution is e. The total momentum imparted by the ball to the floor is [IIT 2002]
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a)p(1+e)
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b) p / (1-e)
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c)
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d)
Explanation
When particle undergoes normal collision with a floor or on wall, with coefficient of restitution eThe speed after collision is e times the speed before collision. Therefore, change in momentum after impact=ep - (-p)=p(1 + e) After second impact , change in momentum would be e(ep) - (-ep)=ep(1 + e) and so on Therefore, total change in momentum of the ball=momentum imparted to floor=p(1 + e) [ 1 + e + e2 +......]=p(1 + e) [ 1 / (1 - e)]=p(1 + e) / (1 - e) Answer: (c)
Q.13
A ball which is at rest i dropped from height h meter. As it bounces off the floor, its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height ..[ kerala 2004]
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a) 0.94h
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b) 0.8h
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c)0.75h
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d)0.64h
Explanation
As v2=0.8v1and Kinetic energy E ∝ v2 ∴ E2=(0.8)2E1 thus mgh2=(0.64)mgh1 h2=0.64h1Answer: (d)
Q.14
A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s . It is subjected to a retarding force x=20m to x=30m. Its final kinetic energy will be [ AIIMS 2005]
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a) 475J
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b) 450 J
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c)275J
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d)250J
Explanation
Initial kinetic energy=½ mv2 Ei=½ 10 (10) 2=500 J Now work done by frictionW=∫ Fdx Thus energy lost to overcome friction=25 J ∴ Final kinetic energy=500-25=475J Answer:(a)
Q.15
A force F acting on a body depends on it displacement S as F ∝ S-1/The power delivered by F will depend on displacement as [IIT screen 2001]
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a) S2/3
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b) S1/2
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c) S
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d) S0
Explanation
F ∝ S-1/3 a ∝ S-1/3Integrating both side sides, we find v2 ∝ S2/3or v ∝ S1/3As P=Fv ∴ P ∝ (S-1/3) (S1/3)=S0Answer: (d)
Q.16
A uniform metal chain is placed on a rough table such that one end of it hangs down over the edge of the table. When one third of its length hangs over the edge, the chain starts sliding. then the coefficient of static friction is [ kerala C.E.T 2005]
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a)3/4
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b) 1/4
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c)2/3
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d)1/2
Explanation
The chain starts sliding , when applied force=force of friction (1/3)mg=µ(2/3)mg µ=1/2Answer: (d)
Q.17
A boy holding a gas filled balloon releases it. The balloon going up then its potential energy will
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a) increase
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b) decrease
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c)First increase and then decrease
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d)remain constant
Explanation
Balloon is going up because up of upward thrust. Not due to work done by external agency Answer: (b)
Q.18
A particle moves in a straight line with retardation proportional to displacement. Its loss of kinetic energy for any displacement x proportional to...
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a) x2
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b) ex
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c)x
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d)log ex
Explanation
Retardation is proportional to displacement Thus Retarding force F ∝ x=kxThen work done by retarding force for small displacement dx is dW=FdxNow W=∫Fd W=∫ kxdx=(k/2)x2 Work done by retarding force ∝ x2Work done by retarding force will bring down kinetic energy Thus loss of kinetic energy ∝ x2 Answer:(a)
Q.19
An electric motor creates a tension of 4500N in a hoisting cable and reels it at the rate of 2 m/s. What is the power of the electric motor
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a) 15kW
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b) 9 kW
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c) 225kW
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d) 900HP
Explanation
Power=force × velocity Power=4500 × 2=9000 W=9.0kW Answer: (b)
Q.20
A body of mass 2kg is projected vertically upwards with speed of 3m/s. The maximum gravitational potential energy of the body is
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a)18J
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b) 4.5J
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c)9 J
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d)2.25J
Explanation
At maximum height velocity is zero thus Kinetic energy=0 and Potential energy max ΔK.E=-ΔP.E.Initial KE=½ m v2 Initial KE=½ 2 (3)2 Initial kinetic energy=9 J Answer: (c)
Q.21
A rod of mass m and length l is lying on a horizontal table. Work done in making it stand on one end will be
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a) mgl
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b) mg (l/2)
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c)mg (l/4)
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d)2 mgl
Explanation
Center of mass of rod is at l/2 Weight of rod is or force=mg Displacement of Center of mass in the process of making it stand on one end=l/2Work done=force × Displacement W=mg(l/2) Answer: (b)
Q.22
A particle A makes a perfectly elastic collision with another particle B at rest. They fly apart in opposite direction with equal speeds. The ratio of their mass mA / mB is [ A.M.U.PMT 1997]
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a)1/2
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b) 1/3
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c)1/4
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d)1/ √3
Explanation
Velocity of first particle after collision is given by formula here e=1 and u2=0 Thus Velocity of second particle after collision is given y formula here also e=1 and u2=0 Thus Given v1=- v2 comparing equations for velocity after collision we get m1 - m2=- 2m1 m2=3m1 Thus m1 / m2=1/3 Answer:(b)
Q.23
The potential energy between two atoms in molecule is given bywhere a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when
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a)
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b)
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c) ZERO
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d)
Explanation
For equilibrium F=-dV/dx=0 ∴ -12ax-13 +6 bx-13=0 6bx-7=12ax-13 x6=12a/6b=2a/b thus option d is correct Answer: (d)
Q.24
A shell fired from a cannon explodes in mid air. As a result
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a)its total momentum increases
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b) its total momentum decreases
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c)its momentum and kinetic energy remain unchanged
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d)its total kinetic energy increases
Explanation
Due to explosion chemical energy converted in to kinetic energy of fragmentsAnswer: (d)
Q.25
A bullet of mass m is fired from below into bob of mass M of long simple pendulum. The bullet stays inside the bob and the bob rises to height h. The initial speed of the bullet will be
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a)
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b)
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c)
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d)
Explanation
Let initial speed of bullet be u Let common velocity of bullet and bob be vApplying law of conservation of momentum mu=(M + m) v Thus v=mu / M + m --(1)After this the motion of bob is controlled by the gravity let maximum height attended by bob is h v2=2gh v=√(2gh) --(2) From equation (1) and (2) we get √(2gh)=mu / (M + m) ∴ u=√(2gh) [ M + m / m]Answer: (d)
Q.26
A block of mass m moving with speed v collides with another block of mass 2m at rest. The lighter block comes to rest after collision. What is the value of coefficient of restitution?
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a) 1/2
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b) 1/3
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c)3/4
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d)1/4
Explanation
Suppose the second block moves at a speed v' after collisionBy law of conservation of momentummv=2v' v'=v/2 Velocity of approach=vby definition e=velocity of separation / velocity of approach=1/2 Answer:(a)
Q.27
A ball heats the floor and rebounds after inelastic collision . In this case
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a) the momentum of the ball after collision is the same as that just before collision
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b) the mechanical energy of the ball remains the same in the collision
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c) the total momentum of the ball and the earth is conserved
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d) the total energy of the ball and the earth is conserved
Explanation
Answer: (c)
Q.28
A body is displaced from x=x1 to x=x2 by a force 2x. The work done is [ CPMT 1993]
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a)2x1 (x2 - x1)
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b) 2x2 (x2 - x1)
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c)
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d)
Explanation
Since force is potion dependentAnswer: (c)
Q.29
A point of application of a force F=3i - 2j + 2k N is displaced by x=3i + 2j +k metres. Then the work done by the force is
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a) 5 J
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b) 7 J
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c)9J
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d)11 J
Explanation
W=F.SW=(3i - 2j + 2k) . ( 3i + 2j + k)W=9 - 4 + 2=7JAnswer: (b)
Q.30
Two identical balls A and B moving with velocities +0.5 m/s and -0.3m/s respectively collied head on , elastically. The velocities of the balls A and B after collision will be respectively [ CBSE 1994]
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a) +0.5 m/s and +0.3 m/s
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b) -0.3 m/s and +0.5 m/s
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c)+0.3 m/s and +0.5 m/s
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d)-0.5 m/s and +0.3 m/s
Explanation
Collision is elastic and masses are same thus velocities get exchanged Answer:(b)
0 h : 0 m : 1 s
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