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Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 9
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Q.1
Two particles of masses m1, m2 move with initial velocities u1 and uOn collision, one of the particles get excited to higher level, after absorbing energy E. If final velocities of particles be v1 and v2 then we must have :
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a)
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b)
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c)
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d)
Explanation
According to law of conservation of energy Energy of system before collision = energy of system after collision Now energy E is absorbed by particle for excitation after collision Answer:(d)
Q.2
Two similar springs P and Q have spring constants KP and KQ, such that KP > KQ. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs WP and WQ are related as, in case (a) and case (b), respectively :
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a) WP < WQ; WQ < WP
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b) WP = WQ; WP > WQ
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c) WP = WQ; WP = WQ
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d) WP > WQ; WQ > WP
Explanation
Work done = Energy of the spring Case a, same displacement Therefore EP > EQ or WP > WQ Case (b) Same force applied Now F=kx WP < WQ Answer:(d)
Q.3
A block of mass 10 kg, moving in x direction with a constant speed of 10ms-1, is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m. Its final KE will be :
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a) 250 J
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b) 475 J
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c) 450 J
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d) 275 J
Explanation
Initial K.E. = ½ mv2 = 500 J Work done by retarding force, retarding force is position dependent Final K.E. = Initial K.E. – Work done by friction Final K.E. = 500 – 25 = 475 J Answer:(b)
Q.4
A ball is thrown vertically downwards from a height of 20 m with an initial velocity u It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity u0 is : (Take g = 10 ms-2)
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a) 10 ms-1
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b) 14 ms-1
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c) 20 ms-1
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d) 28 ms-1
Explanation
Final potential energy after rebound = mgh = m×10×20=200m Final PE. = 50%(Initial K.E+ PE) 800 = v2 + 2× 10×20 v2 = 400 v= 20 m/s Answer:(c)
Q.5
A body of mass 1kg begins to move under the action of a time dependent force F=(2ti+3t2 k)N. What will be power developed by the force
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a) (2t2+3t3 ) W
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b) (2t2+4t4 ) W
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c) (2t3+3t4 ) W
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d) (2t3+3t5 ) W
Explanation
Acceleration Now velocity From (i) Answer:(d)
Q.6
A bullet of mass 10g moving horizontally with a velocity of 400 ms–1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be :-
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a) 120 ms–1
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b) 160 ms–1
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c) 100 ms–1
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d) 80 ms–1
Explanation
According to law of conservation of momentum for block and bullet ⇒ 4 =2v1 + 0.01v2 …(i) Applying work energy theorem for block W = ∆KE Substituting value in (i) 4 =2×1.4 + 0.01v2 v2 = 120 m/s Answer:(a)
Q.7
A particle moves from a point (-2i+ 5j) to (4j + 3k ) when a force of (4i+3j) N is applied. How much work has been done by the force ?
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a) 5 J
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b) 2 J
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c) 8 J
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d) 11 J
Explanation
Displacement S= (4j + 3k ) - (-2i+ 5j) = 2i –j +3k W = F.s = (4i+3j) ∙ (2i – j +3k) = 8 -3 = 5J Answer:(a)
Q.8
This question has statement I and Statement II. Of the four choice given after the statements, choose the one that best describes the two statements. Statement – I : A Point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.
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a) Statement - I is true, Statement - II is true, statement - II is a correct explanation of Statement - I
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b) Statement - I is true, Statement - II is true, statement - II is not a correct explanation of Statement – I
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c) Statement - I is true, Statement - II is false
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d) Statement – I is false, Statement – II is true
Explanation
Maximum energy loss when inelastic collision takes place. So both the particle will stick with each other with common velocity v’ According to law of conservation of momentum mv = (M+m)v’ Initial kinetic Final kinetic energy Thus statement I is wrong Answer:(d)
Q.9
The work done on a particle of mass m by a force (K being a constant of appropriate dimension), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is … [ IIT Advance 2013]
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a)
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b)
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c)
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d) zero
Explanation
Answer:(d)
Q.10
raph A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms−2). [ IIT Advance 2013] Q250A) The speed of the block when it reaches the point Q is ..
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a) 5 ms−1
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b) 10 ms−1
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c) 10√3 ms−1
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d) 20 ms−1
Explanation
mgcosθ + mv2R is normal at Q Answer:(a)
Q.11
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale. [ IIT Advance 2014]
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a)
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b)
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c)
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d)
Explanation
For free fall of ball initial kinetic energy =0 Kinetic energy increases with increase in velocity When ball strike surface. Kinetic energy reduced to zero quickly and it compress the ball. Velocity of ball becomes momentarily zero. Then bounce back Answer:(b)
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