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Physics NEET MCQ
Quiz 9
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Q.1
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.2
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.3
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.4
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.5
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.6
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)
Q.7
A particle of mass 10g is kept on the surface of a uniform sphere of mass 100kg and radius 10cm. Find the work done against the gravitational force between them to take the particle far away from the sphere ( G=6.67 × 10⁻¹¹ Nm2 / kg2 ) [ AIIMS 2008]
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a)3.33 × 10⁻¹⁰ J
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b) 13.34 × 10⁻¹⁰ J
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c)6.67 × 10⁻¹⁰ J
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d)6.67 × 10⁻⁹ J
Explanation
W=GMm /R Answer: (c)
Q.8
A spring 40 mm long is stretched by applying a force. If 10N force is required to stretch the spring through one mm, then work done in stretching the spring through 40mm is [ AIIMS 1998]
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a)24 J
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b) 8 J
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c)56 J
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d)54 J
Explanation
1 mm=0.001 m; 40 mm=4 × 10⁻²mForce constant k=F / x=10 /0.001=104 N /m Work done=Potential energy of springWork done=½(k x2)Work done=½ ( 104 × 16 × 10⁻⁴)Work done=8 JouleAnswer: (b)
Q.9
A child builds a tower from three blocks. The blocks are uniform cubes of side 2cm. The block are initially lying on the same horizontal surface and each block has a mass of 0.1kg. The work done by the child is
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a) 4 J
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b) 0.04 J
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c)6 J
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d)0.06 J
Explanation
Work done by the child=Increase in potential energy of the block Ground block=0 First block work=(0.1g)(2 × 10⁻²) Second block work=(0.1g)(4 × 10⁻²) total work=0.06J Answer:(d)
Q.10
When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by . [Orissa JEE 2002]
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a)100 J
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b) 300 J
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c)200 J
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d)400 J
Explanation
Given when x=2 cm=2 × 10⁻² , enrgy U=100J Thus from formula for potential energy of spring 100=½ k × (2 × 10⁻²) 2 Thus k=50 × 104 Now change in potential energy=½ k ( x'2 - x2) By substituting value of x'=4 cm=4 × 10⁻² and x=2 cm=2 × 10⁻² and We get ΔU=300J Answer:(b)
Q.11
A light inextensible string that goes over a smooth fixed pulley as shown in the figure connected two blocks of masses 0.36kg and 0.72kg. Taking g=10 m/s2, the work done ( in joules) by the string on the block of mass 0.36kg during the first second after the system is released from rest
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a)8 J
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b)4 J
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c)16 J
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d)20 J
Explanation
Tension in string Displacement in 1st secondAcceleration Now initial velocity is zero Thus displacement of 0.3kg mass=½ a t2 s=10/6 Work done=T(s)=(4.8) × (10/6)=8 J Answer:(a)
0 h : 0 m : 1 s
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