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Physics NEET MCQ
Work, Energy And Power Mcq
Quiz 8
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Q.1
A cylinder of mass 10kg is sliding on a plane with an initial velocity of 10m/s. If coefficient of friction between surface and cylinder is 0.5, then before stopping it will describe . [Pb. PMT 2001]
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a) 12.5 m
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b) 7.5 m
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c)5 m
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d)10 m
Explanation
Final velocity is zero. And retardation due to frinction=µg 0=u2 - 2(µg)S S=u22/2μg=(10 × 10)/(2 × 0.5 × 10)=10mAnswer: (d)
Q.2
It is easier to draw up a wooden block along an inclined plane than to haul it vertically, principally because . [CPMT 1977; JIPM]
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a) The friction is reduced
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b) Only a part of the weight has to be overcome
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c)The mass becomes smaller
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d)‘g’ becomes smaller
Explanation
Opposing force in vertical pulling=mg But opposing force on an inclined plane is mg sinθ, which is less than mg. Answer:(b)
Q.3
A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20J in 1s. Then
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a) the tension in the string is Mg
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b) the tension in the string is F
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c) the work done by the tension on the block is 20 J in 1s
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d) the work done by the force of gravity is -20J in 1s
Explanation
As the force F is constant , the tension in the string is F Answer: (b)
Q.4
The force acting on a particle of mass 1kg, starting from rest from the origin, is shown in the figure. The velocity of the particle at x=2 m is
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a)2√2 m/s
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b) √6 m/s
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c)2 m/s
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d)√3 m/s
Explanation
Area under the graph=Work Work done=change in kinetic energy 3=½ × 1 × v2 v=√6 m/s Answer: (b)
Q.5
A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of with the x-axis and displaces it along the x-axis by 4 metres. The work done by the force is . [MP PET 2003]
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a)2.5 J
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b) 40 J
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c)7.25 J
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d)20 J
Explanation
W=Fs cosθ=10 × 4 × cos60=20J Answer:(d)
Q.6
A ball of mass m moves with speed v and strikes a wall having infinite mass and it returns with same speed then the work done by the ball on the wall is . [BCECE 2004]
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a) Zero
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b) m/v.J
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c) mvJ
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d) v/m J
Explanation
The ball rebounds with the same speed. So change in it's Kinetic energy will be zero i.e. work done by the ball on the wall is zero. Answer: (a)
Q.7
A block of mass M is attached with a spring of spring constant k. The whole arrangement is placed on a vehicle as shown in the figure, if the vehicle starts moving towards right with an acceleration a ( there is no friction any where) the
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a)maximum elongation in the spring is (Ma/k)
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b) maximum elongation in the spring is (2Ma/k)
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c) maximum compression in the spring is (2Ma/k)
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d) maximum compression in the spring is (Ma/k)
Explanation
Pseudo force of magnitude Ma will act on the block leftward. for maximum elongation ½ kx2=Max x=2Ma/k ∴ maximum elongation in the spring is (2Ma/k)Answer: (b)
Q.8
A force F=-k (yi + xj) where k is a positive constant, acts on a particle moving in x-y plane starting from origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y -axis to the point (a,a). THe total work done by the force F on the particle is [ JEE 98]
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a) -2ka2
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b) 2ka2
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c)-ka2
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d)ka2
Explanation
From graph equation of line joining origin and ( a,a) is y=x thus dy=dxLet small displacement vector dS=dxi + dyjNow formula for force depends on position thus Answer: (c)
Q.9
A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the position B. The angle θ at which the speed of the bob is half of that at A, satisfies .. [ JEE 2008]
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a)θ=π/4
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b) π/4 < θ < π/2
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c)π/2 < θ < 3π/4
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d)3π/4 < θ < π
Explanation
We know that velocity at A should b v=√(5gl) the only bon will rich at point B Now from conservation of energy Thus option "d" is correct Answer:(d)
Q.10
A particle is moved from (0, 0) to ( a, a) under a force F=( 3i + 4j) from two paths. Path 1 is Op and path 2 is OQP. Let W1 and W2 be the work done by this force in these two paths. Then
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a) W1=W2
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b) W1=2W2
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c) W2=2W1
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d) W2=4W1
Explanation
Here magnitude and direction of force is not changing . hence given force is conservative. So work done in both the cases is same Answer: (a)
Q.11
A block of mass 0.18 kg is attached to a spring of force constant 2N/m. the coefficient of friction between the block and the floor is 0.Initially the block is at rest and the spring is un-starched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06m and comes to rest for the first time. the initial velocity of the block is m/s is V=N/Then N is [ JEE 2011]
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a)1.8
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b) 4
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c)2
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d)3.6
Explanation
From the law of conservation of energy Kinetic energy=Work done by friction + Potential energy of spring N=4Answer: (b)
Q.12
An ideal spring with spring constant k is hung from the celling and a block of mass M is attached to its lower end. The mass is released with the spring initially stretched. Then the maximum extension in the spring is [ JEE 2002]
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a) 4Mg/K
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b) 2Mg/k
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c)Mg/k
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d)Mg/2k
Explanation
Work done by gravity=potential energy of spring Mgx=½ kx2 x=2Mg/kAnswer: (b)
Q.13
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)
Q.14
A rubber ball of mass m and radius r is submerged into water of density ρ to a depth h and released. To what height the ball will jump above the surface of water? Disregard the resistance of water and air?
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a)
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b)
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c)
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d)
Explanation
Taking zero potential energy level at water surface, we get Potential energy of ball at depth h=-mgth and work against buoyant force=(4/3)πr3gh If ball rises to height x then potential energy of ball at height x from water surface=mgx thus Note on the basis of dimensional analysis dimension of option 'a' is only in length Answer: (a)
Q.15
With what minimum speed v must a small ball should be pushed inside a smooth vertical tube from height h so that it may reach the top of the tube? Radius of the tube is R
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a)
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b)
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c)
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d)
Explanation
From conservation of energy ½ mv2=mg(2R - h) v=√[2g(2R - h)]Answer: (d)
Q.16
An object is thrown straight up. Which of the following is true about the sign of work done by the gravitational force while the object moves up and then down?
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a)Work is positive on the way up, work is positive on the way down
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b)Work is negative on the way up, work is negative on the way down
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c)Work is negative on the way up, work is positive on the way down
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d)Work is positive on the way up, work is negative on the way down
Explanation
Answer: (c)
Q.17
A particle is given an initial speed u inside a smooth spherical shell of radius R=1 m that it is just able to complete the circle. Acceleration of the particle when its velocity become vertical is
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a) g√10
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b) g
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c)g√2
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d)g√6
Explanation
To just complete the circle, velocity of the block when it is vertical is v=√(3gR) Answer:(a)
Q.18
Which of the following is not a unit for power?
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a) joule – second
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b) watt
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c) newton-meter per second
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d) horsepower
Explanation
Answer: (a)
Q.19
If the momentum of a body is increased n times, its kinetic energy increases
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a)n times
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b) √n times
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c)2n times
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d)n2 times
Explanation
E=P2/2m i.e. if P is increased n times then E will increase n2 timesAnswer: (d)
Q.20
If the system in the figure is released from rest in the configuration shown, find the velocity of the block Q after it has fallen through a distance 10 metres, given mass of P=mass of Q=10 kg
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a) 8 m/sec
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b) 8.85 m/sec
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c)9.5 m/sec
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d)10 m/sec
Explanation
If Q falls through 10 m, then P goes up by 5 m, similarly VQ=2VP from conservation of energy Answer: (b)
Q.21
The work done against gravity in taking 10 kg mass at 1m height in 1sec will be . [RPMT 2000]
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a)49 J
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b)196 J
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c)98 J
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d)None of these
Explanation
Work done=mgh=10 × 9.8 × 1=98J Answer:(c)
Q.22
A body of mass 10kg at rest is acted upon simultaneously by two forces 4 N and 3N at right angles to each other. The kinetic energy of the body at the end of 10 sec is . [Kerala (Engg.)]
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a) 100 J
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b) 50 J
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c) 300 J
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d) 125 J
Explanation
Resultant force is 5 N acceleration=F/m=5/10=0.5 m/s2 velocity at end of 10 sec v=0 + a t=(0.5) × 10=5 m/s Kinetic energy=½ mv2 Kinetic energy=½ × 10 × 25=125J Answer: (d)
Q.23
A bomb of 12 kg divides in two parts whose ratio of masses is 1:If kinetic energy of smaller part is 288 J, then momentum of bigger part in kgm/sec will be
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a) 37
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b)72
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c)108
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d)Data is incomplete
Explanation
If v is the velocity of smaller particle then velocity of bigger particle will b v/4 from conservation of momentum Kinetic energy of smaller part=½ mv2=288 J Kinetic energy of bigger part Now kinetic energy E=P2 (2 × 4m) Answer: (a)
Q.24
The potential energy of a conservative system is given by U(X)=(x2 - 5x) J.Then the equilibrium position is at....... (where x in m)
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a)x=1.5 m
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b)x=2m
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c)x=2.5 m
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d)x=5 m
Explanation
For a conservation field. Force, F=-dU/dxF=2x - 5At equilibrium position, F=02x - 5=0 x=2.5Answer: (c)
Q.25
Two particles 1 and 2 are allowed to descend on two friction less chord OP and OQ. the ratio of the speeds of the particles 1 and 2 respectively, when they reach on the circumference is
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a) 1/4
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b) 1/2
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c)1
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d)1/2√2
Explanation
½ mv2=mgR(1 - cosθ) ½mv'2=mg(2R)v2 / v'2=(1 - cos60) / 2=1/2 Answer:(b)
Q.26
The potential energy of a particle varies with distance x as shown in the graph. The force acting on the particle is zero at
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a) C
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b) B
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c) B and C
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d) A and D
Explanation
F=-dU/dx From graph it is clear that at point B and C slope of the graph is zero Answer: (c)
Q.27
A man starts walking from a point on the surface of earth (assumed smooth) and reaches diagonally opposite point. What is the work done by him . [DCE 2004]
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a)Zero
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b)Negative
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c)Positive
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d)Nothing can be said
Explanation
As surface is smooth so work done against friction is zero. Also the displacement and force of gravity are perpendicular so work done against gravity is zero.Answer: (a)
Q.28
Two bodies of masses m and 2m have same momentum. Their respective kinetic energies E1 and E2 are in the ratio . [MP PET 1997; KC]
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a)1 : 2
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b)1 : √2
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c)2 : 1
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d)1 : 4
Explanation
E=P2/2m if monemtum is same then E proportional to 1/mAnswer: (c)
Q.29
The kinetic energy of a body moving a straight line varies with time as shown in fiure. The force acting on the body is
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a) zero
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b) constant
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c)directly proportional to velocity
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d)inversly proportional to velocity
Explanation
Kinetic energy ∝ t v ∝ √t or a ∝ 1/√ ( by taking derivative of v with respect to time) Or F ∝ 1/√t=1/v Answer:(d)
Q.30
A heavy particle of mass m makes complete revolution in a smooth circular tube in a vertical plane. Its greatest speed is n times the least speed. If the least speed is u, what is the speed at the instant, when it is moving vertically?
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a)
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b)
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c)
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d)
Explanation
Answer: (b)
0 h : 0 m : 1 s
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