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Physics NEET MCQ
Quiz 7
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Q.1
A body is slowly lowered on a massive platform moving horizontally at a speed of 4m/s , through what distance will the body slide relative to the platform? [ The coefficient of friction is 0.2; g=10 m/s2]
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a) 4 m
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b) 16 m
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c)4 m
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d)20 m
Explanation
The frictional force between the body and the platform=µmg, where m is the mass of the body Initially the relative velocity=4 m/s The relative retardation=µg=0.2 × 10=2 m/s2 If S id the relative displacement before the relative velocity becomes zero, we have for formula v2=u2 + 2as0=42 - 2 × 2 × S S=16/4=4 m Answer: (c)
Q.2
A block slides down an inclined plane of slope θ with constant velocity . It is then projected up the plane with an initial velocity u. How far up the incline will it move before coming to rest.
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b) u2/ 2gsinθ
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c) u2/ 4gsinθ
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d) 2u2/ gsinθ
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a) u²/ gsinθ
Explanation
Constant velocity indicate frictional force acting against gravitational force thus mgsinθ=µmgcosθ For upward motion retardation is=downward gravitional force + frictional force=2gsinθ, now ½ mu2=2mgsinθ × l l=u2/ 4gsinθ Answer: (c)
Q.3
A mass of 1 kg suspended on a thread deviates through an angle 30°. Find the tension of the thread at the moment the weight passes through the position of equilibrium
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a)12.4 N
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b) 15 N
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c)24.8 N
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d)6.2 N
Explanation
At the moment the weight passes through the position of equilibrium the tension of the thread By the conservation of energy mgh=½ m v2v=√2ghBut h=l(1 - cos30) substituting value of mv2 /l in equation for tension we getT=mg [ 1 + 2(1 - cos30)] Given m=1 kg ; g=9.8 m/s2; cos30=√3/2 ∴ T=12.4 N Answer: (a)
Q.4
A block of mass m moving with speed v compresses a spring through a distance x before its speed is halved. What is the value of spring constant
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a)
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b)
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c)
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d)
Explanation
Initial kinetic energy=½ mv2 Final energy=½ m (v/2)2 + ½ k x2 By principle of conservation of energy Answer: (d)
Q.5
The work done against gravity in moving the block of mass m through a distance S up the slope is
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a) mh
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b) mgS
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c)mS
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d)mgh
Explanation
Answer:(d)
Q.6
A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator . The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the friction on the block in time t will be
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a) zero
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b) mgvtcos2θ
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c) mgvtsin2θ
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d) mgvtsinsin2θ
Explanation
Since Block is at rest frictional force f=mgsinθ dispacement=vtsinθ Work W=f.S=mgsinθ.vt.sinθ=mgvtsin2θ Answer: (c)
Q.7
A car is moving with constant speed of 20 m/s against a resistance of 100N. The power exerted by the car is
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a)2 kW
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b) 5 W
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c)200 W
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d)1 kW
Explanation
P=F.vAnswer: (a)
Q.8
Two equal masses are attached to the two ends of the spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each masses is
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a) ½ kx2
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b) - ½ kx2
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c)¼ kx2
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d)- ¼ kx2
Explanation
Energy stored in spring=½ kx2 Total work done by spring=-½ mv2 work done on each mass=by the string=- ¼kx2Answer: (d)
Q.9
The work done by the external force on a system equals the change in
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a) total energy
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b) kinetic energy
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c)potential energy
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d)none of these
Explanation
Answer:(a)
Q.10
Two springs of spring constants 1000 N/m and 2000 N/m are stretched with same source. They will have potential energy in the ratio of
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a) 2:1
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b) 22 : 12
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c) 1 : 2
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d) 12 : 22
Explanation
Potential energy of spring=½ k x2 and F=kx thus x=F/k Potential energy=½ k ( F/k)2 Potential energy=F2/2k F2=E × 2k Given force on both spring is same thus E1 × 2k1=E2 × 2k2 E1 / E2=k2/k1=2000/1000=2:1 Answer: (a)
Q.11
A ball falls under gravity from height 10 m with an initial velocity u. It hits the ground, loses 50% of the energy in collision and it rises to the same height. What is the value of u
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a)14 m/s
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b) 7 m/s
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c)28 m/s
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d)9.8 m/s
Explanation
Let v be the velocity when it hits the ground.Then v2 - u2=2asv2 - u2=2 × 9.8 × 10v2=u2 + 196Now let v' be the velocity after imapact and it reaches the same height 10 m V'2 - 0=2 × 9.8 × 10V'2=196V'=14 m/s Given there is 50% energy loses Ratio of kinetic energy loses thus before impact and after impact=2 Answer: (a)
Q.12
The potential energy of a particle varies with position x according to the relation U(x)=x3 - 4x . The point x=2 is point of
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a) stable equilibrium
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b) Unstable equilibrium
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c)neutral equilibrium
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d)none of the above
Explanation
F=-dU/dx=-3x2 + 4 at x=2 , F=-8 unit and U=0 Since, force is not zero, hence it is not in equilibriumAnswer: (d)
Q.13
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) kinetic energy of the block at time t
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b) work done by friction
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c)total work done on the block
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d)acceleration of block
Explanation
Since both observers have same and uniform velocity they are in inertial frame., the acceleration of the block will be same Answer:(d)
Q.14
A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown
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a) vertically downward
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b) vertically upward
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c) horizontally
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d) the speed does not depend on the initial direction
Explanation
Answer: (d)
Q.15
An ideal massless spring S can be compressed one metre by a force of 100N. The same spring is placed at the bottom of a frictionless inclined plane inclination at 30° to horizontal. A block M of mass 10 kg is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring 2 metre. What is the speed of the mass just before it reaches the spring
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a)√20 m/s
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b) √30 m/s
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c)√10 m/s
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d)√40 m/s
Explanation
Applied force on the spring, F=kxx=F/k=100/1=100 N/m Let the mass M slides a distance S metres along the inclination before hitting the spring. The spring gets compressed by 2 metres. Hence the mass M slides a total distance ( S + 2) metre along the incline . From the geometry of figure height of M=(S + 2)sin30 above the bottom . Thus h=(S + 2)/2 Potential energy of mass M=Mgh=Mg(S + 2)/2 When the spring is compressed, energy is converted in potential energy of spring Mass M has falls at height of Ssin30=S/2=1 m before colliding with spring Gain in K.E=Loss in P.E ½ M v2=Mgh v=√(2gh) v=√( 2 × 10 × 1)=√20 m/sAnswer: (a)
Q.16
A small block of mass 100g is pressed against a horizontal spring fixed at one end and compression is 5cm. The spring constant is 100 N/m. When the block moves horizontally it leaves the spring. Where will it hit the ground 2m below the spring
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a) 1.5 m from free end of the spring
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b) Horizontal distance of 2 m from end of the spring
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c)0.5 m from free end of the spring
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d)Horizontal distance of 1 m from end of the spring
Explanation
Let v be the velocity when it leaves the spring Then kinetic energy of block=potential energy of spring ½ mv2=½ kx2 Time to fall vertical distance of 2 metres from spring Horizontal distance d=velocity × t Answer: (d)
Q.17
The work done by all the forces ( external and internal) on a system equals the change in
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a) total energy
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b) kinetic energy
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c)potential energy
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d)non of these
Explanation
Answer:(b)
Q.18
A body is projected at an angle of 30° to the horizontal with kinetic energy 40 J. What will be the kinetic energy at the top-most point?
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a) 25 J
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b) 40 J
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c) 30 J
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d) 20 J
Explanation
At the topmost point , the horizontal component is ucosθ Initial kinetic energy=½ mu2=40 J Kinetic energy at topmost point=½ m u2 cos2 θ on substituting value of ½mu2 we get Kinetic energy at topmost point=40cos2θ Kinetic energy at topmost point=40 × 3 / 4=30 J Answer: (c)
Q.19
A bob is suspended from a crane by a cable of length l=5m. The crane and load are moving at a constant speed v. the crane is stopped by a bumper and the bob on the cable swings out an angle of 60°. The initial speed v is ( g=9.8 m/s2)
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a)10 m/s
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b) 7 m/s
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c)4 m/s
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d)2 m/s
Explanation
Change in kinetic energy=potential energy of bob ½ mv2=mgl(1 - cos60) Answer: (b)
Q.20
A block of mass 1kg sliding down a curved track that is one quadrant of a circle of radius 1m. Its speed at the bottom is 2m/s. The work done by the frictional force is
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a) -8J
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b) +8J
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c)9J
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d)-9J
Explanation
Here work done by frictional force=change in kinetic energy - loss in potential energy work done by frictional force=( ½ × 1 × 4) - ( 1 × 10 × 1)=-8J Answer: (a)
Q.21
A spring with spring constant k when stretched through 1 cm, the potential energy is U. If it is stretched by 4 cm. The potential energy will be . [Orissa PMT 2004]
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a) 4U
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b) 16 U
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c) 8U
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d) 2U
Explanation
Potential energy U=(1/2)kx2 so U is propotinal to x2 that is why If elongation made 4 times then potential energy will become 16 times. Answer: (b)
Q.22
A weightless rigid rod AB of length l carries two equal masses m , one secured at the end and other at the middle of the rod as shown in figure. The rod can rotate in vertical plane around the hinge at A. The minimum horizontal velocity required to be imparted to the end B of rod so as to make the rod go around in a complete circle is
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a)√(4gl)
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b) √(5gl)
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c)√(24gl/5)
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d)√(24gl/7)
Explanation
If v be the velocity of B, then v/2 is the velocity of C.From conservation of energy Answer: (c)
Q.23
The potential energy of a particle varies with position x according to the relation U(x)=x3 - 4x . The point x=2 is point of
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a) stable equilibrium
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b)unstable equilibrium
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c)neutral equilibrium
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d)none of the above
Explanation
F=-dU/dx=-3x2 + 4 , at x=2, F=-8 unit and U=0 Since force is not zero, hence it is not in equilibriumAnswer: (d)
Q.24
The kinetic energy acquired by a body of mass m is travelling some distance s, starting from rest under the actions of a constant force, is directly proportional to.. [Pb. PET 2000]
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a)mo
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b)m2
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c)m
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d) √m
Explanation
From equation of motionv2=u2 + 2aSsince u=0 v2=2aS but a=F/m v2=2(F/m)S Now kinetic energy=½ mv2 Kinetic energy=½ × m × [ 2(F/m)S] Thus kinetic energy ∝ mo Answer:(a)
Q.25
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to . [AIEEE 2004]
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a) x2
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b) x
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c) ex
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d) logex
Explanation
This condition is applicable for simple harmonic motion. As particle moves from mean position to extreme position its potential energy increases according to expression U=(1/2)kx2 and accordingly kinetic energy decreases Answer: (a)
Q.26
A body at rest may have
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a)Energy
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b)Speed
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c)Momentum
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d)Velocity
Explanation
Answer: (a)
Q.27
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.28
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.29
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.30
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)
0 h : 0 m : 1 s
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