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Quiz 12
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Q.1
Which of the following statements is not true . [CMC Vellore 198]
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a)The coefficient of friction between two surfaces increases as the surface in contact are made rough
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b) Rolling friction is greater than sliding friction
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c)The force of friction acts in a direction opposite to the applied force
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d)The coefficient of friction between wood and wood is less than 1
Explanation
Sliding friction is greater than rolling friction. Answer: (b)
Q.2
A force of 5 N acts on a body of weight 9.8 N. What is the acceleration produced in m/sec[NCERT 1990]
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a)49.00
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b)1.46
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c) 5.00
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d) 0.51
Explanation
As weight = 9.8 N Mass = 1 kg Acceleration = Force/Mass = 5/1 = 5m/s2 Answer: (c)
Q.3
A student attempts to pull himself up by tugging on his hair. He will not succeed. [KCET 2005]
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d)As the force applied is internal to the system.
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a) As the force exerted is small
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b) Newton's law of inertia is not applicable to living beings.
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c)The frictional force while gripping, is small
Explanation
As by an internal force momentum of the system can not be changed. Answer:(d)
Q.4
A man is standing on a weighing machine placed in a lift. When stationary his weight is recorded as 40 kg. If the lift is accelerated upwards with an acceleration of 2m/s2, then the weight recorded in the machine will be . [MP PMT 1994]
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a) 32 kg
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b) 42 kg
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c) 40 kg
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d) 48 kg
Explanation
In stationary lift man weighs 40 kg i.e. 400 N. When lift accelerates upward it's apparent weight=m(g+a)=40(10+2)=480N i.e. 48 kg For the clarity of concepts in this problem kg-wt can be used in place of kg. Answer: (d)
Q.5
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.the weight of the block is . [AIEEE 2003]
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a)2 N
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b) 50 N
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c) 20 N
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d)100 N
Explanation
F=W/μ ; W=μF=0.2 ×10=2N Answer: (a)
Q.6
The maximum static frictional force is
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a) Equal to twice the area of surface in contact
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b) Equal to the area of surface in contact
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c)Independent of the area of surface in contact
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d)None of the above
Explanation
Answer: (c)
Q.7
A force of 10 Newton acts on a body of mass 20kg for 10 seconds. Change in its momentum is. [MP PET 2002]
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a)5 kg m/s
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b) 200 kg m/s
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c)100kg m/s
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d)1000kg m/s
Explanation
dp=F × dt=10 × 10=100 kg m/s Answer:(c)
Q.8
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5m/s2 the reading of the spring balance will be . [AIEEE 2003]
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a) 49 N
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b) 74 N
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c) 24 N
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d) 15 N
Explanation
When the lift is stationary W=mg=49=m*9.8 so m=5kg When the lift is moving downward with an acceleration R=m(9.8-a)=5[9.8-5]=24N Answer: (c)
Q.9
A body of mass 2 kg moving on a horizontal surface with an initial velocity of 4 m/sec comes to rest after 2 sec. If one wants to keep this body moving on the same surface with a velocity of 4 m/sec, the force required is. [NCERT 1977]
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a)8 N
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b) Zero
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c)4 N
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d)2 N
Explanation
from equation of motion v=u +at 0=4 + a ×2 retardation=2m/s2 To keep body moving at constant velocity applied force must be equal to retarding force F=m × a=2 × 2=4NAnswer: (c)
Q.10
A bird weighs 2 kg and is inside a closed cage of 1 kg. If it starts flying, then what is the weight of the bird and cage assembly. [AFMC 1997]
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a)1.5 kg
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b)3 kg
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c)2.5 kg
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d)4 kg
Explanation
When the bird flies, it pushes air down to balance its weight. So the weight of the bird and closed cage assembly remains unchanged.Answer: (b)
Q.11
A wagon weighing 1000 kg is moving with a velocity 50km/h on smooth horizontal rails. A mass of 250 kg is dropped into it. The velocity with which it moves now is . [MP PMT 1994]
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a) 2.5 km/hour
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b) 40 km/hour
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c)20 km/hour
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d)50 km/hour
Explanation
According to principle of conservation of linear momentum 1000 ×50=1250 × v so v=40km/hr Answer:(b)
Q.12
Newton's third law of motion leads to the law of conservation of. [Manipal MEE 199]
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a) Angular momentum
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b) Mass
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c) Energy
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d) Momentum
Explanation
Answer: (d)
Q.13
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is μk=0.5 . [AIEEE 2005]
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a)100 m
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b) 800 m
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c)400 m
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d)1000 m
Explanation
s=u2/2μg=(100)2/(2 ×0.5 ×10)=1000 Answer: (d)
Q.14
A block weighs W is held against a vertical wall by applying a horizontal force F. The minimum value of F needed to hold the block is. [MP PMT 1993]
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a) Less than W
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b) Greater than W
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c)Equal to W
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d)Data is insufficient
Explanation
Here downward force is W and upward force is frictionW=f But f=µN , Here N=F W=µF since µ <1 F > WAnswer: (b)
Q.15
Gravels are dropped on a conveyor belt at the rate of 0.5 kg/sec. The extra force required in newtons to keep the belt moving at 2 m/sec is. [EAMCET 1988]
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a)1
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b)4
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c)2
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d)0.5
Explanation
F=u(dm/dt)=2 ×0.5=1N same amount of force is required to keep the belt moving at 2 m/s Answer:(a)
Q.16
A force of 98 N is required to just start moving a body of mass 100 kg over ice. The coefficient of static friction is
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a) 0.6
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b) 0.2
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c) 0.4
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d) 0.1
Explanation
μ=F/R=F/mg=98/(100 ×9.8)=0.1 Answer: (d)
Q.17
A car turns a corner on a slippery road at a constant speed of 10m/s . If the coefficient of friction is 0.5, the minimum radius of the arc in meter in which the car turns is
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a)20
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b) 5
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c)10
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d)4
Explanation
v=√μrg so r=v2/μg=100/(0.5 × 10)=20 Answer: (a)
Q.18
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is μk=0.5 . [AIEEE 2005]
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a) 100 m
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b) 800 m
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c)400 m
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d)1000 m
Explanation
s=u2/2µg=(100)2/(2*0.5*10)=1000Answer: (d)
Q.19
The limiting friction is
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a) Always greater than the dynamic friction
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b) Equal to the dynamic friction
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c)Always less than the dynamic friction
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d)Sometimes greater and sometimes less than the dynamic friction
Explanation
Answer:(a)
Q.20
A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of 1 m / s, then after 4 sec his displacement relative to the ground will be. [CPMT 1988, 89,]
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a) 5 m
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b) 3.2 m
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c) 4.8 m
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d) 3.0 m
Explanation
Let Vmg be the velocity of man with respect to ground Vtg be the velocity of trolley with respect to groundVmt be the velocity of man with respect to trolley=1 m/s Let M be mass of person=8kgLet M' be the mass of trolley=320kgfrom relative motion Vtg=Vmg - Vmt --eq(1) Initially trolley and man were at rest initial momentum=0 From law of conservation of momentum VmgM + VtgM'=0 Substituting value of Vtg from equation (1) we get VmgM + (Vmg - Vmt)M'=0 Vmg(M+M') - VmtM' Vmg(320+80)=1(320) Vmg=320/400=0.8 m/s Thus displacement of man with respect to observer=Vmg × time=0.8 × 4=3.2 mAnswer: (b)
Q.21
Work done by a frictional force is
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a)Negative
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b)Zero
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c)Positive
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d)All of the above
Explanation
Work done by friction can be positive, negative and zero depending upon the situationAnswer: (d)
Q.22
A particle is moving with a constant speed along a straight line path. A force is not required to. [AFMC 2001]
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a) Increase its speed
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b) Change the direction
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c)Decrease the momentum
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d)Keep it moving with uniform velocity
Explanation
Answer: (d)
Q.23
The coefficient of friction μ and the angle of friction λ are related as
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a)sinλ=μ
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b) tanλ=μ
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c)cosλ=μ
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d)tanμ=λ
Explanation
Answer:(b)
Q.24
A jet engine works on the principle of . [CPMT 1973; MP P]
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a) Conservation of mass
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b) Conservation of linear momentum
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c) Conservation of energy
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d) Conservation of angular momentum
Explanation
Answer: (b)
Q.25
Maximum value of static friction is called . [BHU 1995; RPET]
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a)Limiting friction
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b)Normal reaction
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c)Rolling friction
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d)Coefficient of friction
Explanation
Answer: (a)
Q.26
When two surfaces are coated with a lubricant, then they . [AFMC 1998, 99;]
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a)Stick to each other
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b) Roll upon each other
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c)Slide upon each other
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d)None of these
Explanation
Answer: (c)
Q.27
A man of weight 75 kg is standing in an elevator which is moving with an acceleration of 5 m/s in upward direction the apparent weight of the man will be (g=10) . [Pb. PMT 2004]
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a)1425 N
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b)1250 N
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c)1375 N
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d)1125 N
Explanation
The apparent weight,R=m(g+a)=75(10+5)=1125N Answer:(d)
Q.28
A parachutist of weight ‘w’ strikes the ground with his legs fixed and comes to rest with an upward acceleration of magnitude 3 g. Force exerted on him by ground during landing is. [EAMCET 1988]
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a) w
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b) 3w
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c) 2w
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d) 4w
Explanation
Resultant force is w +3w=4w Answer: (d)
Q.29
Three blocks of mass m, 2m and 3m are placed adjacent to each other on a frictionless horizontal surface as shown in figure. A constant force of magnitude F is applied to the right. Which of the following statement is true
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a)The acceleration of the blocks will very according to their mass
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b) The acceleration of each block will be same F/m
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c)The net force acting on each block is the same
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d)The net force acting on block 3m is three times that the net force acting on m
Explanation
Total mass of three blocks=6m common acceleration of three block=F/6mNet force on block m=m(F/6m)=F/6Net force on block 2m=2m(F/6m)=F/3Net force on block 3m=3m(F/6m)=F/2Answer: (d)
Q.30
A free body of mass 8 kg is travelling at 2 m/s in straight line. At a certain instant, the body splits into equal parts due to internal explosion, which releases 16 joules of energy. Neither part leaves the original line of motion. Finally
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a) both parts continue to move in the same direction as that of the original body
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b) one part comes to rest and the other moves in the same direction asthat of the original body
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c)one part comes to rest and the other moves in the direction opposite to that of the original body
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d)one part moves in the same direction and the other in the direction opposite to that of the original body
Explanation
From law of consevation of momentummv=m1 V1 + m2 V2 8 × 2=4V1 + 4V2V1 + V2=4 --eq(1)but ½ mV2=½ m1 (V1)2 + ½ m2 (V2)2½ × 8 × (2)2=½ ×4×(V1)2 + ½ ×4 ×(V2)2(V1)2 + (V2)2=16 --eq(2) From eq(1) (4 - V2)2 + (V2)2=16 on solving V2=4m/sec Therefore V1=0. Hence one part comes to rest while the other continues to move in the same directionAnswer: (b)
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