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Laws Of Motion Mcq
Quiz 11
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Q.1
Which of the following statement about friction is true?
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a) friction can be reduced to zero
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b) frictional force an accelerate a body
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c)frictional force is proportional to the area of contact between the two surfaces
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d)kinetic friction is always greater than rolling friction
Explanation
Answer: (d)
Q.2
A force of 1N acts upon a mass of 1kg for 1 second and gives it a momentum P and kinetic energy E. The same force acts upon the same mass for displacement of 1m and gives it a momentum P' and kinetic energy E'. Then
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a) P=P' , E=E'
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b)P > P' , E > E'
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c)P < P' , E < E'
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d)P > P' , E < E'
Explanation
case i)a=F/m=1 m/s2v=0 + 1×1=1m/saverage velocity (v + u)/ 2=1/2 Momentum P=mv=1kg×m/seckinetic energy=½ mv2=1/2 jouleCase ii)a=1 m/s2from equation of motion1=+ ½ (t)2 t=2secv=0 + 1×2 Momentum P'=mv=1 × 2=2 Kinetic energy E'=½ m × v2=½ 1×(2)2=2 joule ∴ P' > P and E' > E Correct option is (c) Answer:(c)
Q.3
A 40 kg slab rests on frictionless floor as shown in figure. A 10kg block rests on the top of a slab. The static coefficient of friction between the block and slab is 0.6 while the kinetic friction is 0.The 10kg blcok is acted upon by a horizontal force of 100N. If g=9.8 m/s2, the resultant acceleration of the slab will be
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a) 0.98 m/s2
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b) 1.47 m/s2
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c) 1.52 m/s2
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d) 6.1 m/s2
Explanation
Static friction between the block is less than applied force thus Force on the slab ( m=40 kg )=reaction of frictional force on the upper block ∴ 40a=µk × 10 × g a=0.98 m/s2 Answer: (a)
Q.4
When a bus suddenly takes a turn, the passengers are thrown outwards because of . [AFMC 1999; CPMT]
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a) Inertia of motion
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b) Speed of motion
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c) Acceleration of motion
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d) Both (b) and (c)
Explanation
According to Newton's first law of motion Answer: (a)
Q.5
To avoid slipping while walking on ice, one should take smaller steps because of the. [BHU 1999; BCECE]
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a)Friction of ice is large
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b) Friction of ice is small
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c)Larger normal reaction
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d)Smaller normal reaction
Explanation
Answer: (b)
Q.6
Which one of the following is not used to reduce friction . [Kerala (Engg.)]
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a) Oil
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b) Sand
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c)Ball bearings
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d)Graphite
Explanation
Sand is used to increase the friction.Answer: (b)
Q.7
A bullet of mass M is fired with velocity of 50m/s at an angle of θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob of mass 3M suspended by a massless string of length (10/3) m and gets embedded in the bob. After the collision, the string moves to an angle of 120o. What is the angle θ
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a)cos-1 (4/5)
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b) cos-1(5/4)
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c)sin-1(4/5)
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d)sin-1(5/4)
Explanation
Velocity of bullet at highest point of path=Vcosθ∴ MVcosθ=(3M+M)V'or V'=Vcosθ/4=50cosθ/4 again ½ (M+3M)V'2=Mgl(1-cos120)2V'2=gl [ 1+ (1/2)] or V'=√(3gl)Answer: (a)
Q.8
A diwali rocket is ejecting 0.05 kg of gases per second at a velocity of 400 m/sec. The accelerating force on the rocket is. [NCERT 1979; DPM]
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a) 20 dynes
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b) 22 dynes
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c)20 N
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d)1000 N
Explanation
Use formula F=u(dm/dt)=400*0.05=20NAnswer: (c)
Q.9
An automobile travelling with a speed of 60km/h can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e. 120 km/h, the stopping distance will be . [AIEEE 2004]
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a)20 m
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b)60 m
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c)40 m
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d)80 m
Explanation
From equation of motionv2=u2 + 2as final velocity in both the cases=0 and retardation is constnt thus S ∝ u2 by taking the ratio s/20=(120/60)2 s=80 m Answer:(d)
Q.10
A cork is submerged in water by a spring attached to the bottom of a pail. When the pail is kept in a elevator moving with an acceleration downwards, the spring length . [EAMCET (Engg.)]
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a) Increases
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b) Remains unchanged
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c) Decreases
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d) Data insufficient
Explanation
Decreases Density of cork=d, Density of water=ρ Resultant upward force on cork=V(ρ-d)g This causes elongation in the spring. When the lift moves down with acceleration a, the resultant upward force on cork=V(ρ-d)(g-a) which is less than the previous value. So the elongation decreases. Answer: (c)
Q.11
A 30 gm bullet initially travelling at 120 m/s penetrates 12 cm into a wooden block. The average resistance exerted by the wooden block is. [AFMC 1999; CPMT]
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a)2850N
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b)2000N
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c)2200 N
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d)1800 N
Explanation
From equation of motionV2=u2 + 2asv=0, u=120m/s, s=12cm=12×10-2 m1202=2a×12×10-2a=1202 / 12×10-2a=12×104average retardation=a'=6×104Average force=ma'=30×10-3×6×104 Average Force=1800 N Answer: (d)
Q.12
If a person with a spring balance and a body hanging from it goes up and up in an aeroplane, then the reading of the weight of the body as indicated by the spring balance will . [AIIMS 1998; JIP]
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a)Go on increasing
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b)First increase and then decrease
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c)Go on decreasing
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d)Remain the same
Explanation
Initially due to upward acceleration apparent weight of the body increases but then it decreases due to decrease in gravity.Answer: (b)
Q.13
An object will continue moving uniformly until . [CPMT 1975]
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a)The resultant force acting on it begins to decrease
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b) The resultant force is at right angle to its rotation
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c)The resultant force on it is zero
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d)The resultant force on it is increased continuously
Explanation
According to Newtons first law of motion Answer:(c)
Q.14
A rocket of mass 1000 kg exhausts gases at a rate of 4 kg/sec with a velocity 3000 m/s. The thrust developed on the rocket is. [Orissa JEE 2005]
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a) 12000 N
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b) 800 N
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c) 120 N
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d) 200 N
Explanation
Use formula F=v(dm/dt) here v=3000 m/s and (dm/dt)=4 kg/sec Answer: (a)
Q.15
In doubling the mass and acceleration of the mass, the force acting on the mass with respect to the previous value
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a)Decreases to half
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b)Increases two times
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c)Remains unchanged
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d)Increases four times
Explanation
Force=Mass x Acceleration. If mass and acceleration both are doubled then force will become four times.Answer: (d)
Q.16
A body, whose momentum is constant, must have constant . [AIIMS 2000]
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a)Force
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b)Acceleration
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c)Velocity
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d)All of these
Explanation
For a given mass P proportional to v . If the momentum is constant then it's velocity must have constantAnswer: (c)
Q.17
A cold soft drink is kept on the balance. When the cap is open, then the weight. [AFMC 1996]
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a)Increases
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b) First increases then decreases
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c)Decreases
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d)Remains same
Explanation
Gas will come out with sufficient speed in forward direction, so reaction of this forward force will change the reading of the spring balance Answer:(b)
Q.18
A cricket ball of mass 250 g collides with a bat with velocity 10 m/s and returns with the same velocity within 0.01 second. The force acted on bat is. [CPMT 1997]
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a) 25 N
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b) 250 N
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c) 50 N
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d) 500 N
Explanation
Force=m(dv/dt)change in velocity dv=10 - (-10)=20dt=0.01 sec, m=0.25kg Answer: (d)
Q.19
A block of mass 5kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting. [Pb. PMT 2002]
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a)10 m
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b)6 m
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c)8 m
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d)2 m
Explanation
Force is perpendicular , particle will not accelerate along horizontal, but change its direction, thus horizontal displacement=velocity × times=1.5 × 4=6.0 m vertical initial velocity=0 acceleration=1 s=ut + ½ at2 s=½ (1) (4)2=8 Thus resultant displacement=√(62 + 82)=10 mAnswer: (a)
Q.20
A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball falls . [EAMCET (Med.) 1]
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a)Outside the car
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b)In the car to the side of the person
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c)In the car ahead of the person
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d)Exactly in the hand which threw it up
Explanation
Horizontal velocity of ball and person are same so both will cover equal horizontal distance in a given interval of time and after following the parabolic path the ball falls exactly in the hand which threw it up.Answer: (d)
Q.21
A force of 100 dynes acts on mass of 5 gm for 10 sec. The velocity produced is. [MNR 1987]
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a) 2 cm/sec
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b) 200 cm/sec
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c)20 cm/sec
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d)2000 cm/sec
Explanation
Acceleration a=F/m=100/5=20 cm/s2 Now v=at=20 ×10=200cm/s Answer:(b)
Q.22
A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is . [AIEEE 2003]
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a) Both the scales read M/2 kg each
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b) The scale of the lower one reads M kg and of the upper one zero
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c) Both the scales read M kg each
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d) The reading of the two scales can be anything but the sum of the reading will be M kg
Explanation
As the spring balance are massless therefore both the scales read M kg each. Answer: (c)
Q.23
A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of length of the chain, that can hang over one edge of the table is. [CBSE PMT 1990]
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a)20%
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b) 35%
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c)25%
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d)15%
Explanation
Let L be the length of chainLet M be the mass of chain mass per unit of chain - M/L let 'l' be the length of the chain hang over edge of table and (L-l) is the length of chain on tableNow friction of chain on table=gravitational force on hangover chainAnswer: (a)
Q.24
A body of mass 2 kg is hung on a spring balance mounted vertically in a lift. If the lift descends with an acceleration equal to the acceleration due to gravity ‘g’, the reading on the spring balance will be . [NCERT 1977]
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a) 2 kg
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b) (2×g)kg
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c)(4×g)kg
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d) Zero
Explanation
Reading on the spring balance=m (g – a) and since a=g hence Force=0Answer: (d)
Q.25
Rocket engines lift a rocket from the earth surface because hot gas with high velocity. [AIIMS 1998; JIP]
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a) Push against the earth
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b) React against the rocket and push it up
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c)Push against the air
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d)Heat up the air which lifts the rocket
Explanation
It works on the principle of conservation of momentum. Answer:(b)
Q.26
A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket. [CBSE PMT 1994]
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a) Comes down with uniform acceleration
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b) Moves up with a uniform velocity
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c) Moves round the earth in a geostationary orbit
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d) Moves up with uniform acceleration
Explanation
T = 2π√(l/g) T will decrease, If g increases.It is possible when rocket moves up with uniform acceleration. Answer: (d)
Q.27
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.28
Two carts of masses 200 kg and 300 kg on horizontal rails are pushed apart. Suppose the coefficient of friction between the carts and the rails are same. If the 200 kg cart travels a distance of 36 m and stops, then the distance travelled by the cart weighing 300 kg is . [CPMT 1989; DPMT]
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a) 32 m
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b) 16 m
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c)24 m
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d)12 m
Explanation
s2 / s1=(m1 / m2)2=(200/300)2 ; s2=s1 × (4/9)=36×(4/9)=16OR Same amount of force is applied for same time on both the carts from F=Δp/Δt 200 u1 / Δt=300 u2 / Δt u1/u2=300/200=3/2 retardation=µg from equation of motion u12 / u22=2(µg)36 / 2(µg)S S=16mAnswer: (b)
Q.29
Newton's first law of motion describes the following . [MP PMT 1996]
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a)Energy
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b)Inertia
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c)Work
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d)Moment of inertia
Explanation
Newton’s first law of motion defines the inertia of body. It states that every body has a tendency to remain in its state (either rest or motion) due to its inertia. Answer:(b)
Q.30
A rope of length 5m is kept on frictionless surface and a force of 5N is applied to one of its end. Find tension in the rope at 1m from this end. [RPET 2000]
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a) 1 N
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b) 4 N
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c) 3 N
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d) 5 N
Explanation
Let M be the mass of rope. mass per unit length of rope=M/5 acceleration of rope=Force / mass=5/M Tension pulls 4m rope of mass=4M/5 Tension on=(4M/5) (5/M)=4N Answer: (b)
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