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Laws Of Motion Mcq
Quiz 9
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Q.1
Two persons standing on a floating boat run in succession along its length with a speed 4.2 m/s relative to the boat and dive off from the end. the mass of each man is 80kg and that of boat is 400 kg. If the boat was initially at rest, find the final velocity of the boat. Neglect friction.
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a) 0.6 m/s
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b) 0.7 m/s
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c) 0.1 m/s
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d) 1.3 m/s
Explanation
Mass of boat after first person dive off 480kg Let VMB : Velocity of man with respect to boat VBG = Velocity of boat with respect to ground VMG = Velocity of man with respect to ground According to law of conservation of momentum 80 VMG + 480 VBG = 0 80(VMB+ VBG)+ 480 VBG = 0 560VBG = -80 VMB Given VMB = 4.2 m/s VBG =- 0.6 m/s ( negative sign indicate boat is moving back) Thus momentum of boat = 0.6×480 =- 288 kg m/s Second person dive off, the mass of boat = 400 kg According to law of conservation of momentum 80 VMG + 400 VBG = -288 80(VMB+ VBG)+ 400 VBG = -288 480VBG = -80 VMB-288 Given VMB = 4.2 m/s VBG = -1.3 m/s ( Negative sign indicate velocity is opposite to velocity after first person dive off) Answer: (d)
Q.2
A ball hits the floor and rebounds after inelastic collision. In this case
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a)the momentum of ball just after collision is the same as that just before collision
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b) the mechanical energy of the ball remains the same in the collision
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c)total momentum of the ball and the earth is conserved
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d)the total energy of the ball and the earth is conserved
Explanation
Answer: (c)
Q.3
A particle is describing motion in a horizontal plane in contact with the smooth inside surface of fixed hemispherical bowl with its axis vertical. The radius of the bowl is 25cm and the radius of the orbit of motion of the particle is 15cm. Then the period of revolution in seconds of the particle is
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a) 2π/5
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b) 2π/7
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c)2π/3
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d)π/3
Explanation
From figure Nsinθ=mrω2 and Ncosθ=mg tanθ=ω2r / g and sinθ=15/25=3/5 ∴ tanθ=3/4 Answer: (b)
Q.4
A bom of 12 kg exploded into two pieces of mass 4kg and 8kg. The velocity of 8kg mass is 6m/s. the kinetic energy of the other is
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b) 32 J
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c)24 J
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a) 48 J
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d)288 J
Explanation
Momentum of 8kg mass=8×6=48 kg m/sLet v be the velocity of 4kg massFrom law of conservation of momentum 48 + 4(v)=0 v=12 m/s Thus kinetic energy=(12) mv2 K.E=(1/2) ×4 ×144=288 J Answer:(d)
Q.5
An object initially at rest explodes into three fragments. The momentum of two parts are 20i and 10j. The magnitude of momentum of the third part is
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a) 30 unit
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b) 20 unit
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c) 10√5 unt
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d) 10 unit
Explanation
Applying law of conservation of momentum 20i + 10j + P3=0 P3=-20i - 10j |P|=√(202 +102)=10√5 unit Answer: (c)
Q.6
A machine gun of mass M fires a bullet of mass m. The speed of the bullet with respect to gun is v. The recoil speed of the gun as observed by an observer at rest on the ground is
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a)mv/M
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b) Mv/m
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c)mv/(M+m)
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d)mv/(M-m)
Explanation
Consider u be the recoil speed of gun, then speed of bullet with respect to person on ground=v-uuse law of conservation of momentum to solve Answer: (c)
Q.7
A block of rests on rough horizontal surface. A force of 200N is applied on the body. The block acquires a speed of 4m/s, starting from sets in 2sec. What is the value of coefficient of friction?
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a) 10/√3
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b) √3/10
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c)0.47
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d)0.185
Explanation
Acceleration a=(4-0 )/ 2=2 m/s2 Resultant force=Applied - friction 30×2=200 - f f=140 N friction=µmg 140=µ× 30× 9.8 µ=0.47Answer: (c)
Q.8
A body of mass M just starts sliding down an inclined plane ( rough) with inclination θ. such that tanθ=1/The force acting on body down the plane in this position is
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a) Mg
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b) Mg/3
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c)(2/3)Mg
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d)Mg/√10
Explanation
tanθ=1/3 ∴ sinθ=1/√10Downward force=Mgsinθ Down ward force=Mg(1/√10 Answer:(d)
Q.9
Shell fired from a cannon explodes in mid air. As a result
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a) its total momentum increases
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b) its total momentum decreases
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c) its momentum and kinetic energy remains unchanged
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d) its total kinetic energy increases
Explanation
During explosion , chemical energy is converted in kinetic energy . Hence kinetic energy increases Answer: (d)
Q.10
A cart of mass has a block of mass m attached to it as shown in figure. The coefficient of friction between the block and the cart is µ. What is the minimum acceleration of the cart so that the block m does not fall
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a)µg
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b) g/µ
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c)µ/g
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d)Mµg/m
Explanation
Let a be the acceleration of trolleyFrom the FBD of block N=ma ∴ µN=mg µma=mg a=g/µAnswer: (b)
Q.11
The tension in the cable of 1000 kg elevator is 1000 kg wt, the elevator
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a) is ascending upwards
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b) is descending downwards
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c)may be at rest or acceleration
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d)may be at rest or in uniform motion
Explanation
Answer: (d)
Q.12
A body of mass m moving with a constant velocity v hits another body of the same mass moving with same velocity but in opposite direction and stick to it. the velocity of the compound body after collision is
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a) v
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b) 2v
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c)zero
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d)v/2
Explanation
Answer:(c)
Q.13
A 50 kg man is standing on flat boat at rest in river. He moves 5 m to north and halts. If the boat has mass of 450kg, then the boat moves through
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a)0.5 metres to the south
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b) 0.55 metres to the south
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c)0.5 metres to the north
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d)zero metre
Explanation
Answer:(a)
Q.14
If the system shown in figure is attached to the roof of a vehicle accelerating horizontally with acceleration of 10 m/s2 , then waht will be the acceleration of the masses as seen by the observer sitting in the vehicle
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a) 10√2 m/s2
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b) 10 m/s2
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c) 2.5√2 m/s2
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d) 2.5 m/s2
Explanation
As car acceleration and gravitational acceleration are perpendicular geffective = √(102 +102) = 10√2 from formula Answer:(c)
Q.15
A ship of mass 3×107 kg initially at rest is pulled by force 5×104N through a distance of 3m. Assume that the resistance of water is negligible, the speed of the ship is
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a) 1.5 m/s
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b) 60 m/s
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c) 0.1 m/s
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d) 5 m/s
Explanation
find acceleration using a = F/m Then use v2 = 2as Answer: (c)
Q.16
A man getting down a running bus falls forward because
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a) due to inertia of rest, rod is left behind and man reaches forward
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b) due to inertia of motion upper part of body continues to be in motion in forward direction while feet comes to rest as soon as they touch road
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c) he leans forward as a matter of habit
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d) of the combined effect of all the three factors stated in(a), (b) and (c)
Explanation
Answer: (b)
Q.17
A body of mass 2kg moving on a horizontal surface with initial velocity of 4m/s comes to rest after 2sec. If one wants to keep this body moving on the same surface with velocity of 4 m/sec, the force required is
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a) 8 N
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b) 4 N
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c) zero Newton
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d) 2 N
Explanation
Applied force = resistive force Answer: (b)
Q.18
A wooden box of mass 8kg slides down the inclined plane of inclination 30° to the horizontal with constant acceleration of 0.4 m/sWhat is the coefficient of friction between box and inclined plane? [ g=10 m/s2
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a) 0.53
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b) 0.90
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c) 0.82
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d) 1.0
Explanation
resultant downward force = gravitational force - friction ma = mgsinθ - µ mgcosθ 0.4 = 10sin30 - µgcos30 0.4 = 5 - µ ×10× √3 / 2 0.8 = 10 - µ10√3 µ = 9.2/10√3 µ ≈ 0.53 Answer:(a)
Q.19
A player caught a cricket ball of mass 150gm moving at a rate of 20 m/s. IF the catching process be completed in 0.1s, the force of the blow exerted by the ball on the hands of the player is
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a) 0.3N
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b) 30 N
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c) 300 N
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d) 3000 N
Explanation
Answer: (b)
Q.20
On a stationary sail-boat air is blown at the sail from a fan attached to the boat. The boat will
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a)remain stationary
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b) spin around
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c)move in opposite direction that in which air is blown
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d)move in the direction in which air is blown
Explanation
Air is blown at sails. No external force is acting on boat thus it will remain stationaryAnswer: (a)
Q.21
Two weights w1 and w2 are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up at an acceleration g, the tension in the string will be
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a) 4w1 w2 / (w1 + w2)
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b) 2w1 w2 / (w1 + w2)
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c)w1 - w2 / (w1 + w2)
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d)w1 w2 / [2(w1 + w2)]
Explanation
effective gravity on each mass=g-(-g)=2g m/s2 For m1 m1a=2m1g - T for m2m2 a=T - 2m2g dividing (i) by (ii) Answer: (a)
Q.22
A man is at rest in the middle of a pond on a perfectly mooth ice. HE can get himself to the shore by making use of Netons's
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a) first law
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b) second law
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c)third law
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d)all the laws
Explanation
Answer:(c)
Q.23
A canon after firing recoils due to
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a) conservation of energy
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b) backward thrust of gases produced
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c) Newton's third law of motion
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d) Newton's first law of motion
Explanation
Answer: (c)
Q.24
Inertia is that proprty of a body by virtue of which the body is
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a)unable to change by itself the the state of rest
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b) unable to change by itself the state of uniform motion
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c)unable to change by itself the direction of motion
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d)unable to change by itself the state of rest or of uniform linear motion
Explanation
Answer: (d)
Q.25
A jet plane flies in air because
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a) the gravity does not act on the bodies moving with high speeds
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b) the thrust of the jet compensates for the force of gravity
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c)the flow of air around the wings causes an upward force, which compensates for the force of gravity
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d)the weight of air whose volume equal to the volume of the plane is more than that the weight of plane
Explanation
Answer: (c)
Q.26
You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against surface
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a) by jumping
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b) by spitting or sneezing
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c)by rolling your body on the surface
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d)by running on the plane
Explanation
Answer:(b)
Q.27
A hockey player receives a corner shot at a speed of 15m/s at an angle 30° with y-axis and then shoots the ball along x-axis with a speed of 30m/s. If the mass of the ball be 150g and it remains in contact with the hocky stick for 0.1s the force exerted on the ball along x-axis is
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a) 281 N
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b) 187.5 N
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c) 572 N
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d) 375 N
Explanation
F = ΔP/Δt Answer: (c)
Q.28
A wooden box of mass 8kg slides down the inclined pane of inclination 30° to the horizontal with constant acceleration of 0.4 m/sWhat is the force of friction between box and inclined plane? [ g=10 m/s2)
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a)36.8 N
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b) 76.8 N
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c)65.6 N
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d)None of above
Explanation
Resultant acceleration=downward gravitational acceleration - retardation due to friction 0.4=gsinθ - f/m 0.4=10sin30 - f/8 3.2=80(1/2) -f f=36.8 NAnswer: (a)
Q.29
Which of the following works on the principle of conservation of linear momentum
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a) Jet
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b) Aeroplane
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c)Rocket
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d)All the above
Explanation
Answer: (c)
Q.30
As an inclined plane is made slowly horizontal by reducing the value of an angle θ with horizontal, the component of weight parallel to the plane of a block resting on the inclined plane
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a) decreases
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b) remains same
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c)increases
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d)increases if the plane is smooth
Explanation
Answer:(a)
0 h : 0 m : 1 s
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