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Q.1
A body under the action of a force F=6 i - 8j + 10k.acquires an acceleration of 1 m/s2 The mass of this body must be [ CBSE-PMT 2009]
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a) 10kg
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b) 20 kg
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c) 10√2kg
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d)2√10kg
Explanation
Magnitude of force |F|=(36+64+100) 1/2=10√2 N a=1 m/ s2 F=ma ∴ m=F/a=10√2 / 1 m=10√2 kg Answer: (c)
Q.2
A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 ( vertically) if the acceleration of system is g/8, then the ratio of masses is ... [ CBSE-PMT 2010]
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a)8:1
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b) 9:7
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c)4:3
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d)5:3
Explanation
Let 'a' be resultant acceleration of massesFor mass m1m1a=m1g - T For mass m2m2a=T - m2g now by adding above two equations we get now a=g/8 given therefore Answer:(b)
Q.3
A 60kg man stands on a spring scale in the lift. At same instant he finds, scale reading has changed from 60 kg to 50kg for a while and comes back to the original mark. What should we conclude? [ AFMC 2007]
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a) The lift was in constant motion upwards
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b) The lift was in constant motion downwards
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c)The lift while in constant motion upwards is stopped suddenly
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d)The lift while in constant motion downwards, is suddenly stopped
Explanation
When the lift moves upward, while decreasing its acceleration person experiences pseudo force on upward direction, where as gravitational force was in downward direction, thus decrease in weight and when lift stops scale come back to original Answer: (c)
Q.4
In Newton's second law F =m a , a is the acceleration of the mass 'm' with respect to [ AFMC 2011]
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a) Any observer
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b) Any inertial observer
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c)an observer at rest only
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d)an observer moving with uniform speed only
Explanation
acceleration is in inertial frame. Answer:(b)
Q.5
A motor cycle is going on an over bridge of radius R. The driver maintains a constant speed. As the motor cycle is ascending on the over bridge the normal force on it... [ AFMC 2006]
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a) increases
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b) decreases
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c) remains the same
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d) fluctuates erratically
Explanation
From figure reaction of normal force on the motor cyclist mgcosθ -R=mv2 /r ∴ R=mgcosθ - (mv2) / r As motorcycle ascend on the over bridge, angle of inclination of motorcycle with vertical θ decreases and becomes zero at the highest point so cosθ goes on increasing and becomes 1 at highest point Value of R is continuously increasing while ascending Answer: (a)
Q.6
The limiting friction is ... [ AFMC 2007]
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a)always greater than the dynamic friction
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b) always less than the dynamic friction
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c)equal to the dynamic friction
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d)sometimes greater and sometimes less than the dynamic friction
Explanation
The limiting friction is the maximum frictional force between any two surfaces in contact. Hence, it is always greater than the dynamic frictionAnswer: (a)
Q.7
A person swimming in a fresh water pool obeys [ AFMC 2003]
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a) Newton's second law
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b) Gravitational law
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c)Newton's third law
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d)Newton's first law
Explanation
Answer: (c)
Q.8
When a car moves on a road with uniform speed of 30km/hr, then the resultant force on the car is ... [ AFMC 2004]
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a) The driving force, drives the car in the direction of propagation of car
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b) The resistive force acts opposite to the direction of propagating of car
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c)zero
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d)none of these
Explanation
When a speed of a car is uniform, acceleration is zero, So net resultant force on the car is zero Answer:(c)
Q.9
A cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4m, then coefficient of friction between the cycle tyres and road is ... [ AFMC 2002]
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a) 0.71
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b) 0.61
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c) 0.31
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d) 0.81
Explanation
The friction between tyre and ground provides a centripetal force so µR=mv2 / r µmg=mv2 / r ∴ µ=v2 / rg Here r=4 m g=9.8 m/s2 v=4.9 m/s By substituting values in the above equation and on solving we get µ=0.61 Answer: (b)
Q.10
A spring of length L of uniform cross-section is spread on a smooth plane. One of its end is pulled by a force F as shown in the figure. The Tension in the spring at a distance 'l' from this end is .. [ AFMC 2010]
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a)
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b)
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c)
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d)
Explanation
Tension in the spring at a distance 'l' Answer: (b)
Q.11
A bullet of mass m is fired from a gun of mass M. The recoiling gun compresses a spring of force constant k by a distance d. Then the velocity of bullet is .. [ AFMC..2011]
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a)
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b)
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c)
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d)
Explanation
Let velocity of bullet be 'v', and mass be 'm'. velocity of gun 'V', and mass of gun be M, since no external force acts on system according to law of conservation of momentum Initial momentum = Final momentum 0 = mv + MV V = -mv/ M As spring compresses by d, so According law of conservation of energy Spring Potential energy = Kinetic energy of gun Answer: (c)
Q.12
Which of the following is self adjusting force? [ AFMC 2009]
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a) kinetic friction
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b) static friction
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c) Nuclear force
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d) None of these
Explanation
When a body is at rest in contact with another body even if a force is applied on it, the friction acting between them is called static friction. If the force adjusts itself and does not let itself to be in motion unless the external applied force overcomes it. Therefore, static friction is a self adjusting force. Answer:(b)
Q.13
A body of mass 4.975kg is lying on the frictionless floor. A mass of 25gm moving with 400m/s strikes the mass and gets embedded in it. The velocity of the body combined after collision is ... [ AFMC 2006]
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a) 2 m/s
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b) 4 m/s
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c) 1 m/s
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d) none of these..
Explanation
After collision smaller mass gets embedded into bigger mass. therefore both the masses have common velocity v let smaller mass be m = 25g = 0.25 kg, initial velocity be u =400m/s Let bigger mass be M = 4.975 kg Now applying law of conservation of momentum: mu - M×0 = ( M+m)v v = mu/ (M+m) v = 2 m/s Answer: (a)
Q.14
The rocket engine lift a rocket from the earth, because hot gases: [ AIIMS 1998]
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a)push it against the air with very high velocity
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b) push it against the earth with very high velocity
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c)heat up the air which lifts the rocket with very high velocity
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d)react against rocket and push it up with very high velocity
Explanation
Rocket exerts force on gases, thus according to third law, gases exerts force on rockettAnswer: (d)
Q.15
A man of mass 60kg records his weight on a weighing machine placed inside a lift. the ratio of weight of man recorded when lift ascending up with uniform speed of 2m/s to when it is descending down with uniform speed of 4m/s [ AIIMS 2007}
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a) 0.5
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b) 1.0
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c)2.0
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d)none of these
Explanation
Net force of reaction acts on a body in a lift when it is accelerating. If lift moves up or down with uniform speed then acceleration a=0, therefore no pseudo force acts and weight of the person will remain same in both the cases ascending or descending ratio 1.0Answer: (b)
Q.16
When a horse pulls a wagon, the force that causes the horse to move forward in the force [ AIIMS 2010]
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a) the ground exerts on it
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b) it exerts on the ground
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c)the wagon exerts on it
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d)it exerts on the wagon
Explanation
As per Newton's third law of motion, horse pulls a wagon, the force that causes the horse to move forward is the force that ground exerts on it. Answer:(a)
Q.17
A person standing in an elevator. in which situation he finds his weight less? [ AIIMS 2005]
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a) When the elevator moves upward with constant acceleration.
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b) When the elevator moves downward with constant acceleration
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c) When the elevator moves upward with uniform velocity
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d) When the elevator moves downward with uniform velocity
Explanation
When lift goes down with positive acceleration pseudo acceleration will be upwards for a person as a result net down word acceleration will be less. Answer: (b)
Q.18
A 1kg particle strikes a wall with a velocity 1m/s at an angle 30° and reflects at the same wall in 0.1 second then the force will be: [ AIIMS 1999]
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a)30√3 N
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b) 0
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c)40√3 N
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d)10√3 N
Explanation
sin component of incident and reflected velocity are equal and same direction there for no change while cos component of velocity gets reversedChange in momentum of the ball perpendicular to the wall=m[ v cosθ -(-vcosθ)] =2mvcosθ=2×1×1×cos30 =√3Rate of change of momentum=Change of momentum / time =√3 / 0.1=10√3 NAnswer: (d)
Q.19
A 150 g tennis ball coming at a speed of 40 m/s is hit straight back by a bat to a speed of 60 m/s. the magnitude of the average force F on the ball, when it is in contact for 5 ms is ... [ AIIMS 2011]
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a) 2500 N
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b) 3000 N
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c)3500 N
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d)4000 N
Explanation
The change in momentum Δp=m( vf - vi)Δp=0.15[60 -(-40)]=15NsF=Δp / ΔtF=3×103 NAnswer: (b)
Q.20
A gardener holds a hosepipe through which water is gushing out at a rate of 4 kg/s with speed 2m/s. The moment the speed of water is increased to 3m/s, the gardener will experience a jerk of ... [ AIIMS 2011]
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a)4 N in backward direction
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b)8 N in forward direction
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c)8 N in backward direction
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d)4 N in forward direction
Explanation
Question options are in Newtons which is incorrect, as jerk is defined as third derivative of displacement, it has unit m/s^3 jerk, j can also be given as j = da/dt = (dF/dt)/m, Answer:(c)
Q.21
A bomb of mass 3.0 kg explodes in air into two pieces of masses 2.0kg and 1.0kg. the smallest mass goes at a speed of 80 m/s. the total energy imparted to the two fragments is.. [ AIIMS 2004]
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a) 1.07 kJ
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b) 2.14 kJ
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c) 2.4 kJ
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d) 4.8 kJ
Explanation
Applying law of conservation of momentum 2×v=1×80 v=40 m/s Total energy of system=½(2×402) + ½(1 ×802) Total energy of system=1600+3200=4800J Answer: (d)
Q.22
If the force on rocket, moving with a velocity 500 m/s is 400N, then the rate of combustion of the fuel will be.. [ AIIMS 1997]
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a)0.8 kg/sec
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b) 10.8 kg/sec
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c)8 kg/sec
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d)1.6 kg/sec
Explanation
We know that for variable mass force Fdm/dt is the rate at which fuel burneddm/dt=F/v dm/dt=400 /500=0.8 kg/secAnswer: (a)
Q.23
The velocity of bullet is reduced from 200m/s to 100 m/s while traveling through a wooden block of thickness 10cm. Assuming it is to be uniform, the retardation will be.. [ AIIMS 2001]
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a) 15×104 m/ s2
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b) 10×104 m/ s2
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c)12×104 m/ s2
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d)14.5 m/ s2
Explanation
displacement s=10cm=0.1mUsing formula v2=u2 + 2as1002=2002 - 2a(0.1)on solving we get a=15×104 m / s2Answer: (a)
Q.24
An 80 kg person is parachuting and is experiencing a downward acceleration of 2.8 m/s2 . The mass of the parachute is 5kg. The upward force on the open parachute is .. ( g=9.8 m/s2)
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a)595 N
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b) 675 N
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c)456 N
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d)925 N
Explanation
The upward acceleration is (9.8-2.8)=7 m/s2Total mass=80+5=85 kg So, net upward force is F=85×7=595 N Answer:(a)
Q.25
A parachutist after bailing outfalls 50 m without friction. When parachute opens. If decelerates at 2m/sHe reaches the ground with a speed of 3m/s. At what height, did he bail out? take g=9.8 m/s2 [ AIIMS 2009]
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a) 182 m
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b) 91 m
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c) 111 m
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d) 293 m
Explanation
Initially it was free fall up to 50m there fore velocity at 50m from bailed out will be v=(2gh) 1/2 v=(2×9.8×50)1/2 v=14√5 Now parachutist reaches ground with velocity 3 m/s thus by formula v2=u2 + 2 as here acceleration a=2 m/s2 32=(14√5)2 + 2×2×s on solving we get s=243 m∴ Total height from where ne bailed out=244+50=293 m Answer: (d)
Q.26
A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively.. [ AIEEE2002]
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a)g, g
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b) g-a, g-a
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c)g-a, g
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d)a, g
Explanation
ab : Acceleration of ballam : Acceleration of observerFor the man standing in liftThe acceleration of ball=ab - am=g - a acceleration of mass is a,as lift is moving down with acceleration abFor man standing on ground:man is now at rest am=0The acceleration of ball=ab - am The acceleration of ball=ab - 0=g Answer: (c)
Q.27
A spring balance is attached to the ceiling of lift. A man hangs his bag on the spring and the spring reads 49N, when the lift is stationary. If the lift moves downwards with an acceleration of 5 m/s2, the reading of the spring balance will be [ AIEEE-2003]
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a) 24N
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b) 74N
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c)15N
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d)49N
Explanation
mass of bag=Gravitational force/ gravitational acceleration=49/9.8=5Kg Lift accelerates down ward thus pseudo force F will be upward thus net force will be F'=Gravitational force - F F'=49 - 5×5=49-25=24N Answer: (a)
Q.28
A light string passes over a smooth light pulley connects two blocks of masses m1 and m2 ( vertically). If the acceleration of the system is g/8, then the ratio of the masses is [ AIEEE 2002]
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a) 8:1
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b) 9:7
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c)4:3
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d)5:3
Explanation
For mass m1m1g -T=m1a For mass m2 T- m2 g=m2a On solving above equation we get a=Here a=g/8 substituting value in above equation we getDividing and multiplying rhs of equation by m2 we get Answer:(b)
Q.29
One end of a mass less rope, which passes over a mass less and frictionless pulley P is tied to a hook C while the other end is free. Maximum tension that the rope can bear is 840N. With what value of maximum safe acceleration (in m/s2) can a man of 60 kg climb on the rope [ AIEEE 2002]
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a) 16
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b) 6
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c) 4
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d) 8
Explanation
Given that rope can bear force of 840 N. So maximum tension in rope should not exceed 840N When man starts to climb on rope with acceleration 'a' rope experience downward force along with gravitational force Thus T=mg + ma a=T/m - g a=840 /60 -10 a=14-10=4 m/s2 Answer: (c)
Q.30
Three identical blocks of mass m=2 kg are drawn by a force F=10.2N on a friction less surface, then what is the tension (in N) in the sting between the blocks B and C? [ AIEEE 2002]
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a)9.2
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b) 3.4
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c)4
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d)9.8
Explanation
Now F=(m+m+m)aF=3ma a=F/3m=10.2/6 m/s2Tension between block B and C pull the block C∴ T=ma=2× (10.2/6 )=3.4NAnswer: (b)
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