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Quiz 5
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Q.1
A block of mass m is at rest under the action of force F against a wall as shown in figure. Which of the following statements is correct [ IIT 2005]
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a)f=mg [ where f is the friction force]
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b) F=N [ where N is normal force]
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c)F will not produce torque
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d)N will not produce torque
Explanation
The varies forces acting on the block are shown in figure1)Block remains stationary thus frictional force=weight f=mg 2) Torque due to F and mg is zero as the line of action passes through the center 3)Body is in rotational equilibrium sum of torque is zeroso torque due to friction and normal reaction force (N) must be zeroThus torque due to friction + torque due to Normal=0 Now line of action of f and N may not be passing through center Thus f does not produce torque and N does not produce torques is WRONGAnswer: (d)
Q.2
The position time graph of a particle of mass 0.1kg is shown in figure. The impulse at t=2sec is [AIIMS 2005]
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a) 0.2kg ms-1
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b) 0.02kg ms-1
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c)0.1kg ms-1
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d)0.4kg ms-1
Explanation
From the graph, it is clear that up to t=2.0s, the body moves with a constant velocity=slope of the position time graph=2 m/s After t=2 sec position time graph is parallel to time axis, which means that the body comes to rest Thus change in velocity=dv=2 m/s Impulse=change in momentum Impulse=mdv=0.1 ×2=0.2 kg ms-1Answer: (a)
Q.3
A player caught a cricket ball of mass 150g moving at the rate of 20m/s. If the catching process be completes in 0.1s, the force of the blow exerted by the ball on the hands of the player is [ kerala CET 2005]
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a) 0.3 N
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b) 30 N
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c)300 N
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d)3000 N
Explanation
Force=Change in momentum / time Initial momentum of ball=0.15 ×20=3.0 kg ms-1Final momentum=0 Change in momentum=-3.0 ms-1Thus force on the ball=-3.0/0.1=30 N force on the hand=-Force on the ball force on the hand=-(-30)=30N Answer:(b)
Q.4
Two masses M and M/2 are joined to gather by means of light in-extensible string passed over a frictionless pulley. When the bigger mass is released, the small one will ascend with an acceleration of .. [ keralaCET 2005]
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a) g/3
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b) 3g/2
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c) g
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d) g/2
Explanation
Both the blocks form one system and let acceleration of blocks be a Let T be the tension in spring Then For Block M Ma=Mg - T For Block M/2 (M/2)a=T - (M/2) g Solving above equation for a we get Answer: (a)
Q.5
A book is lying on the table. What is the angle between the action of the block on the table and the reaction of the table on the book [ kerala PMT 2005]
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a)0°
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b) 30°
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c)90°
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d)180°
Explanation
Action of book on table is vertically downward while reaction of the table on the book is vertically upwards. Therefore, angle between the two=180°Answer: (d)
Q.6
A man holds the book weighing 10N between the hands and keep it free from falling by pressing both the hands against the book with force of 25N each. The coefficient of friction between the book and hand would be [ CET 1998]
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a) 0.1
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b) 0.2
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c)0.4
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d)0.6
Explanation
Here down ward force mg=10 N since book is in equilibrium upward friction force f=10 N Force by one hand =25NForce applied by both hands is=2× 25 N = 50N thus Normal force on hand=25 N Now µ=f/ N µ=10/50=0.2Answer: (b)
Q.7
When a bicycle is in motion, the force of friction exerted by ground on the two wheels is such that it acts [ IIT 1990]
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a) in the backward direction on the front wheel and in the forward direction on the rear wheel.
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b)in the forward direction on the front wheel and in the backward direction on the rear wheel
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c)in the backward direction on the both the front and the rear wheel
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d)in the forward direction on the both the front and the rear wheel
Explanation
The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel. The friction force on the rear wheel is more than that on the front wheel if the bicycle accelerates. These two are equal if the bicycle moves with a constant speed. The friction force on the rear wheel is less than that on the front wheel if the bicycle decelerate. Answer:(a)
Q.8
While walking on ice, one should take small steps to avid slipping. This is because, small steps ensure
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a) Large friction
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b) smaller friction
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c) large normal force
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d) smaller normal force
Explanation
Answer: (c)
Q.9
Two cars of unequal masses are having similar tyres. If they are moving with at the same initial speed, the minimum stopping distance
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a)is smaller for the heavier car
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b) is smaller for the lighter car
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c)is same for both cars
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d)cannot be predicted
Explanation
Car stops due to friction between tyre and roadcoefficient of friction will be same for both the carsNow f=µ N for lighter car mass be mbut N=mg f=µmg a=f/m=µg for heavier car mass be M f=µMg a'=f/m=µg from above retardation a and a' is same for both cars and initial velocity is same thus stopping distance will be same from equation -u=2as Answer: (c)
Q.10
in order to stop a car in shortest distance on the horizontal road, one should
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a) apply the brakes very hard, so that the wheels stop rotating
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b) apply the brakes hard enough to just prevent slipping
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c)press and release the brakes
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d)shut the engine off and should not apply brakes
Explanation
car have its momentum when we apply brakes, and wheels are rotatingIf wheels stop rotating then car will slip and driver will loose control , thus option (b) is correctAnswer: (b)
Q.11
A weight W rests on a rough horizontal plane. If the angle of friction be θ, the least force that will move the body along the plane will be [ IIT 1987]
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a) W cosθ
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b) W tanθ
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c) W cotθ
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d) W sinθ
Explanation
angle of friction=angle of repose and tanθ=µ frictional force f=µmg W mg f=Wµ f=W tanθ Answer: (b)
Q.12
A rocket with a lift-off mass 3.5×104kg is blasted upwards with an initial acceleration of 10 m/sThen, initial thrust of blast is [ AIEEE 2003]
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a)3.5×105N
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b) 7.0×105 N
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c)14.0×105 N
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d)1.75×105 N
Explanation
mass m=3.5×104 kg and a=10 m/s2F=ma F=3.5×104 × 10 F=3.5×105Answer: (a)
Q.13
If the force on a rocket moving with velocity of 300 m/s is 210 N, then the rate of combustion of the fuel is [ CBSE 1999]
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a) 0.7 kg/s
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b) 1.4 kg/s
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c)0.07 kg/s
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d)10.7 kg/s
Explanation
Force on rocket is given by equationHere V=300 m/s , F=210 N and dm/dt is rate of combustion of fueldm/dt=F/v=210/300=0.7 kg/sAnswer: (a)
Q.14
A 5000 kg rocket is set for vertical firing. The exhaust speed is 800m/s. To give an initial acceleration of 20 m/s2, the amount of gas ejected per second to supply the needed thrust will be (g=10 m/s2 [CBSE 1998]
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a)127.5 kg s-1
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b) 137.5 kg s-1
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c)155.5 kg s-1
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d)187.5 kg s-1
Explanation
Resultant acceleration of rocket=acceleration due to engine - acceleration due to gravity 20=a' - 10a'=30 m/s2 force acting on rocket F=maF=5000× 30=150000 N From formula dm/dt=150000 / 800=187.5 kg/s Answer:(d)
Q.15
A 100 kg man jumps into a swimming pool from a height of 5m. It takes 0.4s for the water to reduce his velocity to zero. The average force exerted by the water on the man is [ CEt 1998]
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a)25 N
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b) 250 N
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c)2500 N
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d)5000 N
Explanation
Initial velocity of man=0 v2=2gs v2=2×10×5v=10 m/s In water velocity changed to zero in time 0.4 secacceleration a=(v-u) /t a=- 10/0.4=25 m/s2 Force on the man=mass × accelerationF=100 × 25=2500N Answer:(c)
Q.16
A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to 2:What will be the ratio of there nuclear size ( nuclear radius) [ CBSE 1996]
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a) 2 1/3 : 1
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b) 1: 2 1/3
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c) 3 1/2 : 1
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d) 1:3 1/2
Explanation
Let velocity of first part be v1 and mass be m1 Let velocity of second part be v2 and mass be m2 Now according to law of conservation of momentum Final momentum=Initial momentum 0=m1v1 + m2v2 m1v1=- m2v2 v1 / v2=2/1 v1=2v2m12v2=- m2v2 2m1=m2Density of both the parts are same ∴ 2v1=v2∴ 2r13=r2321/3r1=r2r1 / r2=1 : 21/3 Answer: (b)
Q.17
Essential characteristic of equilibrium is [ Roorkee 1989]
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a)momentum equal to zero
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b)acceleration equal to zero
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c)kinetic energy is zero
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d)velocity equal to zero
Explanation
If resultant force is zero object is in equilibrium, thus acceleration zeroAnswer: (b)
Q.18
A glass of water is kept on a table in the dining car of a train. If acceleration of the train is (g/5) in the forward direction, the surface of water is inclined to the horizontal at an angle of [ CET 1994]
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a) tan-1(5/2)
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b) tan-1(2/5)
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c)tan-1(1/5)
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d)tan-1(5)
Explanation
Acceleration of train is shown by green arrow in figure. by red arrow acceleration of water is shown due to acceleration of trainGravitational acceleration is shown by down arrowthus tanθ=g / (g/5)=5θ=tan-15Answer: (d)
Q.19
One end of a spring balance is stretched by the force of 2N and equal and opposite force is applied on its other end. The reading of the spring balance will be
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a) 4N
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b) 2N
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c)0
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d)any value between 0 and 4N
Explanation
When a spring is stretched by 2N then 2N tension will be produced in the spring. Same tension will be exerted by spring on its support, which will be of 2N. thus reading will be 2N Answer:(b)
Q.20
When we kick a stone, we get hurt. Due to which property of the stone it happens?
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a) Inertia
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b) Velocity
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c) Reaction
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d) Momentum
Explanation
Stone exerts equal and opposite force on leg, thus we get hurt Answer: (c)
Q.21
Two skaters of mass 40kg and 60kg respectively stand facing each other at A and B where AB is 5 meters. They pull on a mass less rope stretched between them, then they meet at
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a)25 meters from And B
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b) 3 meters from A and 2 meters from B
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c)2 meters from A and 3 meters from B
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d)3 meters from A and 8 meters from B
Explanation
Both the skater have same force F acceleration of skater at A a=F/40acceleration of skater at B a'=F/60let displacement of skater at A be x thus x=½ at2displacement of skater at b is 5-x=½ a't2From above two equations, by taking ratiox / (5-x)=a / a' x/ (5-x)=60/4040x=300 -60xx=3 m ( 3m from point A and 2 m from point B) Problem can be solved by center of mass since no external force is acting on the systemAnswer: (b)
Q.22
When one swims across a flowing river, maximum energy is spend in [ AIIMS 1985]
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a) first 1/3 of the distance
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b) second 1/3 of the distance
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c)last 1/3 of the distance
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d)equal energy throughout
Explanation
First one-third as the man has to apply greater force to overcome inertia of restAnswer: (a)
Q.23
We can derive Newton's
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a) second and third laws from the first law
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b) first and second laws from the third law
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c)third and first laws from the second law
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d)All the three laws are independent of each other
Explanation
Answer:(c)
Q.24
Three equal weights A, B,C of mass 2 kg each are hanging on a string passing over a fixed pulley which is frictionless as shown in figure. The tension in the string connecting weight B and C is [ JIPMER 1998]
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a) zero
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b) 13 N
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c) 3.3 N
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d) 19.6 N
Explanation
Acceleration of system 'a' Total mass of C and B is 4 kg going down with common acceleration 'a' thus 4a=4g - T mass A is going up with acceleration 'a' thus 2a=T - 2g solving above two equation we get a=g/3 Tension in string C and D C is going down with acceleration 'a' Thus 2a=2g - T' T'=2g -2(g/3) T'=(4/3) g g=9.8 T'=(4/3)×9.8 T'=13.06Answer: (b)
Q.25
With what minimum acceleration can a fireman slide down a rope whose breaking strength is two-third of his weight
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a)(2/)g
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b) (1/3)g
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c)9
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d)zero
Explanation
Let fireman slides down with acceleration a , then pseudo acceleration will be up and gravitational acceleration down When fireman slides down total force on the rope=m(g-a) given that rope can sustain weight or force of (2/3) mg here m is the mass of fireman Thus (2/3)mg=m( g- a) a=g - (2/3) g a=(1/3) g m/s2Answer: (b)
Q.26
The rear side of a truck is open and a box of mass 20kg is placed on the truck 4 metres away from the open end. µ=0.15 and g=10 m/s2.The truck starts from rest with acceleration of 2m/s2 on straight road. The box will fall off the truck when it is at a distance from the starting point equal to ..
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a) 4 metres
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b) 8 metres
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c)16 metres
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d)32 metres
Explanation
Pseudo force acting on the block=ma Frictional force on the block=µmgResultant force acting on the block=ma -µmg Retardation on the block a'=a- µg a'=2.0 - (0.15 ×10 )=0.5 m/s2The time take by the block to travel a distance of 4m is s=½ a' t2 4=0.5 × 0.5 ×t2t=4 sec In this time interval truck travelled a distance s=½ a t2 s=(0.5)×2× (4)2 s=16 mAnswer: (c)
Q.27
Gravels are dropped on a conveyor belt at the rate of 0.5kg/sec. The extra force required in Newton to keep the belt moving at 2m/s is [ EAMCET 1988]
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a) 1 N
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b) 2 N
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c)4 N
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d)0.5N
Explanation
According to formula F=2×0.5=1N Answer:(a)
Q.28
A parachutist of weight 'w' strikes the ground with his legs fixed and come to rest with an upward acceleration of magnitude 3g. Force exerted on him by ground during landing is
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a) w
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b) 2w
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c) 3w
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d) 4w
Explanation
Resultant acceleration upward=3g Acceleration down ward=g Let a be the total acceleration then 3g=a- g a=4g Force F=ma F=(w/g) ( 4g)=4w Answer: (d)
Q.29
Two masses A and B of 10kg and 5kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown in figure. The coefficient of friction A with the table is 0.The minimum mass of C that may be placed on A to prevent it from moving is equal to
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a)15 kg
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b) 10 kg
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c)5 kg
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d)0 kg
Explanation
From figure mBg - T=mBa T=mBg - mBa N=(mA + mC )g T- µ(mA + mC)g=(mA + mC )agiven a=0 ,solve equations (1) and (2) for mC Answer: (a)
Q.30
The law of conservation of momentum is a logical consequence of Newton'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 three laws
Explanation
Answer: (b)
0 h : 0 m : 1 s
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