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Laws Of Motion Mcq
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Q.1
When two surfaces are coated with a lubricant, then they [AFMC 1998]
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a) roll upon each other
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b) slide upon each other
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c)stick to each other
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d)none of these
Explanation
Due to lubricant, friction force will not act. And surface will slideAnswer: (b)
Q.2
While dusting the carpet, we give a sudden jerks or beat it with a stick, because [ AFMC 1981]
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a) inertia of rest keeps the dust in its position and the dirt gets removed as the carpet moves away
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b) inertia of motion removes the dust
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c) no inertia is involved in the process, it is simply due to practical experience
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d) jerk compensates for the force of adhesion between the dust and carpet and the dust is removed
Explanation
Answer: (a)
Q.3
A cyclist moves in a circular track of radius 100m. If the coefficient of friction is 0.2, then the maximum speed with which the cyclist can take a turn without leaning inwards, is [AFMC 1998]
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a) 14.0 m/s
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b) 140 m/s
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c)1.4 m/s
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d)9.8 m/s
Explanation
To take turn centripetal acceleration will be provided by frictional force∴ (mv2 )/ r=μN ∴ v2=rμg -- eq(1) Here radius of circular track 'r'=100 m Coefficient of friction μ=0.2 gravitational acceleration 'g'=9.8 m s-2 Substituting above values in eq(1) we get v=14 m s-1Answer: (a)
Q.4
When a bus suddenly take a turn, the passengers are thrown outwards because of [AFMC 1999]
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a)speed of motion
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b)inertia of motion
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c)acceleration of motion
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d)none of these
Explanation
When bus suddenly turns, passengers are thrown outwards because of inertia of motion Answer:(b)
Q.5
A ball of mass 150 gm moving with an acceleration 20 m / s2 is hit by a force, which acts on it for 0.1 sec. The impulsive force is [1999]
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a) 1.2 N-s
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b) 0.3 N-s
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c) 0.1 N-s
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d) 0.5 N-s
Explanation
Formula for Impulse of force is I=F t ∴ I=(ma)(t) --eq(1) Here mass 'm'=0.15 kgs acceleration 'a'=20 m s-2 time of contact 't'=0.1 secondSubstituting above values in eq(1) we getI=0.3 N-s Answer: (b)
Q.6
A rough vertical board has an acceleration a along the horizontal so that a block of mass M pressing against it does not fall. The coefficient of friction between block and the board is [1999]
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a) >a/g
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b)
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c)=a/g
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d)≥g/a
Explanation
As shown in F.B.D below force mg acting downwards, force frictional µN acting down ward, pseudo force N=ma acting normal to block from figure mg=µN mg=µ(ma)µ=g/aµ≥g/aAnswer: (d)
Q.7
A body of mass a moving with a velocity b . strikes a body of mass c and gets embedded into it. The velocity of the system after collision is [AFMC 2000]
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a) (a+c)/ab
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b) ab/(a+c)
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c)a/(b+c)
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d)a/(a+b)
Explanation
since no external force acts on system we can use law of conservation of momentumAfter collision object having mass 'a' gets embedded in object with mass 'b'so mass of single object after collision will be (a+b) with common velocity let it be 'v' Let initial velocity of object with mass 'a' be u=b object with mass 'b' is stationary its velocity u'=0Now according to law of conservation of momentummomentum before collision=momentum after collision∴ au +cu'=(a+c) v∴ ab=(a+c) v∴ v=ab/(a+c)Answer: (b)
Q.8
A block of mass 60 kg just slides over a horizontal distance of 0.9 m. If the coefficient of friction between their surface is 0.15 then work done against friction will be
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a) 79.4 J
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b)97.54 J
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c)105.25 J
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d)none of these
Explanation
Block just slides and travels a distance 0.9 hence applied force must be equal to frictional force work is done by the object or work is done against frictional force F.B.D. is as shown in figureWork done against friction=(Frictional force)( displacement)W=(F) (d)Here coefficient of friction µ=0.15Normal force 'N'=mgFrictional force=µ NFrictional force=µmgdisplacement 'd' ∴ W=µ(mg)d --eq(1)µ=0.15d=0.9 mm=60 kgg=9.8 ms-2Substituting above value in eq(1) we get W=(0.15)(60)(9.8)(0.9)W=79.38JW ≈79.4 J Answer:(a)
Q.9
A smooth inclined plane of length L having inclination θwith the horizontal is inside a lift which is moving down with a retardation a. The time taken by a body to slide down the inclined plane from rest will be [AFMC 2001]
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a)
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b)
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c)
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d)
Explanation
As shown in figure pseudo acceleration due to downward retarded motion of lift will be in the direction of gravitational acceleration Now components of gravitational acceleration and pseudo acceleration will cause object to slide on the inclined plane acceleration parallel to inclined pane a'=(a+g) sin θ Now according to formula for displacement s=ut + ½a t2 --eq(1) Here s=L; acceleration a=(a+g) sin θ initial velocity u=0 L=½[(a+g) sin θ t2] on solving for 't' we get Answer: (a)
Q.10
A block is moving up at Θ=30° with a velocity 5m/s and stops after 0.5 sec then what is the value of friction(µ)[AFMC 2000]
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a)0.6
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b) 0.5
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c)0.51
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d)0.81
Explanation
Block have some initial velocity 'u'=5 m/s but stops due to friction and component of gravitational acceleration parallel to inclined plane as shown in figureHere retardation is constant a=( final velocity - initial velocity)/ timea=(5-0)/0.5a=10 m -2Retarding force=Frictional force + component of gravitational forcema=µ N + mg sin Θ --eq(1)Here N=mg cos Θ angle 'Θ=30°gravitational acceleration 'g'=10 m -2substituting above values and value of a in eq(1) we get m(10)=µm(10 cos30)+ m10 sin 3010=µ(10) [(√3)/2] + 10 (½)On solving equation for µ we getµ=1/ √3µ=0.6 (approximate)Answer: (a)
Q.11
If cyclist moving with a speed of 4.9 m/s on a level road can take sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyre and the road is [AFMC 2001]
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a)0.71
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b)0.61
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c)0.51
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d)0.81
Explanation
To take turn centripetal acceleration will be provided by frictional force∴ (mv2 )/ r=μN ∴ v2=rμg -- eq(1) ∴ μ=v2/ (rg) --eq(2) Here radius of circular track 'r'=4 m velocity=4.9 m/s gravitational acceleration 'g'=9.8 m s-2 Substituting above values in eq(2) we get μ=0.6125 μ=0.61Answer: (b)
Q.12
A body of mass M, at rest explodes into three masses two of which of mass's M/4 each are thrown off in perpendicular directions with velocity of 3 m/s and 4m/s respectively. The third piece will be thrown off with a velocity of ..[AFMC 2001]
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a)3 m/s
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b)2.5 m/s
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c)2.0 m/s
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d)1.5 m/s
Explanation
Mass of first fraction is M/4 having velocity along x axis is 3i Mass of second fraction is M/4 having velocity along y axis is 4j Mass of third fraction is M/2 having velocity 'v' No external force acts on the body hence we can use law of conservation of momentum Initial momentum=Final momentum 0=(M/4) 3i + (M/4) 4j + (M/2) v v=-1.5i - 2j magnitude of v=[(-1.5)2 + (-2)2] 1/2 v=2.5 m/s Answer:(b)
Q.13
A motor-cyclist moving with a velocity of 72km/hr on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meter. The acceleration due to gravity is 10m -In order to avoid sliding, he must not bend with respect to the vertical plane by an angle greater than.. [AFMC 2001]
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a) Θ=tan -1(4)
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b) Θ=tan -1(25.91)
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c) Θ=tan -1(2)
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d) Θ=tan -1(6)
Explanation
R is the reaction the ground 72Km/hr = 20m/sFrom figure it is clear that R cosθ=mg and Rsinθ=(mv2)/rtanθ=v2 /rg tanθ=(20×20) / (20×10)tanθ=2 θ=tan -1(2)Answer: (c)
Q.14
Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g=10 m/s2) [ CBSE-PMT 1992]
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a) 30m
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b) 40m
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c) 72m
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d) 20m
Explanation
Here initial speed u=72 km/h=20 m/s final speed is v=0 frictional force will stop the carRetarding force=frictional force F=-µmg thus a=-µg=-0.5(10)=-5 m/s2 As v2=u2 + 2as 0=(20)2 +2(-5)s s=40 mAnswer: (b)
Q.15
A spring of force constant k is cut into two pieces whose lengths are in the ratio 1:What is the force constant of the loner piece? [ AFMC 2009]
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a)k/2
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b) 3k / 2
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c)2k
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d)3k
Explanation
Here l2=2l1As for spring, force constant k ∝ 1/l k1 ∝ 1 / l1 k2 ∝ 1 / l2 k ∝ 1 / (l1+ l2)Answer: (b)
Q.16
A rider on a horse back falls forward, when the horse suddenly stops. This is due to [ Karnataka Entrance 1993]
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a)the inertia of the horse
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b) the inertia of the rider
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c)large weight of the horse
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d)losing the balance
Explanation
Answer: (b)
Q.17
Inertia is the property by virtue of which the body is unable to change by itself [ wardha 1982]
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a) the state only
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b) the state of uniform linear motion only
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c)the direction of motion only
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d)the steady state of rest and of uniform linear motion
Explanation
Answer: (d)
Q.18
Starting from rest, a body slides down a 45° inclined plane is twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is [CBSE PMT -1988]
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a) 0.80
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b) 0.75
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c) 0.25
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d) 0.33
Explanation
As shown in figure gravitational force parallel to plane is responsible for downward motion of the object. frictional force resist the downward motion is in upward directionThen resultant force on the object will be=downward gravitational force - upward frictional forceLet a be the resultant acceleration of the object, m be the mass then ma=mgsin θ - f --eq(1)Now f=µN , but N=mg cos θ∴f=µ mg cos θ --eq(2)Substituting eq(2) in eq(1) we getma=mgsin θ - µ mg cos θ ∴ a=gsin θ - µ g cos θ --eq(3)Now let the length of inclined plane be 's'According to equation of motions=ut + ½ at12 --eq(4) here u=0 and a is as per eq(3) substituting values in eq(4) we get2s=(gsin θ - µ g cos θ) t12 eq(5)In absence of frictionsubstituting µ=0 and t2 in above equation we get2s=g sinθt22--eq(6)given that t1=2t2substituting values of time in above equation (5) we get2s=4(gsin θ - µ g cos θ) t22 eq(7)Comparing eq (6) and eq(7) we get sinθ=4 sinθ -4µcos θ∴ µ=(3/4) tan θ substituting value of θ=45 give in above equation we get µ=3/4=0.75Answer: (b)
Q.19
Two bodies of 'masses m and 4m are moving with equal kinetic energies. The ratio of their linear momentum will be [CBSE PMT -1988]
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a) 1:4
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b) 4:1
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c) 1:2
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d) 2:1
Explanation
Formula for kinetic energy in terms of momentum is E=P2 / 2mLet p1 and p2 be the momentum of objectshere kinetic energy of both the object is same thereforeAnswer: (c)
Q.20
A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 :The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of heaviest fragment in m/s will be
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a)7√2
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b) 5√2
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c)3√2
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d)√2
Explanation
Let p1 and p2 be the momentum of fragments having same mass and Let p3 be the momentum of heaviest fragmentSince no external force acts on the system, according to law of conservation of momentump1 + p2 +p3=0 p1 + p2=- p3 angle between p1 and p2 is 90° we getp12 + p22= p32 (1×21)2 + (1×21)2=(3×v)2That is v=7√2 Answer:(a)
Q.21
kg mass and 1 kg are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is [1989]
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a) 1 : 2
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b) 1 : 1
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c) 2:1
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d) 4: 1
Explanation
formula for kinetic energy in terms of momentum is E=P2 / 2m Hence by multiplying both the sides by m we get 2Em=P2 E and 2 are constant Thus momentum ∝ m1/2Hence ratio is 2:1 Answer: (c)
Q.22
A 600 kg rocket is set for a vertical firing. If the exhaust speed is 1000 m/s , the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is [1990]
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a) 117.6 kg /s
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b) 58.6 kg/s
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c) 6kg/s
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d) 76.4 kg/ s
Explanation
exhaust speed is constant, due to burnings of fuel mass of rocket changes, resulting into change in momentum of rocket which should be equal to downward pull due to gravity hence we get following equationon substituting values exhaust speed u=1000 m/s ,mass of rocket m=600 kg and gravitational acceleration g=10 m s-2 and solving we getdm/dt=6 kg/sAnswer: (c)
Q.23
A particle of mass m is moving with a uniform velocity vIt is given an impulse such that its velocity becomes vThe impulse is equal to [CBSE PMT 1990]
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a) m[ |v2|-|v1|]
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b) ½ [ v22 – v12]
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c) m[ v2 +v1]
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d) m[ v2 - v1]
Explanation
Impulse=final momentum - initial momentumI=m[v2 - v1]Answer: (d)
Q.24
heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is [ CBSE-PMT 1991]
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a) 20%
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b) 25%
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c) 35%
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d) 15%
Explanation
The force of friction on the chain lying on the table should be equal to the weight of the hanging chain. Let ρ=mass per unit length of the chain µ=coefficient of friction l=length of the total chain x=length of hanging chain l-x=length of the chain on the table µ=0.25Now, Frictional force on the chain lying on the table=µ(l-x)ρgWeight of hanging chain=xρg∴ µ(l-x)ρg=xρg x=µl /(µ+1)On substituting the value of µ in above equation and solving we getx=0.2lx/l=0.2=20% Answer:(a)
Q.25
Physical independence of force is a consequence of [ CBSE-PMT 1991]
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a) third law of motion
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b) second law of motion
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c) first law of motion
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d) all of these laws
Explanation
Newton's first law of motion is related to physical independence of force. Answer: (c)
Q.26
The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is: [CBSE-PMT 2009]
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a)4 ms-2 upwards
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b) 4 ms-2 downwards
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c)14 ms-2 upwards
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d)30 ms-2 downwards
Explanation
Tension upward force gravitational force downwardNet force F=T -mgma=T - mg2000a=28000 - 20000 a=4 ms-2 upwardsAnswer: (a)
Q.27
A particle of mass M is moving in a horizontal circle of radius R with uniform speed V. When it moves from one point to a diametrically opposite point, its... [ CBSE-PMT 1992]
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a) kinetic energy changes by MV2 /4
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b) momentum does not change
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c) momentum changes by 2MV
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d)kinetic energy changes' by MV2
Explanation
On diametrically opposite points, the velocities have same magnitude but opposite directions. Therefore, change in momentum in MV-(-MV)=2MVAnswer: (c)
Q.28
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is... [ CBSE-PMT 2011]
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a)MV
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b) 1.5MV
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c) 2MV
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d) zero
Explanation
Impulse experienced by the body=change in momentum I=MV -(-MV)=2MV Answer:(c)
Q.29
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 ms-If g=10 ms-2, the tension in the supporting cable is... [ CBSE- PMT 2011]
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a) 8600N
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b) 9680N
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c) 11000 N
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d) 1200N
Explanation
Resultant force acting on system ( Man + lift )a=1 m s-2 total weight M=60+940=1000 kg resultant force=Ma=1000× 1=1000 N Let T be tension in the supporting cable acting upwards Gravitational force=Mg acting downwards Thus Ma=T - Mg T=Ma+ Mg T=1000× 1 + 1000 × 10 T=11000 N Answer: (c)
Q.30
A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them is (µ=0.5) The distance that the box will move relative to belt before coming to rest on it taking g=10ms-2, is [ CBSE-PMT 2011M]
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a) 1.2m
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b) 0.6m
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c) zero
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d) 0.4m
Explanation
Frictional force on box=f=µ mgThis frictional force keeps the object on the belt and do not slips the object, belt is moving with constant velocity If 'a' is the acceleration of box then ma=µ mg a=µg a=5 m/s2 Object stops hence v=0 , according to equation of motion v2=u2 + 2as 0=22 + 2(5)s s=-2/5 with respect to belt s=0.4 mAnswer: (d)
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