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Quiz 7
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Q.1
A body ( mass 0.5kg) is constrained to move Eastwards. A force of 20N acts on the body directed 30° east of North. The acceleration produced in the body due to the force will be
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a) 40 m/s2
0%
b) 20 m/s2
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c) 20√3 m/s2
0%
d) zero
Explanation
The body is constrained to move only in the east direction. Hence only the component of 20N in the East direction can be effective on the body. The component force is 20cos60=10N. This is acting on mass of 0.5kg will produce an acceleration of 10/0.5=20 m/s2 Answer: (b)
Q.2
A trolley is being pulled up an inclined plane by a man on it ( as shown in figure). He applies a force of 250N. IF the combined mass of the man and trolley is 100kg, the acceleration of trolley will be [ in15≥=0.26]
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a)2.4 m/s2
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b) 9.4 m/s2
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c)6.9 m/s2
0%
d)4.9 m/s2
Explanation
Let a be the acceleration of trally force exerted by person is 250N, reactionary force on trolley is upward=250N ∴ ma=T +250 -mgsin15 Now T=2T1 and T1=F Thus T=2F ∴ ma=2F +250 -mgsin15 Answer: (d)
Q.3
A big boulder of mass M has fallen into ditch of width 2d. Two persons are slowly pulling it out using a light rope and two fixed pulleys as shown in figure. Assuming the force exerted by two persons are equal calculate the force when the boulder is at a depth h
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a)
0%
b)
0%
c)
0%
d)
Explanation
From figure 2Tcosθ=Mg Answer: (c)
Q.4
A force of 10N acts alone on a body ( mass 1kg) for 5s. A second force of 20N acts on it now in the opposite direction. The body will
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a) come to rest in 5 s
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b) come to rest in 10 s
0%
c)not come to rest at all
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d)come to rest and then move in the opposite direction
Explanation
Acceleration=F/m=10/1=10 m/s2 final velocity V=at=10×5=50 m/s Retardation=F/m=20 m/s2 resultant acceleration=10-20=-10m/s2 Final velocity V'=0 m/s, initial velocity=50 m/s 0=50 -(10) t t=5 seconds Answer:(a)
Q.5
In figure, the block A, B, C of mass m each, have acceleration a1, a2 and a3 respectively. F1 and F2 are external forces of magnitude 2mg and mg respectively. Then
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a) a1=a2=a3
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b) a1 > a2 > a3
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c) a1=a2, a2 > a3
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d) a1 > a2, a2=a3
Explanation
a1 ma1=2mg -mg a1=g a2 2ma2=2mg -T and ma2=T-mg ∴ 3ma2=mg a2=g/3 a3 ma3=mg+mg -T and ma3=T -mg ∴ 2ma3=mg a3=g/2 ∴ a1 > a2 > a3 Answer: (b)
Q.6
A two kg mass is suspended using two strings AB and CD as shown in figure. A sudden jerk is given to the end D of the sring, then :
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a)part AB of the string breaks
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b) part CD of the string breaks
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c)no part of the string breaks
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d)both the strings will simultaneously break
Explanation
Answer: (c)
Q.7
In above question if the end D of the string is pulled down slowly then
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a) part AB of the string breaks
0%
b) part CD of the string breaks
0%
c)no part of the string breaks
0%
d)both the strings will simultaneously break
Explanation
Answer: (a)
Q.8
A spring toy weighing 1kg on spring balance suddenly jumps upward. A boy standing near notice that the scale of the balance reads 1.05 kg. In this process the maximum acceleration of the toy is ..(g=10m/s2)
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a) 0.05 m/s2
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b) 0.5 m/s2
0%
c) 1.05 m/s2
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d) 1.0 m/s2
Explanation
W=m(g+a) 10.5g=1(g+a) or a=0.05g=0.5 m/s2 Answer:(b)
Q.9
A lift of mass 1000 kg is moving with an acceleration of 1 m/s2 in upward direction. Tension developed in the string, which is connected to the lift, is. [CBSE PMT 2002]
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a)9,800 N
0%
b)10,800 N
0%
c)10,000 N
0%
d)11,000 N
Explanation
T=m(g+a)=1000(9.8+1)=10800N Answer:(b)
Q.10
A toy car consists of three identical compartment A, B, and C. It is being pulled by a constant force F. The ratio of tension in the strings connecting AB and BC is ( the car is pulled from side C)
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a) 1:1
0%
b) 1:2
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c) 2:1
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d) 1:3
Explanation
Let m be the mass of each compartment to tal mass=3m acceleration=F/3m T1=m × a=F/3 T2=2m × a=2F/3 ∴ T1/T2=1 :2 Answer: (b)
Q.11
The ratio of weight of the person in a stationary lift and in a lift accelerating downwards with uniform acceleration a is 3:The acceleration of the lift is
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a)g/3
0%
b) g/2
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c)g
0%
d)2g
Explanation
W1=mg W2=m(g-a) 3/2=mg / m(g-a) 3(g-a)=2g a=g/3Answer: (a)
Q.12
A body of mass m has its position x at a time t, expressed by the equation x=3t3/2 +2t - (1/2) . The instantaneous force F on the body is proportional to
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a) t3/2
0%
b) t
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c)t-1/2
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d)t0
Explanation
Find acceleration of particle by taking derivative of x for time Answer: (c)
Q.13
A toy of mass M1 is pulled along a horizontal frictionless surface by a rope of mass MA force F is applied to the free end of the rope. The force exerted on the toy is
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a)F
0%
b)
0%
c)
0%
d)
Explanation
Answer:(b)
Q.14
Two pulley arrangements are shown in the figure (a) and (b). The rope in both the arrangements is mass-less. The ratio of acceleration of mass m in case (a) and (b) is :
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a) 1:1
0%
b) 1:2
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c) 2:1
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d) 1:3
Explanation
Fig (a) 2ma1=2mg - T ma 1=T-mg 3ma 1=mg a1=g/3 Fig(b) ma 2=F- mg ma2=2mg - mg a2=g a1/a2=1:3 Answer: (d)
Q.15
A block can side on a smooth inclined plane of inclination θ kept on the floor of a lift. When the lift us descending with retardation, a m/s2 the acceleration of the block relative to the inclie will be
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a)(g-a)sinθ
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b) (g+a)sinθ
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c)gsinθ
0%
d)g-a
Explanation
When the lift is descending with retardation of a m/s2 . Thus pseudo acceleration will ebe in down ward direction Hence, total acceleration on the floor=(g+a) Hence, acceleration of block along inclined plane=(g+a) sinθAnswer: (b)
Q.16
A block slides down an inclined plane of slope of an angle θ with constant velocity. It is then projected up the plane with an initial velocity u. The distance up to which it will raise before coming to rest is
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a) u2 / 4gsinθ
0%
b) u / 4gsinθ
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c)u2sinθ / 4g
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d)usinθ/4g
Explanation
The block is projected up the plane will have two forces on down the plane resultant force retarding force ma=mgsinθ + µmgcosθ but µ=tanθ ∴ ma=mgsinθ +mgtanθcosθ a=2gsinθ Let S be the distance moved up before it comes to rest o-u2=-2×2gsinθ×S S=u2 / 4gsinθAnswer: (a)
Q.17
A particle is projected directly up a rough plane of inclination α with velocity u. If after coming to rest the particle returns to the starting point with velocity v, the coefficient of friction between the particle and plane is
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Particle projected up comes to rest after travel of distance l Retardation force ma=mgsinα+ µmgcosθ a=gsinα+ µgcosα Now final velocity=0 0 -u2 - 2×gsinα+ µgcosα×l particle returns to original position and have velocity v resultant acceleration=gsinα -µgcosα Initial velocity u=0 final velocity v, distance travel=l v2 -0=2×gsinα- µgcosα×l Answer:(c)
Q.18
A right circular cone is fixed with its axis vertical and vertex down. A particle in contact with its smooth inside surface describes circular motion in a horizontal plane at height of 20cm above the vertex. Its velocity is m/s is
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a) 1
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b) 1.2
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c) 1.4
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d) 1.6
Explanation
Centripetal force=Nsinθ ∴ Nsinθ=mv2 / R and Ncosθ=mg ∴ tanθ=V2 /Rg and tanθ=h/R ∴ h/R=V2 / Rg v=√gh=√2 m/s=1.4 m/s Answer: (c)
Q.19
A body of mass m1 is placed on a horizontal plank of mass m2 which rests ona smooth horizontal table. The coefficient of friction between the mass m1 and plank is µ. A gradually increasing force F depending on time t as F=kt where k is constant applied to the plank. The time to at which the plank starts sliding under the mass is
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a)m1µg / k
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b) (m1+m2)µg / k
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c)m2µg / k
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d)m1m2µg / k
Explanation
Acceleration of system a=F/(m1 + m2) Force on m1causing acceleration ma=µm1g ∴ a=µg Answer: (b)
Q.20
A given object takes η times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between object and incline is given by
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Acceleration with out friction=gsinθ=gsin45=g/√2 Acceleration with friction=g(sinθ -µcosθ)=g(1-µ)/ √2 Since body starts from rest, distance is equal to (1/2)at2 Thus Answer: (b)
Q.21
How high can a particle rest inside a hollow sphere of radius a if the coefficient of friction is 1/√3?
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a) 0.15a
0%
b) 0.1a
0%
c)0.134a
0%
d)0.2a
Explanation
For not sliding mgsinθ=µmgcosθ tanθ=µ tanθ=1/√3 ∴ θ=30° from figure h=a(1-cosθ)=0.134a Answer:(c)
Q.22
Consider the system shown in figure. The wall is smooth, but the surface of block A and B in contact are rough. The friction on B due to A in equilibrium is
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a) upward
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b) downward
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c) zero
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d) the system cannot remain in equilibrium
Explanation
surface between A and wall is smooth, so the system will with acceleration g Answer: (d)
Q.23
Two blocks A and B of masses 5kg and 3kg respectively rest on a smooth horizontal surface with block B resting over A. The coefficient of friction between A and B is 0.The maximum horizontal force that can be applied to A so that there will be motion of A and B, without separation is ( g=10m/s2)
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a)15 N
0%
b) 25 N
0%
c)40 N
0%
d)50 N
Explanation
The blocks A and B move togather with acceleration a=F/(m1+m1) a=F/8 For masse B , 3a=3µg a=µg F/8=µg F=8µg=0.5×8g=40NAnswer: (c)
Q.24
A bullet of mass m is fired from below into a bob of mass M of a long simple pendulum. The bullet stays inside the bob and the bob rises to a height h. The initial speed of the bullet will be
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a)
0%
b)
0%
c)
0%
d)
Explanation
Let the initial velocity of bullet be u and common velocity of bob and bullet be v Applying law of conservation of momentum mu=(m+M)vv=mu / (m+M) After collision bob move against gravity and its final velocity becomes zero at height hThus v2=2gh From above√(2gh)=mu/(m+M) Thus u=√(2gh) [(m+M)/m]Answer: (d)
Q.25
A body of mass m1 strikes a stationary body of mass mIf the collision is elastic, the fraction of kinetic energy transferred by the first body to the second is
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a)m1m1 / ( m1 +m2)
0%
b) 2m1m2 / ( m1 +m2)
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c)4m1m2 / ( m1 +m2)2
0%
d)2m1m2 / ( m1 +m2)2
Explanation
m1u=m1v1 + m2v2and u=v2 - v1 solvingKinetic energy transferred E=Fraction of energy transferred E'=IFinal kinetic energy of m2 / Initial kinetic energy of m1 Answer:(c)
Q.26
A block is at rest on an inclined plane making an angle α with the horizontal. As the angle α of the inclination is increased, the block just starts slipping when the angle of inclination becomes θ. Then the coefficient of static friction between the block and the surface of the inclination plane is
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a) sinθ
0%
b) cosθ
0%
c) tanθ
0%
d) independent of θ
Explanation
Answer: (c)
Q.27
A rocket of initial mass 5000 kg ejects gas at a constant rate of 60 kg/s with relative speed of 2050 m/s. Acceleration of the rocket 15 second after it is blasted off from the surface of the earth will be ( g=10m/s2)
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a) 10 m/s2
0%
b) 20 m/s2
0%
c) 30 m/s2
0%
d) 40 m/s2
Explanation
After 15 sec mass of rocket=5000 - (60×15)Kg Thrust force on rocket F=v(dm/dt )=(2050×60) N Acceleration of rocket at that instant= Answer: (b)
Q.28
A body moving towards a another body at rest collides with it. It is possible that
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a)both bodies come to rest
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b) both bodies move after collision
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c)stationary body remains stationary, the moving body chnages its velocity
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d)none is possible
Explanation
when the bodies move after collision, momentum will remain conserveredAnswer: (b)
Q.29
A particle of mass is made to move with uniform speed u along the perimeter of a regular polygon of 2n side. the magnitude of impulse applied on the particle at each corner of the polygon is
0%
a) mvsin(π/n)
0%
b) mvsin((π/2n)
0%
c)2mvsin(π/n)
0%
d)2mvsin(π/2n)
Explanation
Angle θ=π - (π/n) Impulse I=Change in momentum Answer: (d)
Q.30
Two bodies A and B of equal mass moving with speeds along X and Y axis respectively of a rectangular coordinate system collide and stick to gather. They will move in a direction inclined to x-axis at an angle
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a) 30°
0%
b) 60°
0%
c)45°
0%
d)90°
Explanation
Conservation of momentum mui + muj=2m( vcosθi + vsinθj on solving θ=45° Answer:(c)
0 h : 0 m : 1 s
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