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Quiz 14
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Q.1
A lightweight, frictionless pulley is placed at the top of a frictionless, 30°incline. A 4 Kg is raised up the incline by means of a rope that simultaneously lowers an 8 Kg block.The blocks will accelerate at ___m/s2
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a)2.5
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b) 3.3
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c)5.0
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d)6.7
Explanation
Answer: (c)
Q.2
The tension in Q393 while accelerating is ___N.
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a)10
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b)20
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c)40
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d)80
Explanation
Answer: (c)
Q.3
If the coefficient of friction between 4Kg block and incline is µ=0.52 the blocks in Q393 will accelerate at ___ m/s2
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a) 2.0
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b) 3.5
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c)5.0
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d)6.5
Explanation
Answer:(b)
Q.4
With the friction mentioned in question #395 the tension in in Q393 the line is ___N.
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a) 14
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b) 28
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c) 42
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d) 52
Explanation
Answer: (d)
Q.5
Which of the following statements about the conservation of momentum is not correct?
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a)Momentum is conserved for a system of objects pushing away from each other.
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b)Momentum is not conserved for a system of objects in a head-on collision.
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c)Momentum is conserved when two or more interacting objects push away from each other.
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d)The total momentum of a system of interacting objects remains constant regardless of forces between the objects
Explanation
Answer: (b)
Q.6
Two objects move separately after colliding, and both the total momentum and total kinetic energy remain constant. Identify the type of collision.
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a)elastic
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b)inelastic
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c)nearly elastic
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d)perfectly inelastic
Explanation
Answer: (a)
Q.7
A rigid bar with weight of 100N is free to rotate about a friction less hinge at a wall and supported in a horizontal position by a spring scale attached to the ceiling at an angle of 30o to the vertical, as shown in figure. What force of tension is indicated by the spring scale?
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a) 100N
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b) 100 √3 N
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c)100/√3
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d)50 N
Explanation
Hinge and spring scale are equidistant from the cetre of mass, so each supports 50 N.Fy-spring scale=50N=Fsprig scalecos30 Fspring scale=50 / [ (√3) / 2]=100 / √3 N Answer:(c)
Q.8
Two objects stick together and move with a common velocity after colliding. Identify the type of collision.
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a) elastic
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b) inelastic
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c) nearly elastic
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d) perfectly inelastic
Explanation
Answer: (d)
Q.9
In an inelastic collision between two objects with unequal masses,
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a)the total momentum of the system will increase.
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b) the total momentum of the system will decrease.
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c)the kinetic energy of one object will increase by the amount that the kinetic energy of the other object decreases.
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d)the momentum of one object will increase by the amount that the momentum of the other object decreases.
Explanation
Answer: (d)
Q.10
A billiard ball collides with a stationary identical billiard ball in an elastic head-on collision. After the collision, which of the following is true of the first ball?
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a) It move in same direction
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b) It has one-half its initial velocity.
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c)It comes to rest.
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d)It moves in the opposite direction.
Explanation
Answer: (c)
Q.11
A rocket with a lift-off mass 3.5×104 is blasted upwards with an initial acceleration of 10m/sThen the initial thrust of the blast
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a) 3.5×105N
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b) 7.0 ×105N
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c)14.0 ×105N
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d)1.74 ×105N
Explanation
Resultant force=Thrust - gravitational force Thrust=Resultant force + gravitational forceThrust=(3.5×104 × 10) + (3.5×104 × 10)Thrust=7.0 ×105N Answer:(b)
Q.12
A horizontal force of 10N is necessary to just hold a block stationary against wall. the coefficient of friction between a wall and block is 0.The weight of the block is
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a) 20 N
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b) 50 N
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c) 100 N
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d) 2N
Explanation
Down force on block=frictional force mg=µ N as weight=mg w=µN here normal force is 10N w=0.2×10 Answer: (d)
Q.13
A marble block of mass 2 kg lying on ice when given a velocity of 6m/s is dropped by friction in 10s. Then the coefficient of friction is
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a)0.02
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b) 0.03
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c)0.06
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d)0.01
Explanation
Change in moment per unit time=Frictional force(2×6)/ 10=1.2=f but f=µN 1.2=µ ( 2×10) µ=0.06Answer: (c)
Q.14
Consider the following statementsA. Linear momentum of a system of particles is zeroB. Kinetic energy of a system of particle is zeroThen
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a) A does not imply B and B does not imply A
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b) A implies B but B does not imply A
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c)A does not imply B but B implies A
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d)A implies B and B implies A
Explanation
Answer: (c)
Q.15
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m.If a force P is applied at the free end of the rope, the force exerted by the rope on the block is
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a) Pm / ( M+m)
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b) Pm( M-m)
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c)P
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d)PM/(M+m)
Explanation
Total mass of system=M+macceleration of system=P/(M+m) Force on block=Mass of block × acceleration Force on block=M[P/(m+M)]=PM/(M+m) Answer:(d)
Q.16
A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is
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a) both the scales read M kg each
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b) the scale of the lower one reads M kg and of the upper one zero
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c) the reading of the two scale can be anything but the sum of the reading will be M kg
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d) the scale of the upper one reads M kg and of the lower one zero
Explanation
Answer: (a)
Q.17
than one correct option Q409) A small block of mass 0.1kg lies on the a fixed inclined plane PQ which makes an angle θ with the horizonatl. A horizontal force of 1N acts on the block through its centre of mass as shown in figure. Block remains stationary if (take g= 10 m/s2
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a) θ = 45°
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b) θ > 45° and a frictional force acts on the block toeards P
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c) θ > 45° and a frictional force acts on the block towards Q
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d) θ < 45° and a frictional force acts on the block tpwards Q
Explanation
The forces are resolved as shown in figure. Now mg = 01×10 = 1N When θ = 45, cosθ = sinθ nad Thus firctional force is zero option a correct If θ > 45° Down direction sinθ > updirection cosθ Object will slide down As object is stationary, friction acts upward or towards Q [ Option c correct] Answer:(a, c)
Q.18
A uniform force of 3i + j newton acts on a particleof mass 2 kg. Hence the particle is displaced fromposition 2i+ k metre to position 4i + 3j - k metre. The work done by the force on the particle is … [ NEET 2013]
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a) 9 J
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b) 6J
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c) 13 J
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d) 15J
Explanation
Final position (4i + 3j -k ) meter Initial position (2i + k )meter Displacement (4i +3 j - k )-(2i + k )=(2i + 3j - 2k ) Force (3i + j ) Newton Work = F ∙ S Work=(3i + j )∙(2i + 3j - 2k )=6+3=9J Answer:(a)
Q.19
A balloon with mass 'm' is descending down with an acceleration 'a' (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration 'a'?
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a)
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b)
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c)
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d)
Explanation
Here we have to consider upward thrust (F) on the balloon is constant. But gravitational force is changing. Case I When balloon moves down with acceleration a ma = mg – F ….(i) Case II When balloon moves up with acceleration a, after removing mass m’ (m-m’)a = F – (m-m’)g …(ii) On adding (i) and (ii) ma+ma-m’a = m’g m’(g+a) = 2m m’ = 2m/(g+a) Answer:(c)
Q.20
A system consists of three masses m1,m2 and m3 connected by a string passing over a pulley P. The mass m1 hangs freely and m2 and m3 are on a rough horizontal table (the coefficient of friction=µ).The pulley is frictionless and of negligible mass. The downward accelerationofmassm1, is ... (Assume m1 =m2 =m3 =m)
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a)
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b)
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c)
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d)
Explanation
For m1 ma = mg – T … (i) For m2 ma = T- µmg – T1 …(ii) For m3 ma = T1 - µmg …(iii) Add ii and iii 2ma = T - 2µmg…(iv) Add (iv) and (i) 3ma = mg(1 -2µ) Answer:(a)
Q.21
A body of mass (4m) is lying in x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds(v). The total kinetic energy generated due to explosion is … [ AIPMT 2014]
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a) 2mv2
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b) 4mv2
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c) mv2
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d) 3/2 mv2
Explanation
Mass of third particle is 2m According to law of conservation of momentum Magnitude of P3 = mv√2 If v’ is the velocity of third particle then 2mv’=mv√2 or v’= v/√2 Kinetic energy of third particle = Total energy = Answer:(d)
Q.22
The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8s is ….[ AIPMT 2014]
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a) 12Ns
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b) 6 Ns
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c) 24 Ns
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d) 20 Ns
Explanation
Change in momentum is equal to area under curve = 6 -6+12 =12 Ns Answer:(a)
Q.23
Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a friction less surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is : …[NEET 2015]
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a) 18 N
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b) 2 N
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c) 6 N
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d) 8 N
Explanation
Total mass = 7 kg, acceleration of each block = 14/7 =2 ms-2 Force on B = 2×(2+1) = 6N Answer:(c)
Q.24
A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The center of mass of the rod is at distance x from A. The normal reaction on A is : …[ NEET 2015]
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a)
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b)
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c)
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d)
Explanation
W is weight i.e. gravitational force Rod is in equilibrium transitional and rotational Transitional equilibrium condition N1 + N2 = W …(i) For Rotational equilibrium consider C.M as fix point N1x= N2(d-x) Normal reaction on A is N1 Substituting value of N2 in (i) Answer:(a)
Q.25
A block A of mass m1 rests on a horizontal table. A lights string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is µk. When the block A is sliding on the table, the tension in the string is : … [AIPMT 2015]
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a)
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b)
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c)
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d)
Explanation
m1 and m2 acceleration is same let be ‘a’ Mass m1 m2a = m2g – T …(i) Mass m2 m1a = T –µkm1g …(ii) multiply (i) by m1 and (ii) by m2 m1m2a = m1m2g –m1T …(iii) m1m2 a = m2T – µkm1m2 g …(iv) (iii)-(iv) m1m2g –m1 T = m2T – µkm1m2 g (m1 + m2)T = m1m2g + µkm1m2 g Answer:(d)
Q.26
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30°, the box starts to slip and slides 4.0 m down the plank in 4.0s.The coefficients of static and kinetic friction between the box and the plank will be, respectively:
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a) 0.4 and 0.3
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b) 0.6 and 0.6
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c) 0.6 and 0.5
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d) 0.5 and 0.6
Explanation
tan30=μs μs=0.6 h=ut+ ½ at2 4= ½ a(4)2 a= 0.5 is resultant acceleration Resultant acceleration a is due to down ward acceleration of gravity – retardation due to friction fk Now fk =μkmgcosθ, as gcosθ is perpendicular to plank Answer:(c)
Q.27
On a frictionless surface, a block of mass. M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v/The second block's speed after the collision is :-
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a)
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b)
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c)
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d)
Explanation
Collision is elastic according law of conservation of kinetic energy K.E. before collision = kinetic energy after collision Answer:(b)
Q.28
A rigid ball of mass m strikes a rigid wall at 60° and gets reflected without loss of speed as shown in the figure below. The value of impulse imparted by the wall on the ball will be … [ NEET II – 2016]
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a)
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b)
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c) mV
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d) 2mV
Explanation
Impulse Answer:(c)
Q.29
Two identical balls A and B having velocities of0.5 m/s and –0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be …[NEET II -2016]
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a) –0.3 m/s and 0.5 m/s
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b) 0.3 m/s and 0.5 m/s
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c) –0.5 m/s and 0.3 m/s
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d) 0.5 m/s and –0.3 m/s
Explanation
Since both bodies are identical and collision is elastic. Therefore velocities will be interchanged after collision. vA = –0.3 m/s and vB = 0.5 m/s Answer:(d)
Q.30
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively…[ NEET 2017]
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a) g, g/3
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b) g/3, g
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c) g, g
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d) g/3,g/3
Explanation
After cutting string between A and B. Block B will follow free fall and acceleration of block B = g On removal of mg force on block A, it will start to accelerate in upward direction. Before string is cut force which pulls up the spring is 4mg = kx And down ward force after cut is 3mg Net force = 4mg – 3mg Therefore 3ma =mg a = g/3 Answer:(a)
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