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Physics NEET MCQ
Quiz 14
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
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)
Q.25
than one correct option Q423) In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is µ1 and that between the floor and the ladder is µThe normal reaction of the wall on the ladder is N1 and that of the floor is NIf the ladder is about to slip, then[ IIT Advance 2013]
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a)
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b)
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c)
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d)
Explanation
i) Linear equilibrium mg = N2 + µ1N1 From figure µ2N2 = N1 mg = N2 + µ1µ2N2 (µ1µ2 +1 ) N2 = mg ∴ Option c correct From figure N1 tanθ=N2 ii) From rotational equilibrium Let l be the length of ladder Clock wise motion torque = mg (l/2) cosθ Anti-clock wise motion torque = N1lsinθ Option d correct Answer:(c, d)
Q.26
A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see figure) on aninclined plane with angle of inclination θ. Various values of θ are given in List I. The coefficient of friction between the block m1 and the plane is always zero. The coefficientof static and dynamic friction between the block m2 and the plane are equal to µ = 0. In List II expressions for the friction on block m2 are given. Match the correct expression ofthe friction in List II with the angles given in List I, and choose the correct option. Theacceleration due to gravity is denoted by g. [Useful information : tan (5.5°) θ ≈ 0.1; tan(11.5°) ≈ 0.2; tan(16.5°) ≈ 0.3]
List I
List II
P. θ = 5°
m
2
g sinθ
Q. θ = 10°
(m
1
+ m
2
) g sinθ
R. θ = 15°
µm
2
g cosθ
S. θ = 20°
µ (m
1
+ m
2
) g cosθ
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a) P-1, Q-1, R-1, S-3
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b) P-2, Q-2, R-2, S-3
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c) P-2, Q-2, R-2, S-4
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d) P-2, Q-2, R-3, S-3
Explanation
For equilibrium condition f = m1gsinθ + m2gsinθ …(i) friction f = µN = μm2gcosθ ..(ii) µm2gcosθ = m1gsinθ + m2gsinθ µm2g =( m1g + m2g) tanθ 0.3 × 2 = (1+2) tanθ tanθ = 0.2 If angle less or equal to 11.5 Masses will not slip List I P ,θ = 5°< 11.5° List I Q ,θ = 10°< 11.5° equilibrium conditions given in (i) hold good Thus P → 2, Q →2 List I R. θ = 15° List I S. θ = 20° Equilibrium condition given in (i) not followed masses will slip And friction will be as per (ii) Thus R → 3, S →3 P → 2, Q →2, R → 3, S →3 Answer: d Answer:(d)
Q.27
A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.
List I
List II
P. Lift is accelerating vertically up.
d = 1.2 m
Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.
d > 1.2 m
R. Lift is moving vertically up with constant speed.
d < 1.2 m
S. Lift is falling freely.
No water leaks out of the jar
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a) P-2, Q-3, R-2, S-4
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b) P-2, Q-3, R-1, S-4
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c) P-1, Q-1, R-1, S-4
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d) P-2, Q-3, R-1, S-1
Explanation
Let level of water in jar be h, and be at height H When lift was not moving velocity of water ejected from hole Time taken for water to reach floor, vertical velocity initial is zero Horizontal distance So long resultant g ≠ 0 P. Lift is accelerating vertically up. Resultant g’ > g, so d is intendant of g , d = 1.2 m Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. Resultant g’ < g, so d is intendant of g , d = 1.2 m R. Lift is moving vertically up with constant speed. Resultant g’ =g, so d is intendant of g , d = 1.2 m S. Lift is falling freely. Resultant g’ =0, so no water will come out of jar P, Q, R → 1 and S → 4 Answer:(c)
Q.28
A uniform wooden stick of mass 1.6 kg and length “l” rests in an inclined manner on a smooth, vertical wall of height h(
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a)
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b)
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c)
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d)
Explanation
Linear equilibrium Nsin30 + N = mg And fr= Ncos30 Rotational equilibrium at point A Option d correct Answer:(d)
Q.29
than one correct option Q428) A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true ?
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a) The resistive force experienced by the plate is proportional to v
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b) The pressure difference between the leading and trailing faces of the plate is proportional to uv.
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c) The plate will continue to move with constant non-zero acceleration, at all times
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d) At a later time the external force F balances the resistive force.
Explanation
Let m be the mass of gas molecules. Plate is moving with velocity v. Speed of gas molecules with respect to leading plate is u+v. gas molecule under goes a collision with plate change in momentum of plate due to one molecule= 2m(u+v). In time ∆t, volume swept by plate of unit area =vΔt If n is the number of molecules per unit volume can be considered as constant as pressure is low. Then number of molecules collide with surface = nv∆t Thus momentum gain by plate per unit area = nvΔt × 2m(u+v). = 2mn (vu+v2)Δt Force on leading face per unit area = PL = change in momentum/Δt = 2mn (vu+v2) For rear face velocity of gas with respect to plate is v-u For rear face force per unit area = PR = 2mn (v2 -uv) Thus force difference per unit area = pressure difference = PL- PR 2mn (vu+v2) - 2mn (v2-uv ) = 2mn (2uv) = 4mn(uv) Thus option [C] correct Force per unit area due to gas = 4mn(uv) thus proportional to v [ option a correct] Net force on plate = F(applied )- F (resistive) Net force on plate = F – 4mn(uv) Here F is constant thus at a particular velocity of plate F = 4mn(uv) and acceleration of plate will be zero [ Option c is wrong and option d correct] Answer:(a, b, d)
Q.30
A block A of mass m1 rests on a horizontal table. A light 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 a block slide on the table what is the tension in the string
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Explanation
Here m2 > m1 .Both the blocks will move with same acceleration 'a' Forces are shown in figure For block m2 m2a = m2g – T … (1) For block m1 m1a = T –fk … (2) To find T we have to eliminate a as we have assumed it (1) + (2) m2a+ m1a = m2g -fk Now fk = µkN and N = m1g ; substituting value of fk = µk m1g ...(3) Answer : (c)
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