JEE Physics Laws Of Motion MCQs Set 19

Physics Objective Questions and Answers: Laws Of Motion

Access targeted multiple-choice questions for Laws Of Motion designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Laws Of Motion MCQs with Answers

View or download the dedicated Laws Of Motion MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: A block of mass m is placed on a surface whose vertical cross-section is given by \( y=\frac{x^3}{6} \). If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is:
(a) \( \frac16 \) m
(b) \( \frac13 \) m
(c) \( \frac12 \) m
(d) \( \frac23 \) m
Show Answer & Explanation

Answer: (a) \( \frac16 \) m

Explanation:
1. The block is on the verge of slipping when \( \tan\phi=\mu \), where \( \tan\phi \) is the slope of the surface.
2. Slope \( =\frac{dy}{dx}=\frac{x^2}{2}=0.5 \), so \( x=1 \) m.
3. Height \( y=\frac{1^3}{6}=\frac16 \) m.

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Question 2: A block of weight W hangs from a spring balance that reads W1, while the lower surface of the block rests on a weighing machine that reads W2 (all in the same unit). Which statements are correct? (I) \( W=W_1+W_2 \) if the system is at rest. (II) \( W>W_1+W_2 \) if the system moves down with some acceleration. (III) \( W_1>W_2 \) if the system moves up with some acceleration. (IV) No relation between W, W1 and W2 can be obtained.
(a) I and III only
(b) I and II only
(c) II and III only
(d) IV only
Show Answer & Explanation

Answer: (b) I and II only

Explanation:
1. The block is acted on by W downward, the spring pull W1 upward and the normal force W2 upward.
2. At rest: \( W_1+W_2=W \) (I true). Moving down with acceleration a: \( W-(W_1+W_2)=\frac{W}{g}a>0 \), so \( W>W_1+W_2 \) (II true).
3. Nothing fixes how the load is shared between the spring and the machine, so W1 > W2 cannot be concluded (III false), and IV is false because the relation in I and II exists.

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Question 3: A particle O of mass m is attached to a vertical rod by two inextensible strings AO (upper) and BO (lower) of equal length \( \ell \). The distance between the points of suspension A and B on the rod is also \( \ell \). The set-up rotates about the rod with angular frequency \( \omega \). Then:
(a) The tension in BO is greater
(b) The tensions in the two strings are equal
(c) The tension in AO is greater
(d) The tension in AO or BO is greater according as the rotation is anticlockwise or clockwise
Show Answer & Explanation

Answer: (c) The tension in AO is greater

Explanation:
1. The triangle AOB is equilateral, so each string makes \( 30^\circ \) with the horizontal; AO slopes upward from O and BO slopes downward.
2. Vertical equilibrium: \( T_{AO}\sin30^\circ-T_{BO}\sin30^\circ=mg \).
3. So \( T_{AO}-T_{BO}=2mg>0 \): the upper string AO has the greater tension, whatever the sense of rotation.

Teacher's Note:
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Question 4: A block of mass 10 kg is placed in a box which moves with a constant acceleration of 5 m/s² directed at \( 53^\circ \) above the horizontal. The force exerted by the box on the block in the vertical direction is (g = 10 m/s², \( \tan53^\circ=4/3 \)):
(a) 40 N
(b) 50 N
(c) 150 N
(d) 140 N
Show Answer & Explanation

Answer: (d) 140 N

Explanation:
1. The vertical component of the acceleration is \( a\sin53^\circ=5\times0.8=4\ \text{m/s}^2 \) upward.
2. Vertical equation: \( N-mg=m(4) \).
3. \( N=10(10+4)=140 \) N.

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Question 5: The force required to just move a body up an inclined plane is double the force required to just prevent the body from sliding down. If the coefficient of friction is \( \mu \), the inclination \( \theta \) of the plane is:
(a) \( \tan^{-1}(3\mu) \)
(b) \( \tan^{-1}(\mu) \)
(c) \( \tan^{-1}(\mu/2) \)
(d) \( \tan^{-1}(2\mu) \)
Show Answer & Explanation

Answer: (a) \( \tan^{-1}(3\mu) \)

Explanation:
1. Force to move up: \( F_1=mg(\sin\theta+\mu\cos\theta) \). Force to prevent sliding: \( F_2=mg(\sin\theta-\mu\cos\theta) \).
2. \( F_1=2F_2 \) gives \( \sin\theta+\mu\cos\theta=2\sin\theta-2\mu\cos\theta \).
3. \( \sin\theta=3\mu\cos\theta \), so \( \theta=\tan^{-1}(3\mu) \).

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Question 6: Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in a circle of radius \( \frac r2 \) and the lighter one in a circle of radius r. They experience the same centripetal force. The tangential speed of the lighter stone is n times that of the heavier stone. The value of n is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. Lighter stone: \( F=\frac{mv_1^2}{r} \). Heavier stone: \( F=\frac{2m\,v_2^2}{r/2}=\frac{4mv_2^2}{r} \).
2. Equating: \( v_1^2=4v_2^2 \), so \( v_1=2v_2 \).
3. Hence \( n=2 \).

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Question 7: Three equal masses m are connected by a string over a smooth pulley: mass \( m_1 \) hangs freely while \( m_2 \) and \( m_3 \) both lie on a rough horizontal table (coefficient of friction \( \mu \)), connected to m1 through the string. The downward acceleration of m1 is:
(a) \( \frac{g(1-g\mu)}{9} \)
(b) \( \frac{2g\mu}{3} \)
(c) \( \frac{g(1-2\mu)}{3} \)
(d) \( \frac{g(1-2\mu)}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{g(1-2\mu)}{3} \)

Explanation:
1. Driving force \( mg \); friction on the two masses on the table \( =2\mu mg \).
2. Total mass moving \( =3m \).
3. \( a=\frac{mg-2\mu mg}{3m}=\frac{g(1-2\mu)}{3} \).

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Question 8: An explosion breaks a rock into three parts in a horizontal plane. Two of them fly off at right angles to each other: the first of mass 1 kg with speed 12 m/s and the second of mass 2 kg with speed 8 m/s. If the third part flies off with speed 4 m/s, its mass is:
(a) 7 kg
(b) 17 kg
(c) 3 kg
(d) 5 kg
Show Answer & Explanation

Answer: (d) 5 kg

Explanation:
1. Momenta of the first two: 12 kg m/s and 16 kg m/s at right angles. Resultant \( =\sqrt{12^2+16^2}=20 \) kg m/s.
2. The rock was at rest, so the third part has momentum 20 kg m/s in the opposite direction.
3. \( m_3=\frac{20}{4}=5 \) kg.

Teacher's Note:
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Question 9: A person of mass 60 kg is inside a lift of mass 940 kg. The lift starts moving upward with an acceleration of 1.0 m/s². If g = 10 m/s², the tension in the supporting cable is:
(a) 11000 N
(b) 8600 N
(c) 9680 N
(d) 1200 N
Show Answer & Explanation

Answer: (a) 11000 N

Explanation:
1. Treat the lift and person as one system of mass \( 60+940=1000 \) kg.
2. \( T-1000g=1000\times1 \).
3. \( T=1000\times11=11000 \) N.

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Question 10: Sand is dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s is:
(a) Mv/2 newton
(b) Mv newton
(c) zero
(d) 2Mv newton
Show Answer & Explanation

Answer: (b) Mv newton

Explanation:
1. Each second, a mass M of sand is brought from rest to speed v.
2. Force \( =\frac{dp}{dt}=v\frac{dm}{dt} \).
3. \( F=Mv \) newton (half the work done goes into heat, but the force is Mv).

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Question 11: A man of mass 80 kg stands on a weighing scale in a lift moving upward with a uniform acceleration of 5 m/s². The reading of the scale in newton is (g = 10 m/s²):
(a) zero
(b) 400 N
(c) 1200 N
(d) 800 N
Show Answer & Explanation

Answer: (c) 1200 N

Explanation:
1. \( N-mg=ma \).
2. \( N=80(10+5) \).
3. \( N=1200 \) N.

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Question 12: A block of mass 1 kg is placed on a truck with coefficient of friction 0.6. The truck moves with an acceleration of 5 m/s². The frictional force on the block is (g = 10 m/s²):
(a) 6 N
(b) 5.88 N
(c) 8 N
(d) 5 N
Show Answer & Explanation

Answer: (d) 5 N

Explanation:
1. Maximum friction \( =0.6\times1\times10=6 \) N.
2. The block needs \( ma=5 \) N to move with the truck, which is less than 6 N, so it does not slip.
3. Friction \( =5 \) N (static, equal to the force needed).

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Question 13: In a rocket, fuel burns at the rate of 1 kg/s and is ejected with a velocity of 60 km/s. The force this exerts on the rocket is:
(a) 60000 N
(b) 6000 N
(c) 60 N
(d) 600 N
Show Answer & Explanation

Answer: (a) 60000 N

Explanation:
1. Thrust \( =v\frac{dm}{dt} \).
2. \( v=60\ \text{km/s}=6\times10^4 \) m/s.
3. \( F=6\times10^4\times1=60000 \) N.

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Question 14: Forces F1, F2 and F3 act on a particle of mass m such that F2 and F3 are mutually perpendicular and the particle stays at rest. If F1 is now removed, the acceleration of the particle is:
(a) \( F_2F_3/(mF_1) \)
(b) \( F_1/m \)
(c) \( (F_2-F_3)/m \)
(d) \( F_2/m \)
Show Answer & Explanation

Answer: (b) \( F_1/m \)

Explanation:
1. At rest, the net force is zero, so F2 and F3 together balance F1.
2. After F1 is removed, the net force is the resultant of F2 and F3, which equals \( F_1 \) in magnitude.
3. \( a=F_1/m \).

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Question 15: Two forces are such that the sum of their magnitudes is 18 N and their resultant, of magnitude 12 N, is perpendicular to the smaller force. The magnitudes of the forces are:
(a) 12 N, 6 N
(b) 10 N, 8 N
(c) 13 N, 5 N
(d) 16 N, 2 N
Show Answer & Explanation

Answer: (c) 13 N, 5 N

Explanation:
1. Let the forces be P (smaller) and Q. The resultant R is perpendicular to P, so Q is the hypotenuse: \( Q^2=P^2+R^2 \).
2. \( Q^2-P^2=144 \) and \( Q+P=18 \), so \( Q-P=8 \).
3. \( Q=13 \) N, \( P=5 \) N.

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Question 16: A block of mass M is pulled along a horizontal frictionless surface by a uniform rope of mass m. A force P is applied at the free end of the rope. The force exerted by the rope on the block is:
(a) \( \frac{Pm}{M+m} \)
(b) \( \frac{PM}{M-m} \)
(c) \( P \)
(d) \( \frac{PM}{M+m} \)
Show Answer & Explanation

Answer: (d) \( \frac{PM}{M+m} \)

Explanation:
1. Acceleration of the whole system: \( a=\frac{P}{M+m} \).
2. The rope pulls the block with the force that gives it this acceleration: \( F=Ma \).
3. \( F=\frac{PM}{M+m} \).

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Question 17: A horizontal force of 10 N is necessary to just hold a block stationary against a vertical wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is:
(a) 2 N
(b) 20 N
(c) 50 N
(d) 100 N
Show Answer & Explanation

Answer: (a) 2 N

Explanation:
1. Normal force \( N=10 \) N.
2. Friction balances the weight at the limit: \( W=\mu N \).
3. \( W=0.2\times10=2 \) N.

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Question 18: A machine gun fires bullets of mass 40 g with a velocity of 1200 m/s. The man holding it can exert a maximum force of 144 N on the gun. The maximum number of bullets he can fire per second is:
(a) one
(b) three
(c) two
(d) four
Show Answer & Explanation

Answer: (b) three

Explanation:
1. Momentum of each bullet \( =0.04\times1200=48 \) kg m/s.
2. Force on the gun \( =n\times48\le144 \).
3. \( n\le3 \).

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Question 19: A block is kept on a frictionless inclined surface of inclination \( \alpha \). The incline is given a horizontal acceleration a to keep the block stationary relative to it. Then a is equal to:
(a) \( g \)
(b) \( g\csc\alpha \)
(c) \( g\tan\alpha \)
(d) \( g/\tan\alpha \)
Show Answer & Explanation

Answer: (c) \( g\tan\alpha \)

Explanation:
1. Forces on the block: weight mg and the normal reaction N perpendicular to the incline.
2. Vertical: \( N\cos\alpha=mg \). Horizontal: \( N\sin\alpha=ma \).
3. \( a=g\tan\alpha \).

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Question 20: A car moves on a straight road with a speed of 100 m/s. The minimum distance in which the car can be stopped is (\( \mu_k=0.5 \), g = 10 m/s²):
(a) 400 m
(b) 100 m
(c) 800 m
(d) 1000 m
Show Answer & Explanation

Answer: (d) 1000 m

Explanation:
1. Maximum retardation \( =\mu_kg=5\ \text{m/s}^2 \).
2. \( s=\frac{v^2}{2a}=\frac{10000}{10} \).
3. \( s=1000 \) m.

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Question 21: A body of mass 3.513 kg moves along the x-axis with a speed of 5.00 m/s. The magnitude of its momentum, recorded with the correct number of significant figures, is:
(a) 17.6 kg m/s
(b) 17.565 kg m/s
(c) 17.56 kg m/s
(d) 17.57 kg m/s
Show Answer & Explanation

Answer: (a) 17.6 kg m/s

Explanation:
1. \( p=mv=3.513\times5.00=17.565 \) kg m/s.
2. The speed has 3 significant figures, the mass has 4. The product can only carry 3.
3. Rounded to 3 significant figures: 17.6 kg m/s.

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Question 22: Masses of 10 kg and 20 kg are connected by a massless spring on a frictionless horizontal surface. A force of 200 N acts on the 20 kg mass. At the instant considered, the 10 kg mass has an acceleration of 12 m/s². The acceleration of the 20 kg mass is:
(a) \( 6\ \text{m/s}^2 \)
(b) \( 4\ \text{m/s}^2 \)
(c) \( 8\ \text{m/s}^2 \)
(d) \( 12\ \text{m/s}^2 \)
Show Answer & Explanation

Answer: (b) \( 4\ \text{m/s}^2 \)

Explanation:
1. Spring force on the 10 kg mass \( =10\times12=120 \) N.
2. The spring pulls back on the 20 kg mass with 120 N.
3. \( a=\frac{200-120}{20}=4\ \text{m/s}^2 \).

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Question 23: A spring toy of weight 1 kg rests on a weighing machine. The toy suddenly jumps up and the machine reads 11 N at that instant. The acceleration of the toy, in m/s², just after jumping is (g = 10 m/s²):
(a) \( 11\ \text{m/s}^2 \)
(b) \( 10\ \text{m/s}^2 \)
(c) \( 1\ \text{m/s}^2 \)
(d) \( 0.1\ \text{m/s}^2 \)
Show Answer & Explanation

Answer: (c) \( 1\ \text{m/s}^2 \)

Explanation:
1. The machine pushes the toy up with 11 N (equal to the force the toy applies on it) while gravity pulls down 10 N.
2. \( 11-10=1\times a \).
3. \( a=1\ \text{m/s}^2 \).

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