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
(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.
Teacher's Note:
1. L
2. i
3. m
4. i
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7. n
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9.
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150. .
(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.
Teacher's Note:
1. W
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153. .
(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:
1. R
2. e
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144. .
(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.
Teacher's Note:
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149. .
(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 \( \theta=\tan^{-1}(3\mu) \).
Teacher's Note:
1. F
2. r
3. i
4. c
5. t
6. i
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10. r
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104. .
(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 \( n=2 \).
Teacher's Note:
1. E
2. q
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138. .
(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. \( a=\frac{mg-2\mu mg}{3m}=\frac{g(1-2\mu)}{3} \).
Teacher's Note:
1. F
2. o
3. r
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156. .
(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:
1. I
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123. .
(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.
Teacher's Note:
1. T
2. h
3. e
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6. a
7. b
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(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).
Teacher's Note:
1. T
2. h
3. e
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5. f
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(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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(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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137. .
(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.
Teacher's Note:
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(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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(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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(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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(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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(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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118. .
(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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109. .
(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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93. .
(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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72. .
(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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118. .
(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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Download Chapter MCQs: JEE Physics Laws Of Motion
JEE Physics Laws Of Motion Objective Test Questions
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NCERT-Aligned Objective Questions and Solutions
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FAQs
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