Multiple Choice Questions (MCQs) for JEE Physics: Kinematics
Review structured MCQ sets for JEE Physics Kinematics. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.
Practice Kinematics MCQs for JEE Physics
Access the complete set of multiple-choice questions for Kinematics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.
Multiple Choice Questions
(a) 1 s
(b) 2 s
(c) 4 s
(d) 10 s
Show Answer & Explanation
Answer: (a) 1 s
Explanation:
1. Take upward positive. Bullet: \( y_1=100t-\tfrac12gt^2 \); block: \( y_2=100-\tfrac12gt^2 \).
2. They meet when \( y_1=y_2 \): \( 100t=100 \), so \( t=1 \) s.
Teacher's Note:
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164. .
(a) 2h
(b) h
(c) \( \sqrt2\,h \)
(d) 4h
Show Answer & Explanation
Answer: (a) 2h
Explanation:
1. Time of fall: \( h=\tfrac12gt^2 \) gives \( t=\sqrt{2h/g} \).
2. Horizontal distance \( =\sqrt{2gh}\times\sqrt{2h/g}=2h \).
Teacher's Note:
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138. .
(a) 400 m
(b) 850 m
(c) 1250 m
(d) 2500 m
Show Answer & Explanation
Answer: (c) 1250 m
Explanation:
1. In the frame of A, B starts at rest and accelerates at 1 m/s\(^2\).
2. Relative displacement in 50 s \( =\tfrac12(1)(50)^2=1250 \) m.
3. The guard of B, at the rear of B, reaches the driver of A at the front of A after covering this relative distance, so the initial gap is 1250 m.
Teacher's Note:
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127. .
(a) 2 m/s\(^2\)
(b) about 2.7 m/s\(^2\)
(c) 1.8 m/s\(^2\)
(d) 3.8 m/s\(^2\)
Show Answer & Explanation
Answer: (b) about 2.7 m/s\(^2\)
Explanation:
1. Centripetal acceleration \( a_c=v^2/r=900/500=1.8 \) m/s\(^2\).
2. Tangential acceleration \( a_t=2 \) m/s\(^2\).
3. Total \( a=\sqrt{1.8^2+2^2}=\sqrt{7.24}\approx2.7 \) m/s\(^2\).
Teacher's Note:
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108. .
(a) 5 m
(b) 15 m
(c) 20 m
(d) 10 m
Show Answer & Explanation
Answer: (d) 10 m
Explanation:
1. At height \( H/2 \): \( v^2=u^2-2g(H/2)=u^2-gH \).
2. Since \( u^2=2gH \), \( v^2=gH \).
3. \( 100=10H \), so \( H=10 \) m.
Teacher's Note:
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118. .
(a) 8 m
(b) 4 m
(c) 16 m
(d) 6 m
Show Answer & Explanation
Answer: (a) 8 m
Explanation:
1. With constant retardation, \( s=u^2/2a \), so \( s\propto u^2 \).
2. Doubling the speed makes the distance four times: \( 4\times2=8 \) m.
Teacher's Note:
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(a) 7.0 m/s
(b) 14.0 m/s
(c) 21.0 m/s
(d) 10.5 m/s
Show Answer & Explanation
Answer: (b) 14.0 m/s
Explanation:
1. \( x(1)=4.5+3.5=8 \) m and \( x(3)=4.5+31.5=36 \) m.
2. Average velocity \( =(36-8)/2=14 \) m/s.
Teacher's Note:
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161. .
(a) 7 m/s
(b) 1 m/s
(c) 3.5 m/s
(d) 5 m/s
Show Answer & Explanation
Answer: (d) 5 m/s
Explanation:
1. \( v^2=2gh=9 \), so \( 2gh=9 \).
2. For the second body \( v^2=4^2+2gh=16+9=25 \), so \( v=5 \) m/s.
Teacher's Note:
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84. .
(a) \( \dfrac{\alpha\beta\,t}{\alpha+\beta} \)
(b) \( \dfrac{(\alpha+\beta)t}{\alpha\beta} \)
(c) \( \dfrac{(\alpha+\beta)t}{2} \)
(d) \( \dfrac{\alpha t}{\alpha+\beta} \)
Show Answer & Explanation
Answer: (a) \( \dfrac{\alpha\beta\,t}{\alpha+\beta} \)
Explanation:
1. Let the maximum velocity be v. Acceleration time \( t_1=v/\alpha \); deceleration time \( t_2=v/\beta \).
2. \( t_1+t_2=t \), so \( v\left(\dfrac1\alpha+\dfrac1\beta\right)=t \).
3. Hence \( v=\dfrac{\alpha\beta t}{\alpha+\beta} \).
Teacher's Note:
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123. .
(a) 600 m
(b) 2400 m
(c) 1200 m
(d) 1960 m
Show Answer & Explanation
Answer: (c) 1200 m
Explanation:
1. Speed \( =216\times\tfrac{5}{18}=60 \) m/s. Fall time: \( 1960=\tfrac12(9.8)t^2 \) gives \( t=20 \) s.
2. \( AB=60\times20=1200 \) m.
Teacher's Note:
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102. .
(a) 54 m/s
(b) 50 m/s
(c) 70 m/s
(d) 45 m/s
Show Answer & Explanation
Answer: (a) 54 m/s
Explanation:
1. \( v_x=20 \) m/s and \( v_y=gt=50 \) m/s.
2. \( v=\sqrt{20^2+50^2}=\sqrt{2900}\approx54 \) m/s.
Teacher's Note:
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103. .
(a) 60 km/h
(b) 56 km/h
(c) 48 km/h
(d) 72 km/h
Show Answer & Explanation
Answer: (a) 60 km/h
Explanation:
1. Average speed for equal distances is the harmonic mean: \( \dfrac{2\cdot40\,v}{40+v}=48 \).
2. \( 80v=1920+48v \), so \( v=60 \) km/h.
Teacher's Note:
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138. .
(a) 28.5 cm/s
(b) 18.0 cm/s
(c) 30.4 cm/s
(d) 22.5 cm/s
Show Answer & Explanation
Answer: (a) 28.5 cm/s
Explanation:
1. \( x(3)-x(2)=1.5(27-8)=28.5 \) cm in 1 s.
2. Average velocity \( =28.5 \) cm/s.
Teacher's Note:
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134. .
(a) 18.0 cm/s
(b) 28.5 cm/s
(c) 9.0 cm/s
(d) 13.5 cm/s
Show Answer & Explanation
Answer: (a) 18.0 cm/s
Explanation:
1. \( v=dx/dt=4.5t^2 \).
2. At \( t=2 \) s, \( v=4.5\times4=18 \) cm/s.
Teacher's Note:
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111. .
(a) 5.71 m/s\(^2\)
(b) 2.85 m/s\(^2\)
(c) 6.5 m/s\(^2\)
(d) 4.2 m/s\(^2\)
Show Answer & Explanation
Answer: (a) 5.71 m/s\(^2\)
Explanation:
1. \( v^2-u^2=2as \): \( 54^2-33^2=2a(160) \).
2. \( 2916-1089=1827=320a \), so \( a=5.71 \) m/s\(^2\).
Teacher's Note:
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99. .
(a) 95.4 m
(b) 47.7 m
(c) 190.8 m
(d) 160 m
Show Answer & Explanation
Answer: (a) 95.4 m
Explanation:
1. \( s=v^2/2a=1089/(2\times5.71) \).
2. \( s\approx95.4 \) m.
Teacher's Note:
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138. .
(a) 0.1 m/s
(b) 0.5 m/s
(c) 0 m/s
(d) 1.15 m/s
Show Answer & Explanation
Answer: (a) 0.1 m/s
Explanation:
1. Use \( s=ut-\tfrac12at^2 \). \( 2u-2a=2 \) and \( 6u-18a=4.2 \).
2. From the first, \( u=1+a \); then \( 6+6a-18a=4.2 \), so \( a=0.15 \) m/s\(^2\) and \( u=1.15 \) m/s.
3. \( v(7)=1.15-0.15\times7=0.1 \) m/s.
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137. .
(a) about 10.4 m
(b) about 25 m
(c) about 5 m
(d) about 20.8 m
Show Answer & Explanation
Answer: (a) about 10.4 m
Explanation:
1. Distance 250 m in 10 s and 500 m in 30 s: \( 250=10u-50a \) and \( 500=30u-450a \).
2. Solving, \( a=\tfrac{5}{6} \) m/s\(^2\) and \( u=\tfrac{175}{6}\approx29.17 \) m/s.
3. Total stopping distance \( =u^2/2a\approx510.4 \) m, so the extra distance is about 10.4 m.
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142. .
(a) 50 m
(b) 150 m
(c) 0 m (it hits the obstacle)
(d) 25 m
Show Answer & Explanation
Answer: (a) 50 m
Explanation:
1. Time to stop: \( t=20/2=10 \) s, so the car is already at rest at 15 s.
2. Stopping distance \( =20^2/(2\times2)=100 \) m.
3. Distance from the obstacle \( =150-100=50 \) m.
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147. .
(a) 5.10 s and 441.8 m
(b) 20.4 s and 1767 m
(c) 10.20 s and 883.7 m
(d) 10.20 s and 1020 m
Show Answer & Explanation
Answer: (c) 10.20 s and 883.7 m
Explanation:
1. \( T=2u\sin\theta/g=2(100)(0.5)/9.8=10.20 \) s.
2. \( R=u^2\sin2\theta/g=10^4(0.866)/9.8=883.7 \) m.
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