JEE Physics Motion In A Straight Line MCQs Set 28

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Chapter-wise Objective Questions: Motion In A Straight Line

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Multiple Choice Questions

Question 1: An object is projected upwards with a velocity of 100 m/s. It will strike the ground after approximately (take \( g=10 \) m/s\(^2\))
(a) 10 s
(b) 20 s
(c) 15 s
(d) 5 s
Show Answer & Explanation

Answer: (b) 20 s

Explanation:
1. Time of flight for return to the same level is \( 2u/g \).

Teacher's Note:
1. Key relation used: \( T=2u/g=20 \) s.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 2: A stone dropped from the top of a tower touches the ground in 4 s. The height of the tower is about (\( g=10 \) m/s\(^2\))
(a) 80 m
(b) 40 m
(c) 20 m
(d) 160 m
Show Answer & Explanation

Answer: (a) 80 m

Explanation:
1. Free fall from rest.

Teacher's Note:
1. Key relation used: \( h=\tfrac12gt^2=\tfrac12\times10\times16=80 \) m.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 3: A mass \( m \) slips along the wall of a semicircular surface of radius \( R \) (from the rim). The velocity at the bottom of the surface is
(a) \( \sqrt{Rg} \)
(b) \( \sqrt{2Rg} \)
(c) \( 2\sqrt{\pi Rg} \)
(d) \( \sqrt{\pi Rg} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{2Rg} \)

Explanation:
1. Height drop is \( R \).

Teacher's Note:
1. Key relation used: Energy conservation: \( mgR=\tfrac12mv^2 \).
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 4: A body is slipping from an inclined plane of height \( h \) and length \( l \). If the angle of inclination is \( \theta \), the time taken by the body to come from the top to the bottom of this inclined plane is
(a) \( \sqrt{2h/g} \)
(b) \( \sqrt{2l/g} \)
(c) \( \dfrac1{\sin\theta}\sqrt{2h/g} \)
(d) \( \sin\theta\sqrt{2h/g} \)
Show Answer & Explanation

Answer: (c) \( \dfrac1{\sin\theta}\sqrt{2h/g} \)

Explanation:
1. Acceleration along a smooth incline is \( g\sin\theta \).

Teacher's Note:
1. Key relation used: \( l=\tfrac12(g\sin\theta)t^2 \) and \( l=h/\sin\theta \) give \( t=\dfrac1{\sin\theta}\sqrt{\dfrac{2h}g} \).
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 5: A particle is projected up with an initial velocity of 80 ft/s. The ball will be at a height of 96 ft from the ground after (\( g=32 \) ft/s\(^2\))
(a) 2.0 and 3.0 s
(b) only at 3.0 s
(c) only at 2.0 s
(d) after 1 and 2 s
Show Answer & Explanation

Answer: (a) 2.0 and 3.0 s

Explanation:
1. It passes 96 ft once going up and once coming down.

Teacher's Note:
1. A quadratic gives both times.

Question 6: A body falls from rest. Its velocity at the end of the first second is (\( g=32 \) ft/s\(^2\))
(a) 16 ft/s
(b) 32 ft/s
(c) 64 ft/s
(d) 24 ft/s
Show Answer & Explanation

Answer: (b) 32 ft/s

Explanation:
1. Free fall from rest.

Teacher's Note:
1. Key relation used: \( v=gt=32\times1 \).
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 7: Two stones of different masses are dropped simultaneously from the top of a building.
(a) the smaller stone hits the ground earlier
(b) the larger stone hits the ground earlier
(c) both stones reach the ground simultaneously
(d) which stone reaches earlier depends on the composition of the stone
Show Answer & Explanation

Answer: (c) both stones reach the ground simultaneously

Explanation:
1. Mass cancels in the equations of motion.

Teacher's Note:
1. Key relation used: Free-fall acceleration is the same for all masses (no air resistance).
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 8: A body thrown with an initial speed of 96 ft/s reaches the ground after (\( g=32 \) ft/s\(^2\))
(a) 3 s
(b) 6 s
(c) 12 s
(d) 8 s
Show Answer & Explanation

Answer: (b) 6 s

Explanation:
1. Time of flight \( 2u/g \).

Teacher's Note:
1. Key relation used: Thrown vertically upward from the ground: \( T=2u/g=6 \) s.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 9: A stone is dropped from a certain height which can reach the ground in 5 s. If the stone is stopped after 3 s of its fall and then allowed to fall again, the time taken by the stone to reach the ground for the remaining distance is
(a) 2 s
(b) 3 s
(c) 4 s
(d) none of these
Show Answer & Explanation

Answer: (c) 4 s

Explanation:
1. Restart from rest.
2. \( \tfrac12gt^2=8g \) gives \( t=4 \) s.

Teacher's Note:
1. After being stopped the speed is zero again.

Question 10: A man in a balloon rising vertically with an acceleration of 4.9 m/s\(^2\) releases a ball 2 s after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (\( g=9.8 \) m/s\(^2\))
(a) 14.7 m
(b) 19.6 m
(c) 9.8 m
(d) 24.5 m
Show Answer & Explanation

Answer: (a) 14.7 m

Explanation:
1. \( v=at=9.8 \), \( h=\tfrac12at^2=9.8 \).
2. After release the ball rises \( \dfrac{v^2}{2g}=4.9 \) m more.

Teacher's Note:
1. The ball keeps the balloon's upward velocity at release.

Question 11: A particle is dropped under gravity from rest from a height \( h \) (\( g=9.8 \) m/s\(^2\)) and it travels a distance \( 9h/25 \) in the last second. The height \( h \) is
(a) 100 m
(b) 122.5 m
(c) 145 m
(d) 167.5 m
Show Answer & Explanation

Answer: (b) 122.5 m

Explanation:
1. Distance in the last second \( =\tfrac12g[t^2-(t-1)^2] \).
2. \( 9t^2-50t+25=0 \), \( t=5 \) s (the root \( 5/9 \) is less than 1 s).

Teacher's Note:
1. Reject the root smaller than 1 s.

Question 12: An aeroplane is moving with a velocity \( u \). It drops a packet from a height \( h \). The time \( t \) taken by the packet to reach the ground will be
(a) \( \sqrt{2g/h} \)
(b) \( 2u/g \)
(c) \( \sqrt{h/2g} \)
(d) \( \sqrt{2h/g} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{2h/g} \)

Explanation:
1. \( h=\tfrac12gt^2 \).

Teacher's Note:
1. Key relation used: The vertical motion is free fall from rest.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 13: An aeroplane is moving with horizontal velocity \( u \) at height \( h \). The velocity of a packet dropped from it on the earth's surface will be (\( g \) is the acceleration due to gravity)
(a) \( \sqrt{u^2+2gh} \)
(b) \( \sqrt{2gh} \)
(c) \( 2gh \)
(d) \( \sqrt{u^2-2gh} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{u^2+2gh} \)

Explanation:
1. The resultant speed is the vector sum.

Teacher's Note:
1. Key relation used: Horizontal component \( u \), vertical component \( \sqrt{2gh} \).
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 14: A rocket is fired upward from the earth's surface such that it creates an acceleration of 19.6 m/s\(^2\). If after 5 s its engine is switched off, the maximum height of the rocket from the earth's surface would be
(a) 245 m
(b) 490 m
(c) 980 m
(d) 735 m
Show Answer & Explanation

Answer: (d) 735 m

Explanation:
1. \( g=9.8 \) m/s\(^2\) after cutoff.

Teacher's Note:
1. Key relation used: Powered stage: \( v=98 \) m/s, \( h_1=245 \) m. Then rises \( v^2/2g=490 \) m more; total 735 m.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 15: A bullet is fired with a speed of 1000 m/s in order to hit a target 100 m away. If \( g=10 \) m/s\(^2\), the gun should be aimed
(a) directly towards the target
(b) 5 cm above the target
(c) 10 cm above the target
(d) 15 cm above the target
Show Answer & Explanation

Answer: (b) 5 cm above the target

Explanation:
1. Horizontal and vertical motions are independent.

Teacher's Note:
1. Key relation used: \( t=0.1 \) s; drop \( =\tfrac12\times10\times0.01=5 \) cm.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 16: A body starts to fall freely under gravity. The distances covered by it in the first, second and third second are in the ratio
(a) 1 : 3 : 5
(b) 1 : 2 : 3
(c) 1 : 4 : 9
(d) 1 : 5 : 6
Show Answer & Explanation

Answer: (a) 1 : 3 : 5

Explanation:
1. \( s_n=\tfrac g2(2n-1) \).

Teacher's Note:
1. Key relation used: Distances in successive seconds are in the ratio of odd numbers.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 17: P, Q and R are three balloons ascending with velocities \( U \), \( 4U \) and \( 8U \) respectively. If stones of the same mass are dropped from each at the same height, then
(a) they reach the ground at the same time
(b) stone from P reaches the ground first
(c) stone from R reaches the ground first
(d) stone from Q reaches the ground first
Show Answer & Explanation

Answer: (b) stone from P reaches the ground first

Explanation:
1. \( t\) grows with the upward launch speed.

Teacher's Note:
1. Key relation used: Each stone starts upward with the balloon's speed, so the smallest upward speed lands first.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 18: A body is projected up with a speed \( u \) and the time taken by it is \( T \) to reach the maximum height \( H \). Pick out the correct statement.
(a) it reaches \( H/2 \) in \( T/2 \) s
(b) it acquires velocity \( u/2 \) in \( T/2 \) s
(c) its velocity is \( u/2 \) at \( H/2 \)
(d) same velocity at \( 2T \)
Show Answer & Explanation

Answer: (b) it acquires velocity \( u/2 \) in \( T/2 \) s

Explanation:
1. At \( T/2 \) the height reached is \( 3H/4 \), not \( H/2 \).

Teacher's Note:
1. Key relation used: \( v=u-gt \) with \( T=u/g \): at \( T/2 \), \( v=u/2 \).
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 19: A ball is dropped from a height \( h \) with an initial speed zero and strikes the ground with a velocity of 3 km/h. Another ball of the same mass is dropped from the same height \( h \) with an initial speed of 4 km/h downward. The final velocity of the second ball with which it strikes the ground is
(a) 3 km/h
(b) 4 km/h
(c) 5 km/h
(d) 12 km/h
Show Answer & Explanation

Answer: (c) 5 km/h

Explanation:
1. From the first ball \( 2gh=9 \) (km/h)\(^2\).

Teacher's Note:
1. Key relation used: \( v^2=u^2+2gh \): \( 4^2+3^2=25 \).
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 20: A ball of mass \( m_1 \) and another ball of mass \( m_2 \) are dropped from equal height. If the time taken by the balls are \( t_1 \) and \( t_2 \) respectively, then
(a) \( t_1=t_2/2 \)
(b) \( t_1=t_2 \)
(c) \( t_1=4t_2 \)
(d) \( t_1=t_2/4 \)
Show Answer & Explanation

Answer: (b) \( t_1=t_2 \)

Explanation:
1. Neglect air resistance.

Teacher's Note:
1. Key relation used: Fall time is independent of mass.
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 21: A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top?
(a) 1 s
(b) 2 s
(c) 4 s
(d) 16 s
Show Answer & Explanation

Answer: (b) 2 s

Explanation:
1. Constant acceleration from rest.

Teacher's Note:
1. Key relation used: \( s\propto t^2 \): \( t'=4\sqrt{1/4}=2 \) s.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

Question 22: A ball is dropped downwards. After 1 second another ball is dropped downwards from the same point. What is the distance between them after 3 seconds? (\( g=10 \) m/s\(^2\))
(a) 25 m
(b) 20 m
(c) 50 m
(d) 9.8 m
Show Answer & Explanation

Answer: (a) 25 m

Explanation:
1. \( \tfrac12\times10\times9-\tfrac12\times10\times4=25 \).

Teacher's Note:
1. Key relation used: First has fallen 45 m and the second 20 m: 25 m.
2. Check the direction of each proportionality (direct or inverse) before substituting numbers.

Question 23: A stone is thrown with an initial speed of 4.9 m/s from a bridge in the vertically upward direction. It falls down in water after 2 s. The height of the bridge is (\( g=9.8 \) m/s\(^2\))
(a) 4.9 m
(b) 9.8 m
(c) 19.8 m
(d) 24.7 m
Show Answer & Explanation

Answer: (b) 9.8 m

Explanation:
1. Upward positive; displacement is negative.

Teacher's Note:
1. Key relation used: \( -h=4.9\times2-\tfrac12\times9.8\times4=-9.8 \), so \( h=9.8 \) m.
2. State the governing law first, then substitute; this avoids sign and power-of-ten slips.

Question 24: A body freely falling from rest has a velocity \( v \) after it falls through a height \( h \). The distance it has to fall down for its velocity to become double is
(a) \( 2h \)
(b) \( 4h \)
(c) \( 6h \)
(d) \( 8h \)
Show Answer & Explanation

Answer: (b) \( 4h \)

Explanation:
1. \( v^2=2gh \).

Teacher's Note:
1. Key relation used: \( v^2\propto h \), so doubling v needs \( 4h \).
2. After finding the value, test it against the options by a quick order-of-magnitude estimate.

Question 25: The time taken by a block of wood (initially at rest) to slide down a smooth inclined plane 9.8 m long (angle of inclination \( 30^\circ \)) is
(a) \( \tfrac12 \) s
(b) 2 s
(c) 4 s
(d) 1 s
Show Answer & Explanation

Answer: (b) 2 s

Explanation:
1. Smooth incline acceleration is \( g\sin\theta \).

Teacher's Note:
1. Key relation used: \( a=g\sin30^\circ=4.9 \) m/s\(^2\); \( t=\sqrt{2\times9.8/4.9}=2 \) s.
2. Reduce the data to ratios and keep every quantity in SI units before calculating.

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