CBSE Class 10 Physics Motion MCQs Set 24

Answer: (a) 44.1 m

Explanation:
1. h = 0.5 x 9.8 x 3\( ^2 \) = 44.1 m.

Teacher's Note:
1. Use \( s = \tfrac12 gt^2 \) for a body dropped from rest.

Answer: (a) 10 m

Explanation:
1. Distance in nth second = \( u + a(2n-1)/2 = 3 + 2 \times 7/2 = 10 \) m.

Teacher's Note:
1. Check by subtraction: s(4) = 12 + 16 = 28 m, s(3) = 9 + 9 = 18 m, difference 10 m.

Answer: (b) uniform

Explanation:
1. For uniform acceleration the velocity changes linearly with time, so the average is \( (u+v)/2 \).

Teacher's Note:
1. This shortcut holds only for uniform acceleration.
2. For variable acceleration use total displacement over total time.

Answer: (c) \( \sqrt{\dfrac{2h}{g}} + \dfrac{h}{v} \)

Explanation:
1. Time to fall = \( \sqrt{2h/g} \).
2. Time for sound to travel up = h/v.
3. Total = \( \sqrt{2h/g} + h/v \).

Teacher's Note:
1. Add the fall time and the sound travel time.
2. This is the basis of the well-depth experiment.

Answer: (a) 15 m/s down

Explanation:
1. v = u - gt = 35 - 50 = -15 m/s.
2. The negative sign means the object is moving downward.

Teacher's Note:
1. Take upward as positive.
2. The stone reached the top at 3.5 s and is already falling.

Answer: (a) opposite

Explanation:
1. The cycle slows down, so acceleration (retardation) points against the motion.

Teacher's Note:
1. Slowing down means acceleration is opposite to velocity.
2. Speeding up means they point the same way.

Answer: (d) 0 m/s

Explanation:
1. Acceleration = (40 - 20)/(4 - 2) = 10 m/s\( ^2 \).
2. Initial velocity u = 20 - 10 x 2 = 0.

Teacher's Note:
1. Use two readings to get acceleration, then go back to t = 0.

Answer: (a) The car is moving on a straight road

Explanation:
1. The two are equal only when the body moves along a straight line without reversing.
2. In the other three cases the path is curved or reverses.

Teacher's Note:
1. Compare the path length with the straight line joining start and end.
2. Any turning or curving makes distance larger.

Answer: (d) 50 m

Explanation:
1. Net northward displacement = 50 - 20 = 30 m.
2. Eastward displacement = 40 m.
3. Magnitude = \( \sqrt{30^2+40^2} = 50 \) m.

Teacher's Note:
1. Displacement is the straight line from start to end, not the path length.
2. Resolve into perpendicular parts and use Pythagoras.

Answer: (a) 36 km/h

Explanation:
1. Total distance = 2D.
2. Total time = D/30 + D/45 = D/18.
3. Average speed = 2D / (D/18) = 36 km/h.

Teacher's Note:
1. Average speed uses total distance, so the return trip adds to it.
2. Average velocity for the round trip would be zero.

Answer: (c) 206 m/s

Explanation:
1. The stone starts with the balloon's velocity, 10 m/s downward.
2. v = 10 + 9.8 x 20 = 206 m/s.

Teacher's Note:
1. A released object keeps the velocity of its carrier at the instant of release.
2. Both velocity and g are downward, so they add.

Answer: (b) m/s, m/s\( ^2 \)

Explanation:
1. Each term must have the unit of s (metre).
2. at: a = m / s.
3. \( bt^2 \): b = m/s\( ^2 \).

Teacher's Note:
1. Check units term by term.
2. Terms added together must have the same unit.

Answer: (a) \( \dfrac{\pi}{2\sqrt{2}} \)

Explanation:
1. Distance = quarter circumference = \( \pi r/2 \).
2. Displacement = chord = \( \sqrt{2}\,r \).
3. Ratio = \( \dfrac{\pi r/2}{\sqrt{2}\,r} = \dfrac{\pi}{2\sqrt{2}} \).

Teacher's Note:
1. The chord of a quarter circle is \( r\sqrt{2} \).
2. Distance divided by displacement is always at least 1: here it is about 1.11.

Answer: (b) 28 m/s

Explanation:
1. v = u + at = 20 + 4 x 2 = 28 m/s.

Teacher's Note:
1. Use the first equation of motion.
2. The acceleration is in the same direction as the velocity.

Answer: (b) 160 m ; \( 80\pi \) m

Explanation:
1. Displacement = diameter = 2 x 80 = 160 m.
2. Distance = half the circumference = \( \pi r = 80\pi \) m.

Teacher's Note:
1. Half a circle gives displacement equal to the diameter.
2. Distance follows the curved path, so it equals half the circumference.

Answer: (d) equal to or less than 1

Explanation:
1. Displacement is the shortest path between two points, so its magnitude never exceeds the distance.
2. It equals the distance only for straight-line motion without turning back.

Teacher's Note:
1. Distance is always at least as large as the magnitude of displacement.
2. The ratio is 1 only for motion along a straight line in one direction.

Answer: (d) 34.3 m

Explanation:
1. Distance in 4 s = 0.5 x 9.8 x 16 = 78.4 m.
2. Distance in 3 s = 0.5 x 9.8 x 9 = 44.1 m.
3. Difference = 34.3 m.

Teacher's Note:
1. Equivalent: nth-second distance = \( g(2n-1)/2 \) = 9.8 x 7/2.

Answer: (a) zero

Explanation:
1. Acceleration = change in velocity / time = 0.

Teacher's Note:
1. Constant velocity means no acceleration.
2. Both magnitude and direction must stay the same.

Answer: (a) 0

Explanation:
1. Displacement = 0 because the body ends where it started.
2. Distance = \( 2 \times 2\pi R = 4\pi R \).
3. Ratio = 0.

Teacher's Note:
1. Complete circles always give zero displacement.
2. Distance keeps adding with every turn.

Answer: (d) 0 m/s\( ^2 \) ; 300 m

Explanation:
1. The graph is horizontal, so the velocity is constant and the slope is zero: a = 0.
2. Distance = area under graph = 20 x 15 = 300 m.

Teacher's Note:
1. Slope of a v-t graph gives acceleration.
2. Area under a v-t graph gives displacement.

Answer: (d) 17 : 13

Explanation:
1. Distance along the sides = 12 + 5 = 17 m.
2. Displacement = diagonal = \( \sqrt{12^2+5^2} = 13 \) m.
3. Same time, so ratio = 17 : 13.

Teacher's Note:
1. Speed uses path length, velocity uses the straight-line displacement.
2. Both are divided by the same time, so the ratio is just distance to displacement.