CBSE Class 10 Physics Motion MCQs Set 13

Answer: (a) 68.6

Explanation:
1. Time for the first 50 km = 50/60 = 0.833 h.
2. Time for the next 50 km = 50/80 = 0.625 h.
3. Average speed = 100 / 1.458 = 68.6 km/h.

Teacher's Note:
1. Equal distances at different speeds give the harmonic mean, not the simple average.
2. The simple average (70) is a common wrong answer.

Answer: (b) it covers equal distances in equal intervals of time

Explanation:
1. Constant speed means equal distances in equal times.
2. Position changes, and direction may change (for example on a circle), so acceleration need not be zero.

Teacher's Note:
1. Speed measures only how fast, not which way.
2. Constant speed with changing direction still means non-zero acceleration.

Answer: (b) acceleration is not zero but velocity is zero

Explanation:
1. At the highest point the body momentarily stops, so v = 0.
2. Gravity still acts, so a = g downward.

Teacher's Note:
1. Zero velocity does not mean zero acceleration.
2. Acceleration due to gravity is the same at every point of the flight.

Answer: (d) the car

Explanation:
1. The person does not change position relative to the car.
2. Tree, cyclist and building all change their position relative to the person.

Teacher's Note:
1. Rest and motion are relative terms.
2. Always name the reference body before calling something at rest.

Answer: (b) 2 m/s

Explanation:
1. Displacement for half a revolution = diameter = 80 m.
2. Average velocity = 80 / 40 = 2 m/s.

Teacher's Note:
1. Average velocity uses displacement, which is the diameter for half a circle.
2. The average speed would be \( \pi \) m/s.

Answer: (d) cm/s\( ^2 \)

Explanation:
1. CGS uses centimetre and second, so acceleration is in cm/s\( ^2 \).

Teacher's Note:
1. SI unit is m/s\( ^2 \).
2. 1 m/s\( ^2 \) = 100 cm/s\( ^2 \).

Answer: (b) displacement = \( \sqrt{2}\,r \), distance = \( \dfrac{3\pi r}{2} \)

Explanation:
1. Distance = \( \tfrac34 \times 2\pi r = \dfrac{3\pi r}{2} \).
2. Start and end points are at right angles to each other about the centre, so displacement = \( \sqrt{r^2+r^2} = \sqrt{2}\,r \).

Teacher's Note:
1. Three quarters of a circle leaves the end point a quarter turn away from the start.
2. Use Pythagoras for the chord.

Answer: (c) 6 m

Explanation:
1. u = 43.2 km/h = 12 m/s.
2. \( s = u^2/(2a) = 144/24 = 6 \) m.

Teacher's Note:
1. Convert the speed to m/s first.
2. Use \( v^2 = u^2 - 2as \) with v = 0.

Answer: (a) 1/4

Explanation:
1. Area of first triangle = 0.5 x 2 x 10 = 10 m.
2. Rectangle = 2 x 10 = 20 m.
3. Last triangle = 10 m. Total = 40 m.
4. Ratio = 10/40 = 1/4.

Teacher's Note:
1. Break the trapezium into two triangles and a rectangle.
2. Distance is the area under a v-t graph.

Answer: (a) At a height of 20 m above the ground after 4 s

Explanation:
1. The gap closes at the relative speed 25 m/s (gravity acts equally on both).
2. Time = 100 / 25 = 4 s.
3. The falling stone has dropped 0.5 x 10 x 16 = 80 m, so the meeting height is 100 - 80 = 20 m.

Teacher's Note:
1. Relative acceleration is zero, so relative speed stays constant.
2. Check with the upward stone: 25(4) - 5(16) = 20 m.

Answer: (d) constant speed

Explanation:
1. Distance = speed x time.
2. Proportionality means the speed is constant.

Teacher's Note:
1. Distance is a scalar, so the constant quantity is speed.
2. Velocity would need constant direction as well.

Answer: (b) -9.8 m/s\( ^2 \)

Explanation:
1. Gravity acts downward throughout the flight, so with upward positive a = -9.8 m/s\( ^2 \).

Teacher's Note:
1. The acceleration stays the same while rising, at the top and while falling.

Answer: (c) 320 m

Explanation:
1. Distance in first 8 s = \( \tfrac12 \times 5 \times 64 = 160 \) m.
2. Velocity reached = 5 x 8 = 40 m/s.
3. Next 4 s at 40 m/s = 160 m. Total = 320 m.

Teacher's Note:
1. Split the journey into the accelerated and the uniform parts.
2. The velocity at the end of the first part is the speed of the second.

Answer: (d) 7/11

Explanation:
1. From rest, distance in nth second is proportional to (2n - 1).
2. 4th: 7, 6th: 11. Ratio = 7/11.

Teacher's Note:
1. From rest the nth-second distances are in the ratio 1 : 3 : 5 : 7 : 9 : 11.

Answer: (b) \( 4u^2/g \)

Explanation:
1. Take downward as positive. \( v^2 = u^2 + 2gH \): \( 9u^2 = u^2 + 2gH \).
2. \( H = 4u^2/g \).

Teacher's Note:
1. The sign of u does not matter in \( v^2 = u^2 + 2gH \) because u is squared.

Answer: (c) 9.8 m

Explanation:
1. Both fall the same distance in the same time, so the separation is unchanged.
2. It remains 9.8 m.

Teacher's Note:
1. Relative acceleration between two freely falling bodies is zero.
2. The gap stays constant until one lands.

Answer: (c) 78.4 m

Explanation:
1. Distance in the last second = \( g(T - \tfrac12) = 34.3 \) gives T = 4 s.
2. Height = 0.5 x 9.8 x 16 = 78.4 m.

Teacher's Note:
1. Use the nth-second formula to find the total time first.

Answer: (b) \( \dfrac{2n-1}{n^2} \)

Explanation:
1. Distance in nth second = \( \tfrac12 a(2n-1) \).
2. Distance in n seconds = \( \tfrac12 a n^2 \).
3. Ratio = \( (2n-1)/n^2 \).

Teacher's Note:
1. Distance in the nth second is the difference of distances in n and n - 1 seconds.

Answer: (c) 2 s

Explanation:
1. Time of flight = 2u/g = 20/10 = 2 s.

Teacher's Note:
1. Time up equals time down.
2. The time up is u/g = 1 s.

Answer: (d) 5.3 m/s

Explanation:
1. Total distance = 32 m.
2. Total time = 6 s.
3. Average speed = 32/6 = 5.3 m/s.

Teacher's Note:
1. Do not average the two speeds (4 and 8).
2. Always divide total distance by total time.

Answer: (c) total displacement is zero

Explanation:
1. The ball ends at the thrower's hand, so net change in position is zero.
2. Distance = 40 + 40 = 80 m.

Teacher's Note:
1. Displacement depends only on start and end points.
2. A round trip always has zero displacement.