CBSE Class 7 Science Advanced Physics Kinematics MCQs Set 03

Answer: (d) Both (a) and (c)

Explanation:
1. The propeller spins about its axis, which is rotatory motion.
2. It also moves along the straight path with the helicopter, which is translatory motion.
3. So it has both: (d).

Teacher's Note:
1. A body can have two kinds of motion at the same time.
2. Think of a wheel of a moving car: it spins and moves forward.

Answer: (a) A bus covering 10 km in 10 minutes.

Explanation:
1. (a) 10 km in 10 min = 10 km in 1/6 h = 60 km h⁻¹.
2. (b) 40 km h⁻¹.
3. (c) 2 m s⁻¹ = 2 × 18/5 = 7.2 km h⁻¹.
4. (d) 200 ÷ 6 = 33.3 km h⁻¹.
5. The largest speed is 60 km h⁻¹, so (a).

Teacher's Note:
1. Change every speed to the same unit before comparing.
2. To change m s⁻¹ to km h⁻¹, multiply by 18/5.

Answer: (d) 1

Explanation:
1. Four oscillations take 8 s, so the time period T = 8 ÷ 4 = 2 s.
2. One extreme to the other extreme is half an oscillation = T/2 = 1 s.

Teacher's Note:
1. One to-and-fro motion is one full oscillation.
2. Extreme to extreme is T/2; extreme to mean is T/4.

Answer: (c) 0.95

Explanation:
1. T = 38 ÷ 10 = 3.8 s.
2. Extreme to mean position is a quarter of an oscillation = T/4 = 3.8 ÷ 4 = 0.95 s.

Teacher's Note:
1. Mean to extreme is T/4.
2. 0.475 s would be T/8, which is not a natural part of the swing.

Answer: (b) 50

Explanation:
1. Distance = 10,300 − 10,150 = 150 km.
2. Time = 9 a.m. − 6 a.m. = 3 h.
3. Average speed = 150 ÷ 3 = 50 km h⁻¹.

Teacher's Note:
1. An odometer records distance travelled, so subtract the two readings.
2. Average speed = total distance ÷ total time.

Answer: (a) 10

Explanation:
1. The stone is dropped, so u = 0, and v = 100 m s⁻¹, t = 10 s.
2. a = (v − u) ÷ t = (100 − 0) ÷ 10 = 10 m s⁻².

Teacher's Note:
1. Dropped means the initial velocity is zero.
2. 10 m s⁻² is the acceleration due to gravity.

Answer: (a) − 3

Explanation:
1. u = 90 m s⁻¹, v = 0, t = 30 s.
2. a = (0 − 90) ÷ 30 = −3 m s⁻².
3. The negative sign shows retardation.

Teacher's Note:
1. Braking gives a negative acceleration.
2. Final velocity is zero when the bike halts.

Answer: (a) S_A > S_B > S_C

Explanation:
1. All three cars cover the same distance (the dashed horizontal line).
2. The times taken are 100 s, 200 s and 300 s, so t_A < t_B < t_C.
3. Speed = distance ÷ time, so the car that takes the least time is fastest: S_A > S_B > S_C.

Teacher's Note:
1. A steeper line on a distance-time graph means a higher speed.
2. Same distance, less time: more speed.

Answer: (c) The average velocity of the robot is 5 m s⁻¹.

Explanation:
1. Distance = 5 × 10 + 5 × 5 = 75 m and displacement = 50 − 25 = 25 m, total time = 15 s.
2. Average speed = 75 ÷ 15 = 5 m s⁻¹ (true) and distance > displacement (true).
3. Average velocity = 25 ÷ 15 = 1.67 m s⁻¹, not 5 m s⁻¹, so statement (c) is wrong.
4. The speed is always 5 m s⁻¹, so (d) is true. The wrong statement is (c).

Teacher's Note:
1. Average velocity uses displacement; average speed uses distance.
2. Reversing direction reduces displacement but not distance.

Answer: (b) 150

Explanation:
1. Distance = speed × time = 100 km h⁻¹ × 1.5 h = 150 km.

Teacher's Note:
1. 1½ h means 1.5 h.
2. Write the formula before putting numbers in.

Answer: (a) A → c; B → a; C → b; D → e; E → d

Explanation:
1. A pendulum of a clock repeats its motion: periodic motion (A → c).
2. Fastness or slowness is speed (B → a).
3. Uniform velocity has magnitude and direction both constant (C → b).
4. Uniform speed means equal distances in equal intervals of time (D → e).
5. Foot s⁻¹ is the FPS unit of velocity (E → d).
6. Answer (a).

Teacher's Note:
1. Velocity is a vector, so direction is part of it.
2. Foot per second is the FPS counterpart of m s⁻¹.

Answer: (a) BCAD

Explanation:
1. First take the pendulum and the stopwatch (B).
2. Displace the bob and release it (C).
3. Start the stopwatch and time 20 oscillations (A).
4. Divide the time by 20 to get T (D).
5. The order is BCAD.

Teacher's Note:
1. Set up, release, measure, then calculate.
2. Timing many oscillations reduces the error in T.

Answer: (d) ACBDE

Explanation:
1. Take the apparatus (A).
2. Roll the ball and start the watch together (C).
3. Note the time when the ball stops (B).
4. Measure the distance rolled (D).
5. Calculate speed = distance ÷ time (E).
6. The order is ACBDE.

Teacher's Note:
1. Do the activity first, then the calculation.
2. The stopwatch starts the moment the ball is rolled.

Answer: (d) Both (b) and (c)

Explanation:
1. The wheels turn about their axles: rotatory motion.
2. They also move forward along the track with the train: translatory motion.
3. So the answer is (d) Both (b) and (c).

Teacher's Note:
1. A rolling wheel does two motions together.
2. Oscillatory motion is a to-and-fro motion, which a wheel does not do.

Answer: (c) A tuning fork vibrating with a frequency of 100 Hz.

Explanation:
1. Frequency is the number of vibrations per second.
2. The frequencies are 10 Hz, 25 Hz, 100 Hz and 50 Hz.
3. The largest is 100 Hz, so the tuning fork moves (vibrates) fastest: (c).

Teacher's Note:
1. More vibrations per second means a faster to-and-fro motion.
2. Compare the frequencies in the same unit.

Answer: (c) 4

Explanation:
1. Extreme to mean position is T/4 = 1 s.
2. So T = 4 s.

Teacher's Note:
1. Extreme to mean is a quarter of a full oscillation.
2. Four such quarters make one time period.

Answer: (b) 0.4

Explanation:
1. T = 32 ÷ 20 = 1.6 s.
2. Mean to extreme = T/4 = 1.6 ÷ 4 = 0.4 s.

Teacher's Note:
1. Find T first, then take a quarter.
2. 0.8 s would be half of a time period.

Answer: (a) 50

Explanation:
1. From 7 a.m. to 7 a.m. next day = 24 h, and 7 a.m. to 1 p.m. = 6 h more, so the time is 30 h.
2. Average speed = 1500 ÷ 30 = 50 km h⁻¹.

Teacher's Note:
1. Count the full hours carefully across the two days.
2. Average speed = total distance ÷ total time.

Answer: (b) − 10 m s⁻²

Explanation:
1. u = 50 m s⁻¹, v = 0, t = 5 s.
2. a = (0 − 50) ÷ 5 = −10 m s⁻².
3. The negative sign means the velocity decreases (gravity acts downwards).

Teacher's Note:
1. Going up, the velocity decreases, so a is negative.
2. This equals the acceleration due to gravity.

Answer: (d) 900

Explanation:
1. u = 0, v = 150 km h⁻¹, t = 1/6 h.
2. a = (150 − 0) ÷ (1/6) = 150 × 6 = 900 km h⁻².

Teacher's Note:
1. Dividing by 1/6 is the same as multiplying by 6.
2. Keep the unit km h⁻² as asked.

Answer: (d) The speed of P is greater than that of Q and R.

Explanation:
1. P is a horizontal line: its position does not change, so it is at rest and its velocity is zero.
2. Q is steeper than R, so Q has the greater velocity, and at time t the displacement of R is less than that of Q.
3. P is at rest, so its speed is zero, not greater than that of Q and R. The wrong option is (d).

Teacher's Note:
1. A horizontal line on a position-time graph means rest.
2. A steeper line means a greater speed.

Answer: (d) The average velocity of the fighter plane is zero.

Explanation:
1. The speed stays 500 km h⁻¹ all the time, so (a) and (c) are true.
2. The direction changes, so the velocity is not uniform: (b) is true.
3. The plane ends up away from the starting point, so the displacement is not zero and the average velocity is not zero. Statement (d) is wrong: (d).

Teacher's Note:
1. Velocity changes when the direction changes, even at constant speed.
2. Average velocity is zero only when the body returns to its starting point.

Answer: (c) 40

Explanation:
1. Distance = speed × time = 20 × 2 = 40 km.
2. His displacement is zero because he is back at the start, but the distance covered is 40 km.

Teacher's Note:
1. Distance counts the whole path; displacement does not.
2. Do not choose 0, which is the displacement.

Answer: (a) A → b; B → a; C → d; D → c

Explanation:
1. A guitar string vibrates: vibratory motion (A → b).
2. A speedometer reads speed in km h⁻¹ (B → a).
3. An odometer shows the distance moved (C → d).
4. Rate of change of velocity is acceleration (D → c).
5. Answer (a).

Teacher's Note:
1. Speedometer = speed; odometer = distance.
2. Acceleration is the rate of change of velocity.

Answer: (b) CABD

Explanation:
1. First find the length from the data (C).
2. Let the time period be T (A).
3. Use T = 2π√(ℓ/g) (B).
4. Finally frequency n = 1/T (D).
5. The order is CABD.

Teacher's Note:
1. Frequency is the last step because it needs T.
2. Write the given data first.

Answer: (c) ABC

Explanation:
1. A dropped stone starts from rest, so u = 0 (A).
2. Write a = (v − u)/t (B).
3. Rearrange to v = u + at (C).
4. The order is ABC.

Teacher's Note:
1. Dropped means u = 0.
2. Write the formula before rearranging it.

Answer: (a) To describe speed, direction is necessary.

Explanation:
1. Speed has only magnitude, so no direction is needed: statement (a) is wrong.
2. Along a straight line in one direction, distance and displacement have the same magnitude: (b) is true.
3. Time has no direction, so it is a scalar: (c) is true.
4. The wrong statement is (a).

Teacher's Note:
1. Direction is needed for velocity, not for speed.
2. Scalars: distance, speed, time. Vectors: displacement, velocity.

Answer: (a) BCA

Explanation:
1. Time period for each trial = time ÷ 20 (B): 0.75, 0.8, 0.75, 0.85 s.
2. Average of these values = 0.7875 s (C).
3. Frequency = 1 ÷ average time period (A).
4. The order is BCA.

Teacher's Note:
1. Reciprocal of T gives frequency, so it comes last.
2. Averaging over several trials reduces the error.

Answer: (d) Both (a) and (b)

Explanation:
1. T = 5 ÷ 20 = 1/4 s, so (a) is correct.
2. 1/4 s = (1/4) ÷ 60 min = 1/240 min, so (b) is correct.
3. (c) is 5/20 min = 15 s, which is wrong.
4. The answer is (d).

Teacher's Note:
1. The same time can be written in different units.
2. To change seconds to minutes, divide by 60.

Answer: (b) Frequency

Explanation:
1. \( \frac{1}{T} = \frac{1}{2\pi}\sqrt{\frac{g}{\ell}} \).
2. The reciprocal of the time period is the frequency.

Teacher's Note:
1. Frequency = 1/T.
2. Look at the formula and invert it.

Answer: (d) s²

Explanation:
1. Unit of displacement = m and unit of acceleration = m s⁻².
2. Ratio = m ÷ (m s⁻²) = s².

Teacher's Note:
1. Divide the units just as you divide numbers.
2. The metres cancel.

Answer: (d) \( \frac{3(4+\sqrt{10})}{50} \)

Explanation:
1. AB = 3 m, BC = 9 m and the angle at B is 90°, so AC = √(9 + 81) = 3√10 m.
2. Total distance = 3 + 9 + 3√10 = 3(4 + √10) m.
3. Total time = 5 + 20 + 25 = 50 s.
4. Average speed = \( \frac{3(4+\sqrt{10})}{50} \) m s⁻¹: (d).

Teacher's Note:
1. Use the whole path ABCA for distance.
2. Add the times written in brackets or on each side.

Answer: (d) Both (a) and (c)

Explanation:
1. Uniform velocity means equal displacement in equal times, so the s – t graph is a straight line.
2. (a) and (c) are straight lines (one rising, one falling), while (b) is a curve.
3. The answer is (d).

Teacher's Note:
1. A straight line on an s – t graph means constant velocity.
2. A falling straight line is uniform velocity in the opposite direction.

Answer: (c) 2π

Explanation:
1. Distance in one rotation = 2πr = 2π × 0.5 = π m.
2. Distance in 50 rotations = 50π m, in 25 s.
3. Speed = 50π ÷ 25 = 2π m s⁻¹.

Teacher's Note:
1. Speed uses the total path length (distance).
2. Circumference = 2πr.

Answer: (c) A → b; B → c; C → a; D → d

Explanation:
1. Rest and motion are relative (A → b).
2. A simple pendulum shows oscillatory motion (B → c).
3. A fan shows rotatory motion (C → a).
4. A car on the road shows translatory motion (D → d).
5. Answer (c).

Teacher's Note:
1. Whether a body is at rest depends on the observer.
2. Link each motion with one everyday example.

Answer: (c) A is true, but R is false.

Explanation:
1. For a bus moving along a straight line without turning back, distance = displacement, so average speed equals the magnitude of average velocity. A is true.
2. Average speed is total distance ÷ total time, not displacement ÷ time. R is false.
3. A is true but R is false: (c).

Teacher's Note:
1. Displacement ÷ time gives velocity, not speed.
2. Check the formula in the reason first.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. T = 2π√(ℓ/g), so T is smaller when g is larger.
2. g on the earth is six times g on the moon, so the same pendulum has a smaller T on the earth. Both statements are true.
3. R is the reason for A: (a).

Teacher's Note:
1. T is inversely proportional to √g.
2. On the moon the same pendulum swings more slowly.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A plucked wire moves up and down and repeats its motion at regular intervals. A is true.
2. R is the definition of periodic motion and explains why A is called periodic.
3. Answer (a).

Teacher's Note:
1. Periodic motion repeats after equal intervals of time.
2. Examples: pendulum, plucked string, hands of a clock.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. T = 2π√(ℓ/g), so T is directly proportional to √ℓ (R is true).
2. So when ℓ increases, T increases (A is true) and R is the reason.
3. Answer (a).

Teacher's Note:
1. T does not depend on the mass of the bob.
2. Four times the length gives twice the time period.

Answer: (c) A is true, but R is false.

Explanation:
1. Velocity has both magnitude and direction, so it is a vector. A is true.
2. R says velocity has no direction, which is false.
3. A is true but R is false: (c).

Teacher's Note:
1. Speed is a scalar; velocity is a vector.
2. Velocity = speed together with direction.