CBSE Class 7 Science Advanced Physics Kinematics MCQs Set 02

Answer: (b) 10π

Explanation:
1. Radius of the rod R = 5 cm = 0.05 m.
2. One turn has a length equal to the circumference = 2πR = 2π × 0.05 = 0.1π m.
3. Length wound in 2 s = 200 × 0.1π = 20π m.
4. Length wound in 1 s = \( \frac{20\pi}{2} \) = 10π m.

Teacher's Note:
1. Each turn of the wire is one circle around the rod.
2. Remember to change the diameter into a radius and cm into m.

Answer: (b) 1

Explanation:
1. From the graph, the time period of oscillation T = 2 s.
2. T = 2π√(ℓ/g), so T² = 4π² ℓ/g and ℓ = \( \frac{gT^2}{4\pi^2} \).
3. ℓ = \( \frac{10 \times 4}{4 \times 10} \) = 1 m.

Teacher's Note:
1. First read T from the graph, then use ℓ = gT²/(4π²).
2. A seconds pendulum (T = 2 s) is about 1 m long, which is a quick check.

Answer: (d) All the above

Explanation:
1. Distance in 6 rounds = 6 × 2πr = 6 × 2 × \( \frac{22}{7} \) × 14 = 528 m.
2. Time = \( \frac{528}{11} \) = 48 s.
3. 48 s = \( \frac{48}{60} \) = 0.8 min = \( \frac{48}{3600} \) = 0.0133 h.
4. All three values are the same time, so (d) All the above.

Teacher's Note:
1. One round is the circumference of the track.
2. The same time can be written in seconds, minutes or hours.

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

Explanation:
1. Radius = 7 m, so the distance along the semicircle = πr = 7π = 22 m, not 14 m. (a) is false.
2. Displacement PQ = diameter = 14 m. Average velocity = \( \frac{14}{2\pi} \) = 2.23 m s⁻¹. (b) is true.
3. Average speed = \( \frac{7\pi}{2\pi} \) = 3.5 m s⁻¹. (c) is true.
4. So (d) Both (b) and (c).

Teacher's Note:
1. Average speed uses the curved distance; average velocity uses the straight-line displacement.
2. 14 m is the displacement, not the distance.

Answer: (a) 25

Explanation:
1. The ball falls straight down, so the displacement = 125 m.
2. Average velocity = \( \frac{\text{displacement}}{\text{time}} \) = \( \frac{125}{5} \) = 25 m s⁻¹.

Teacher's Note:
1. Average velocity is total displacement divided by total time.
2. This is not the speed with which the ball hits the ground.

Answer: (a) 100π

Explanation:
1. Distance in one rotation = 2πr = 2π × 0.5 = π m.
2. Distance in 100 rotations (1 s) = 100π m.
3. Speed = \( \frac{100\pi}{1} \) = 100π m s⁻¹.

Teacher's Note:
1. The tip of the wing moves in a circle of radius 0.5 m.
2. Multiply the circumference by the number of rotations per second.

Answer: (b) 6

Explanation:
1. Let the total distance be 2d. Time for the first half = \( \frac{d}{2} \) = t₁.
2. Let the speed in the second half be x, then t₂ = \( \frac{d}{x} \).
3. Average speed = \( \frac{2d}{t_1 + t_2} \) = 3, so 2d = 3(\( \frac{d}{2} + \frac{d}{x} \)).
4. 2 = \( \frac{3}{2} + \frac{3}{x} \), so \( \frac{3}{x} = \frac{1}{2} \) and x = 6 m s⁻¹.

Teacher's Note:
1. Average speed is total distance over total time, not the mean of the two speeds.
2. Write the two times in terms of d and solve the equation for x.

Answer: (a) 0.5, \( \frac{4.5}{17} \)

Explanation:
1. Each step has a rise of ¼ m and a run of ¼ m. The path has 9 horizontal parts and 8 vertical parts (4 up and 4 down), each ¼ m long.
2. Total distance = (9 + 8) × ¼ = \( \frac{17}{4} \) m. Speed = \( \frac{17/4}{8.5} \) = 0.5 m s⁻¹.
3. Displacement PQ = 2.25 m. Velocity = \( \frac{2.25}{8.5} = \frac{4.5}{17} \) m s⁻¹.

Teacher's Note:
1. Count the horizontal and vertical pieces of the path carefully.
2. Speed uses the whole length of the path; velocity uses only the straight line from P to Q.

Answer: (b) \( \frac{5}{7} \)

Explanation:
1. The path PQRSTUVP is made up of the full circles of radius 2 m and 3 m.
2. Total distance = 2π × 2 + 2π × 3 = 10π m = 10 × \( \frac{22}{7} \) m.
3. Average speed = \( \frac{10 \times 22}{7 \times 44} \) = \( \frac{5}{7} \) m s⁻¹.

Teacher's Note:
1. The athlete ends where he started, so the displacement is zero, but the question asks for speed.
2. Add up the circumferences of both circles for the distance.

Answer: (c) 26.66

Explanation:
1. From the figure, the train of length 200 m covers half the circular path, so the length of the whole path (circumference) = 2 × 200 = 400 m.
2. Speed = 54 km h⁻¹ = 54 × \( \frac{5}{18} \) = 15 m s⁻¹.
3. Time = \( \frac{400}{15} \) = 26.67 s, which is 26.66 s in the given option.

Teacher's Note:
1. Change the speed to m s⁻¹ before dividing.
2. The figure shows the 200 m train covering only half of the circle.

Answer: (a) 20

Explanation:
1. Average velocity = \( \frac{\text{displacement}}{\text{time}} \), so displacement = average velocity × time.
2. Height of the tower = 10 × 2 = 20 m.

Teacher's Note:
1. The displacement of the falling body is the height of the tower.
2. Rearrange the average velocity formula to find the displacement.

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

Explanation:
1. On a straight path, distance and displacement are equal, so the magnitudes of speed and velocity are equal. (a) is true.
2. Average velocity may be zero when the displacement is zero, but average speed is zero only when the body is at rest. So (b) is false.
3. Velocity is speed in a specified direction, so (c) is true.
4. Hence (d) Both (a) and (c).

Teacher's Note:
1. Average velocity can be zero while average speed is not (for example, a round trip), but not the other way around.
2. Always include direction when you talk about velocity.

Answer: (d) All the above

Explanation:
1. Speed and velocity are both measured in m s⁻¹, so (a) is true.
2. 50 m s⁻¹ = 50 × \( \frac{18}{5} \) = 180 km h⁻¹, so (b) is true.
3. Speedometer measures speed, so (c) is true.
4. (d) All the above.

Teacher's Note:
1. Convert 50 m s⁻¹ by multiplying with 18/5.
2. On a straight line the magnitude of velocity equals the speed.

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

Explanation:
1. The length of a pendulum is measured from the point of suspension to the middle (centre) of the bob, not to its bottom.
2. So the thread is shorter than the pendulum length by the radius R of the bob: thread = (100 − R) cm. Assertion A is true.
3. Reason R is false because the length is measured up to the centre of the bob, not the bottom most point.
4. So A is true but R is false.

Teacher's Note:
1. Length of pendulum = length of thread + radius of the bob.
2. Pay attention to the exact words in the reason: bottom most point versus mid-point.

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

Explanation:
1. Velocity is speed together with its direction, so R is true.
2. A car with speed 20 m s⁻¹ moving east has a velocity of 20 m s⁻¹ towards the east, so A is true.
3. R correctly explains A.
4. So both A and R are true and R is the correct explanation of A.

Teacher's Note:
1. Speed plus direction gives velocity.
2. R is the definition that directly gives A.

Answer: (a) BCA

Explanation:
1. First find the time period in each trial = time ÷ number of oscillations (B).
2. Then take the average of these time periods (C): (1 + 1.05 + 1 + 0.95 + 1) ÷ 5 = 1 s.
3. Finally the frequency is the reciprocal of the average time period (A): 1 ÷ 1 = 1 Hz.
4. The order is B, C, A: (a) BCA.

Teacher's Note:
1. You cannot take a reciprocal of the average until you have the average.
2. Frequency = 1/T is always the last step.

Answer: (a) ABCD

Explanation:
1. Find the distance (A) and the time (B).
2. Write the formula (C).
3. Substitute the values (D).
4. Order: A, B, C, D, so (a) ABCD.
5. (Distance = 4 × 10 = 40 m in 3600 s.)

Teacher's Note:
1. Data first, formula next, substitution last.
2. The substitution step cannot come before the data steps.

Answer: (d) 3 ½

Explanation:
1. From the graph, the bob completes 3½ oscillations in 21 s, that is, one oscillation takes 6 s.
2. Number of oscillations in 21 s = \( \frac{21}{6} \) = 3 ½.

Teacher's Note:
1. Find the time period from the graph first.
2. Divide the total time by T.

Answer: (a) 15

Explanation:
1. The maximum displacement of the bob from its mean position is called the amplitude.
2. From the graph, this maximum displacement is 15 cm, so the amplitude is 15 cm.

Teacher's Note:
1. Amplitude is measured from the mean line to a crest or a trough.
2. Do not use the full height from crest to trough.

Answer: (a) \( \frac{1}{6} \)

Explanation:
1. The bob completes one oscillation every 6 s.
2. Number of oscillations in 1 s = \( \frac{1}{6} \), which is the frequency of the pendulum.

Teacher's Note:
1. Frequency is the number of oscillations per second.
2. n = 1/T.

Answer: (a) 2π

Explanation:
1. Distance covered in 100 rotations = 100 × 2πr = 100 × 2π × 0.5 = 100π m.
2. Time = 50 s.
3. Speed = \( \frac{100\pi}{50} \) = 2π m s⁻¹.

Teacher's Note:
1. Each rotation is one circle of radius 0.5 m.
2. Divide the total distance by the total time (50 s).

Answer: (a) 0.2

Explanation:
1. By the Pythagoras theorem, PR = \( \sqrt{PQ^2 + QR^2} = \sqrt{16 + 9} \) = 5 m.
2. Total distance = 4 + 3 + 5 = 12 m and total time = 20 + 10 + 30 = 60 s.
3. Speed = \( \frac{12}{60} \) = 0.2 m s⁻¹.

Teacher's Note:
1. The third side of a 3-4-5 triangle is 5 m.
2. Speed uses the total path length, even though the person ends where he started.

Answer: (b) frequency

Explanation:
1. Unit of acceleration = m s⁻² and unit of velocity = m s⁻¹.
2. Ratio = m s⁻² ÷ m s⁻¹ = s⁻¹.
3. s⁻¹ is the unit of frequency.

Teacher's Note:
1. Cancel the metres and subtract the powers of seconds.
2. s⁻¹ means 'per second'.

Answer: (a) \( \frac{3}{4} \)

Explanation:
1. From the graph, the total distance moved by the body = OF = 12 m.
2. Total time = 16 s.
3. Average speed = \( \frac{12}{16} \) = \( \frac{3}{4} \) m s⁻¹.

Teacher's Note:
1. Use the final distance and final time only.
2. Simplify 12/16 to 3/4.

Answer: (b) 6

Explanation:
1. Let the total distance be 2d. Then t₁ = \( \frac{d}{2} \) and t₂ = \( \frac{d}{x} \).
2. \( \frac{2d}{t_1 + t_2} \) = 3 gives 2d = 3 × (\( \frac{d}{2} + \frac{d}{x} \)).
3. Dividing by d: 2 = \( \frac{3}{2} + \frac{3}{x} \), so \( \frac{3}{x} = \frac{1}{2} \) and x = 6 m s⁻¹.

Teacher's Note:
1. Use average speed = total distance ÷ total time.
2. Keep the distance as d so that it cancels.

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

Explanation:
1. A → b: rest and motion are relative.
2. B → c: the bob of a simple pendulum has oscillatory as well as periodic motion.
3. C → d: the wheels of a moving bicycle have both translatory and rotatory motion.
4. D → a: a car taking a turn has variable velocity because its direction changes.
5. So (a) is correct.

Teacher's Note:
1. Match the entries you are sure of first, then use elimination.
2. Option (b) uses D twice, so it cannot be right.

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

Explanation:
1. Every to-and-fro motion repeats at equal intervals, so every oscillatory motion is periodic. (a) is true.
2. The rotation of the earth around the sun is periodic but not oscillatory, so (b) is false.
3. The pendulum bob's motion repeats at regular intervals, so (c) is true.
4. So (d) Both (a) and (c).

Teacher's Note:
1. Periodic is the bigger group; oscillatory is a part of it.
2. Remember one example of periodic motion that is not oscillatory.

Answer: (d) All the above

Explanation:
1. The SI unit of acceleration is m s⁻², so (a) is true.
2. Speed = distance ÷ time, so (b) is true.
3. Displacement is the shortest distance between the initial and final positions, so its magnitude can never be more than the distance travelled. (c) is true.
4. (d) All the above.

Teacher's Note:
1. Displacement is a straight line, the shortest route between two points.
2. Distance equals the magnitude of displacement only for straight-line motion without turning back.

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

Explanation:
1. Quartz clocks contain an electric circuit and are more accurate than pendulum clocks. So A is true.
2. Pendulum clocks, not quartz clocks, use the periodic motion of a simple pendulum. So R is false.
3. A is true but R is false.

Teacher's Note:
1. Old wall clocks use pendulums; modern quartz clocks use an electric circuit.
2. A true statement can be paired with a wrong reason, so check R separately.

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

Explanation:
1. Acceleration = change in velocity ÷ time. This is the definition, so R is true.
2. If the velocity is constant, the change in velocity is zero, so the acceleration is zero. A is true.
3. R explains A, so both A and R are true and R is the correct explanation of A.

Teacher's Note:
1. Constant velocity means no change in velocity.
2. No change in velocity means no acceleration.

Answer: (c) BCAD

Explanation:
1. First note the given values of T and g (B).
2. Write the formula T = 2π√(ℓ/g) (C).
3. Rearrange it to get ℓ = T²g/(4π²) (A).
4. Substitute the data to get ℓ (D).
5. The order is B, C, A, D: (c) BCAD.
6. (ℓ = 16 × 9.8 ÷ (4 × 9.87) ≈ 3.97 m.)

Teacher's Note:
1. Rearranging the formula is done before substituting the numbers.
2. Collect the data, then the formula, then the rearranged formula, then the numbers.

Answer: (d) CBAD

Explanation:
1. Start by noting the initial velocity u = 0 and the acceleration (C).
2. Note the final velocity and change it into m s⁻¹: 54 km h⁻¹ = 15 m s⁻¹ (B).
3. Write the relation v = u + at (A).
4. Substitute and find t = (15 − 0) ÷ 3 = 5 s (D).
5. The order is C, B, A, D: (d) CBAD.

Teacher's Note:
1. Convert to SI units before you substitute.
2. The relation and substitution come after all the data has been noted.

Answer: (b) 5

Explanation:
1. From the graph, the bob completes 1 oscillation in 2 s.
2. Number of oscillations in 10 s = \( \frac{10}{2} \) = 5.

Teacher's Note:
1. Divide the total time by the time period.
2. You can also count the waves in the graph.

Answer: (b) 1

Explanation:
1. From the graph, T = 2 s.
2. ℓ = \( \frac{gT^2}{4\pi^2} = \frac{10 \times 4}{4 \times 10} \) = 1 m.

Teacher's Note:
1. Read T from the graph and use ℓ = gT²/(4π²).
2. A pendulum with T = 2 s is a seconds pendulum, about 1 m long.

Answer: (a) 0.5

Explanation:
1. From the graph, the time period T = 2 s.
2. Frequency = \( \frac{1}{T} = \frac{1}{2} \) = 0.5 Hz.

Teacher's Note:
1. Frequency is the reciprocal of the time period.
2. T = 2 s means half an oscillation per second.

Answer: (b) 10π

Explanation:
1. Radius = 5 cm = 0.05 m. One turn = 2πr = 0.1π m.
2. In 2 s the wire wound = 200 × 0.1π = 20π m.
3. In 1 s = 20π ÷ 2 = 10π m.

Teacher's Note:
1. Each turn equals the circumference of the rod.
2. Divide by 2 for the time in seconds.

Answer: (a) 0.5, \( \frac{4.5}{17} \)

Explanation:
1. Each step is ¼ m wide and ¼ m high. The path has 9 horizontal and 8 vertical pieces, so the total distance = 17 × ¼ = \( \frac{17}{4} \) m.
2. Average speed = \( \frac{17/4}{8.5} \) = 0.5 m s⁻¹.
3. Displacement = 2.25 m, so the average velocity = \( \frac{2.25}{8.5} = \frac{4.5}{17} \) m s⁻¹.

Teacher's Note:
1. Speed uses the length of the whole zig-zag path.
2. Velocity uses only the straight horizontal distance from P to Q.

Answer: (c) m s⁻²

Explanation:
1. From the formula, g = \( 4\pi^2\frac{\ell}{T^2} \).
2. So the expression is the acceleration due to gravity, whose unit is m s⁻².

Teacher's Note:
1. Rearrange the formula for g.
2. The unit of acceleration is m s⁻².

Answer: (a) 2

Explanation:
1. From the graph, the distance moved by the body = 20 m in a time of 10 s.
2. Average speed = \( \frac{20}{10} \) = 2 m s⁻¹.

Teacher's Note:
1. Take the end values of distance and time from the graph.
2. The line is straight, so the speed is uniform.

Answer: (b) 13.3

Explanation:
1. 72 km h⁻¹ = 20 m s⁻¹ and 36 km h⁻¹ = 10 m s⁻¹.
2. If the distance is x, then t₁ = \( \frac{x/2}{20} = \frac{x}{40} \) s and t₂ = \( \frac{x/2}{10} = \frac{x}{20} \) s.
3. Total time = \( \frac{3x}{40} \) s, so the average velocity = \( \frac{x}{3x/40} = \frac{40}{3} \) = 13.3 m s⁻¹.

Teacher's Note:
1. Do not just average 20 and 10; the time spent at each speed is different.
2. Find the total time and divide the total distance by it.

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

Explanation:
1. A → d: covering 5 m in every second is constant (uniform) speed.
2. B → b: equal distances in unequal intervals of time is non-uniform speed.
3. C → a: a lunar month is based on the periodic motion of the moon around the earth.
4. D → c: the age of a person is expressed in years.
5. So (b) is correct.

Teacher's Note:
1. Match the easy pairs first (D → c is the quickest).
2. Use the options to eliminate wrong pairs one by one.