CBSE Class 7 Science Advanced Physics Kinematics MCQs Set 04

Answer: (b) \( \frac{20}{3} \)

Explanation:
1. Let each half be x. Times: x/5 and x/10.
2. Total time = x/5 + x/10 = 3x/10 and total distance = 2x.
3. Average speed = 2x ÷ (3x/10) = 20/3 m s⁻¹: (b).
4. 7.5 is the plain average of 5 and 10, which is wrong because the times are different.

Teacher's Note:
1. For equal distances, use total distance ÷ total time, not the plain average.
2. The shortcut is 2ab/(a + b).

Answer: (a) 1

Explanation:
1. From the graph, the displacement crosses zero at 0, 1, 2, … s, and two zero crossings (one full oscillation) take 2 s, so T = 2 s.
2. T = 2π√(ℓ/g) gives ℓ = gT² ÷ 4π² = 10 × 4 ÷ 40 = 1 m.

Teacher's Note:
1. Read T from the graph first.
2. A pendulum with T = 2 s is called a seconds pendulum and is about 1 m long.

Answer: (d) 0.5

Explanation:
1. From the graph, T = 2 s.
2. Frequency = 1/T = 1/2 = 0.5 Hz.

Teacher's Note:
1. Frequency is the reciprocal of the time period.
2. Hz means oscillations per second.

Answer: (b) 5

Explanation:
1. The graph completes one full oscillation (up, down and back to zero) at 5 s.
2. So the time period is 5 s.
3. Half an oscillation is 2.5 s.

Teacher's Note:
1. One complete wave on the graph is one oscillation.
2. Do not read the half-way point (2.5 s) as T.

Answer: (b) 4

Explanation:
1. T = 5 s, so frequency = 1/5 = 0.2 oscillations per second.
2. Oscillations in 20 s = 20 × 0.2 = 4.

Teacher's Note:
1. Number of oscillations = total time ÷ T.
2. 20 ÷ 5 = 4.

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

Explanation:
1. Both distance and displacement are lengths, measured in metres: (a) is true.
2. 5 m s⁻¹ = 5 × 18/5 = 18 km h⁻¹: (b) is true.
3. The answer is (c).

Teacher's Note:
1. 1 m s⁻¹ = 3.6 km h⁻¹.
2. Check each statement separately.

Answer: (a) CBA

Explanation:
1. Note the given values (C): 20 oscillations in 2 s.
2. Find the frequency = 20 ÷ 2 = 10 Hz (B).
3. Time period = 1 ÷ frequency = 0.1 s (A).
4. The order is CBA.

Teacher's Note:
1. Here the frequency is found first and T is its reciprocal.
2. The first step is always to note down the given data.

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

Explanation:
1. T = 1/20 = 0.05 s.
2. 100 oscillations take 100 × 0.05 = 5 s.
3. 5 s = 5/60 min = 1/12 min, so (b) and (c) are both right: (d).

Teacher's Note:
1. 0.05 s is only the time for one oscillation.
2. Convert seconds to minutes by dividing by 60.

Answer: (a) Hz

Explanation:
1. \( T = 2\pi\sqrt{\ell/g} \), so \( \frac{1}{2\pi}\sqrt{g/\ell} = \frac{1}{T} \), the frequency.
2. The SI unit of frequency is the hertz (Hz): (a).

Teacher's Note:
1. The expression is the reciprocal of T.
2. Minute is not an SI unit.

Answer: (d) Frequency

Explanation:
1. m s⁻¹ ÷ m = s⁻¹.
2. s⁻¹ is the unit of frequency (Hz): (d).

Teacher's Note:
1. Cancel the metres.
2. Frequency is measured in s⁻¹ = Hz.

Answer: (b) 0

Explanation:
1. The person comes back to P, the starting point.
2. Displacement = 0, so average velocity = 0 ÷ 60 s = 0.
3. The distance is 60 m, but that is not used for velocity.

Teacher's Note:
1. Average velocity = total displacement ÷ total time.
2. A closed path always has zero displacement.

Answer: (c)

Explanation:
1. Rest after some displacement means the position first changes and then stays constant.
2. The graph must first rise and then become horizontal: graph (c).
3. (a) has no motion at all, (b) is uniform motion and (d) is uniform motion towards the origin.

Teacher's Note:
1. A horizontal part of an s – t graph means the body is at rest.
2. The slanted part shows motion.

Answer: (c) 0.47

Explanation:
1. After 1½ rotations the particle is at the opposite end of the diameter.
2. Displacement = 2 × 0.7 = 1.4 m.
3. Average velocity = 1.4 ÷ 3 = 0.47 m s⁻¹.

Teacher's Note:
1. Velocity uses displacement, not the length of the circular path.
2. Half a rotation takes the particle to the opposite point.

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

Explanation:
1. Velocity: m s⁻¹ (A → b).
2. Acceleration: m s⁻² (B → a).
3. Displacement: m (C → c).
4. Speed: m s⁻¹ (D → b).
5. Answer (c).

Teacher's Note:
1. Speed and velocity have the same unit.
2. Work out each unit from its definition.

Answer: (b) 3.5

Explanation:
1. Let each half of the time be t. Distances: 3t and 4t.
2. Total distance = 7t in total time 2t.
3. Average velocity = 7t ÷ 2t = 3.5 m s⁻¹.

Teacher's Note:
1. For equal times, the average is the simple mean.
2. For equal distances it is not.

Answer: (b) 2.5

Explanation:
1. The graph shows one oscillation in every 2 s (T = 2 s).
2. Oscillations in 5 s = 5 ÷ 2 = 2.5.

Teacher's Note:
1. Number of oscillations = time ÷ T.
2. Half oscillations can be counted too.

Answer: (c) 0.05

Explanation:
1. Amplitude is the maximum displacement from the mean position.
2. From the graph it is 5 cm = 0.05 m: (c).

Teacher's Note:
1. Change cm to m by dividing by 100.
2. Amplitude is the largest displacement, not the total range.

Answer: (b) 5

Explanation:
1. One full oscillation is complete at 5 s on the graph.
2. So the time period is 5 s.

Teacher's Note:
1. Look for where the curve completes one full cycle.
2. Do not use 2.5 s, which is only half a cycle.

Answer: (d) 6.25

Explanation:
1. From the graph, T = 5 s.
2. ℓ = gT² ÷ 4π² = 10 × 25 ÷ (4 × 10) = 6.25 m.

Teacher's Note:
1. Square the time period before dividing.
2. Substitute π² = 10 directly.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Distance has only magnitude, so it is a scalar. A is true.
2. Distance (length) is a fundamental quantity. R is true.
3. Being fundamental does not make a quantity a scalar, so R does not explain A: (b).

Teacher's Note:
1. Check whether R really explains A.
2. A scalar has only magnitude; that is the reason for A.