CBSE Class 7 Science Advanced Physics Kinematics MCQs Set 01

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

Explanation:
1. On a straight path (without turning back) the distance and the displacement are equal, so the magnitudes of speed and velocity are the same. Statement (a) is true.
2. Average speed is zero only when no distance is covered. Then the displacement is also zero, so average velocity is also zero. Statement (b) is false.
3. Velocity is speed in a specified direction, so direction is necessary. Statement (c) is true.
4. Statements (a) and (c) are true, so the answer is (d) Both (a) and (c).

Teacher's Note:
1. Check every statement separately before choosing the combined option.
2. Remember that velocity = speed + direction, which is why the two are equal in magnitude on a straight path.

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. The unit s⁻¹ (hertz) is the unit of frequency.

Teacher's Note:
1. Divide the units just like numbers and cancel the common ones.
2. A unit of s⁻¹ always points to something happening per second, such as frequency.

Answer: (d) All the above

Explanation:
1. Speed = distance ÷ time, so any unit of distance divided by any unit of time is a unit of speed.
2. m min⁻¹, km h⁻¹ and km s⁻¹ are all of this form, so all of them are units of speed.

Teacher's Note:
1. The SI unit is m s⁻¹, but other combinations are also used in daily life.
2. Choose the unit that suits the situation, for example km h⁻¹ for cars.

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

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

Teacher's Note:
1. Read only the end point of the graph for the average speed: the last distance and the last time.
2. Do not average the speeds of the separate parts.

Answer: (c) 5.3 × 10⁻³

Explanation:
1. The tip makes one full circle of radius 5 cm = 0.05 m in 60 s.
2. Distance = 2πr = 2 × 3.14 × 0.05 = 0.314 m.
3. Speed = 0.314 ÷ 60 = 5.2 × 10⁻³ m s⁻¹, which is closest to 5.3 × 10⁻³ m s⁻¹.

Teacher's Note:
1. Change the length to metres before you start.
2. A second hand takes 60 s for one round, so the time is 60 s.

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

Explanation:
1. The time period T = 2π√(l/g) depends only on the length l (and g).
2. It does not depend on the mass, size or shape of the bob.
3. So it is independent of the mass and the shape of the bob: (d).

Teacher's Note:
1. In the formula T = 2π√(l/g), mass does not appear.
2. Only a change in length changes the time period at a given place.

Answer: (b) 20 minute

Explanation:
1. Time = distance ÷ speed = 20 km ÷ 60 km h⁻¹ = \( \frac{1}{3} \) h.
2. \( \frac{1}{3} \) h = \( \frac{1}{3} \) × 60 min = 20 minutes.

Teacher's Note:
1. Keep the units together: km with km h⁻¹ gives hours.
2. Convert the fraction of an hour to minutes at the end.

Answer: (a) at rest

Explanation:
1. A line parallel to the time axis means the distance stays the same as the time passes.
2. If the distance does not change, the object is not moving, so it is at rest.

Teacher's Note:
1. The slope of a distance-time graph tells the speed; a flat line has zero slope.
2. A body covering no distance in any time is at rest.

Answer: (c) 30

Explanation:
1. 20 min = \( \frac{1}{3} \) h.
2. Distance in the first 20 min = 60 × \( \frac{1}{3} \) = 20 km.
3. Distance in the next 20 min = 30 × \( \frac{1}{3} \) = 10 km.
4. Total distance = 20 + 10 = 30 km.

Teacher's Note:
1. Convert minutes to hours before multiplying with km h⁻¹.
2. Add the distances of the two parts; do not add the speeds.

Answer: (d) Both OA and BC

Explanation:
1. Uniform speed means equal distances in equal times, so the distance-time graph is a straight line.
2. OA and BC are straight lines, while AB is curved.
3. So the uniform speed is represented by both OA and BC.

Teacher's Note:
1. Straight line = uniform speed; curved line = non-uniform speed.
2. Two different straight lines can have different slopes, but each is still uniform on its own.

Answer: (a) 10

Explanation:
1. The speed is uniform, so the average speed equals the speed itself.
2. Check: distance = 10 × 2 = 20 km, time = 2 h, average speed = 20 ÷ 2 = 10 km h⁻¹.

Teacher's Note:
1. Average speed = total distance ÷ total time, not distance alone.
2. For uniform speed the average speed is the same as the given speed.

Answer: (d) Graph is not possible.

Explanation:
1. The distance covered by a moving vehicle can only increase (or stay the same) with time; it can never decrease.
2. In the given graph the distance falls from 60 m to 0 as the time goes on.
3. Hence such a distance-time graph is not possible.

Teacher's Note:
1. Distance is the length of the path travelled, so it never becomes smaller.
2. A falling line can appear on a displacement-time graph, but not on a distance-time graph.

Answer: (b) \( \frac{5}{18} \)

Explanation:
1. 1 km = 1000 m and 1 h = 3600 s.
2. 1 km h⁻¹ = \( \frac{1000}{3600} \) m s⁻¹ = \( \frac{5}{18} \) m s⁻¹.

Teacher's Note:
1. To change km h⁻¹ into m s⁻¹ multiply by 5/18.
2. To change m s⁻¹ into km h⁻¹ multiply by 18/5.

Answer: (d) All the above

Explanation:
1. Both speed and velocity are measured in m s⁻¹ (or km h⁻¹), so (a) is true.
2. 50 m s⁻¹ = 50 × \( \frac{18}{5} \) = 180 km h⁻¹, and on a straight path the magnitude of velocity equals speed, so (b) is true.
3. A speedometer records the speed of a vehicle, so (c) is true.
4. All three are correct: (d) All the above.

Teacher's Note:
1. Test each statement on its own, then decide on the combined option.
2. Practise the 18/5 conversion; it is used in many numerical questions.

Answer: (a) 0.2

Explanation:
1. PQ = 4 m, QR = 3 m and so PR = \( \sqrt{4^2 + 3^2} \) = 5 m.
2. Total distance = 4 + 3 + 5 = 12 m.
3. Total time = 20 + 10 + 30 = 60 s.
4. Speed = 12 ÷ 60 = 0.2 m s⁻¹.

Teacher's Note:
1. Use the Pythagoras theorem for the third side of the right-angled triangle.
2. The person returns to P, so the displacement is zero, but the speed still uses the whole distance.

Answer: (c) BCADE

Explanation:
1. First measure ℓ₁ (B) and the diameter of the bob (C).
2. Then find the radius R from the diameter (A).
3. Next the length of the pendulum is ℓ = ℓ₁ − R (D).
4. Finally put ℓ in the formula T = 2π√(ℓ/g) (E).
5. The order is B, C, A, D, E: (c) BCADE.

Teacher's Note:
1. R must be known before ℓ₁ − R can be found, so A comes before D.
2. The formula step always comes last, after the length is known.

Answer: (a) ABCD

Explanation:
1. First read the displacements of the two parts from the data (A).
2. Add them to get the total displacement (B), and find the total time (C).
3. Use the formula and substitute (D).
4. Order: A, B, C, D, which is (a) ABCD.
5. (Check: average velocity = 40 ÷ 6 = 6.67 m s⁻¹.)

Teacher's Note:
1. Collect all the data first and then apply the formula.
2. In such arranging questions, the formula step is normally the last one.

Answer: (b) 6

Explanation:
1. One oscillation is one complete wave of the graph (a crest and a trough).
2. In the graph, one complete wave takes 6 s (the graph crosses the axis at 3 s, 6 s, 9 s, ... and three and a half waves are completed in 21 s).
3. Time period = 21 ÷ 3.5 = 6 s.

Teacher's Note:
1. The time period is the time for one full oscillation, not half of it.
2. Check with the whole graph: 21 s ÷ 3½ oscillations = 6 s.

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

Explanation:
1. The time period of the bob is 6 s, so it completes 1 oscillation in 6 s.
2. Number of oscillations in 1 s (frequency) = \( \frac{1}{6} \).

Teacher's Note:
1. Frequency is the reciprocal of the time period.
2. Here n = 1/T = 1/6 Hz.

Answer: (a) 15

Explanation:
1. Amplitude is the maximum displacement of the bob from its mean position.
2. From the graph, the maximum displacement is 15 cm on either side.
3. So the amplitude is 15 cm.

Teacher's Note:
1. Read the highest point of the wave measured from the axis, not from the bottom of the wave.
2. Amplitude is half of the peak-to-peak height.

Answer: (d) 3 ½

Explanation:
1. Time period T = 6 s.
2. Number of oscillations in 21 s = \( \frac{21}{6} \) = 3.5.
3. So the bob completes 3 ½ oscillations.

Teacher's Note:
1. Divide the total time by the time period.
2. The answer may be a fraction because the bob can be in the middle of an oscillation.

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

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

Teacher's Note:
1. Oscillatory is a smaller group inside the bigger group of periodic motions.
2. Use the earth going around the sun as the standard example of periodic but not oscillatory motion.

Answer: (c) m s⁻²

Explanation:
1. Squaring the formula gives T² = 4π² ℓ/g, so g = 4π² ℓ/T².
2. So 4π² ℓ/T² is the acceleration due to gravity g.
3. Its unit = m ÷ s² = m s⁻².

Teacher's Note:
1. Rearrange the formula first to find which quantity the expression stands for.
2. The unit of acceleration is always m s⁻² in SI.

Answer: (b) negative

Explanation:
1. Speed = distance ÷ time. Distance and time are never negative.
2. So the speed can be positive or zero (for a body at rest), but it can never be negative.

Teacher's Note:
1. Speed has only magnitude, so it has no direction and no negative sign.
2. A negative sign can appear only for velocity or displacement, where direction matters.

Answer: (a) the average speed of the body is 2 m s⁻¹.

Explanation:
1. From the graph, the body moves 20 m in 10 s along a straight line.
2. Average speed = \( \frac{20}{10} \) = 2 m s⁻¹.
3. The graph is a straight line so the speed is uniform and the body does not change direction; (b) and (c) are false.

Teacher's Note:
1. A straight sloping line on a distance-time graph means uniform speed.
2. Average speed = total distance ÷ total time.

Answer: (c) 5

Explanation:
1. Speed = distance ÷ time = 10 m ÷ 2 s = 5 m s⁻¹.

Teacher's Note:
1. Divide the distance by the time; do not multiply them.
2. Always write the unit with the answer.

Answer: (a) 1

Explanation:
1. Distance in one revolution = 2πr = 2π × 1 = 2π m.
2. Time = distance ÷ speed = 2π ÷ 2π = 1 s.

Teacher's Note:
1. One revolution means the full circumference of the circle.
2. The π cancels, which is a hint that the answer is a simple number.

Answer: (b) 20

Explanation:
1. Distance = 6960 − 6700 = 260 km.
2. Time from 9 a.m. to 10 p.m. = 13 h.
3. Average speed = 260 ÷ 13 = 20 km h⁻¹.

Teacher's Note:
1. An odometer measures distance, so the difference of the readings is the distance travelled.
2. From 9 a.m. to 10 p.m. is 13 hours, not 1.

Answer: (a) 0.5

Explanation:
1. Time for 120 oscillations = 1 minute = 60 s.
2. Time period = \( \frac{60}{120} \) = 0.5 s.

Teacher's Note:
1. Time period = total time ÷ number of oscillations.
2. Change minutes to seconds first.

Answer: (b) 36 km

Explanation:
1. Time = 1 h = 3600 s.
2. Distance = speed × time = 10 × 3600 = 36,000 m.
3. 36,000 m = 36 km.

Teacher's Note:
1. Use consistent units: m s⁻¹ with seconds gives metres.
2. Change metres to km by dividing by 1000.

Answer: (b) 2

Explanation:
1. Going from one extreme to the other is only half an oscillation.
2. So one full oscillation takes 2 × 1 = 2 s.
3. Time period = 2 s.

Teacher's Note:
1. One oscillation = extreme to extreme and back again.
2. So the time period is double the extreme-to-extreme time.

Answer: (b) non-uniform motion

Explanation:
1. When a body is thrown upwards, its speed keeps decreasing, becomes zero at the highest point and then increases as it falls.
2. The speed is not constant, so it is non-uniform motion.
3. The motion does not repeat itself, so it is neither periodic nor oscillatory.

Teacher's Note:
1. Look at whether the speed stays the same.
2. A ball thrown up changes its speed all the time due to gravity.

Answer: (a) 0.08

Explanation:
1. From the graph, the total distance = 20 m and the total time = 4 minutes = 4 × 60 = 240 s.
2. Average speed = \( \frac{20}{240} \) = 0.083 m s⁻¹ ≈ 0.08 m s⁻¹.

Teacher's Note:
1. The time axis is in minutes, so change it to seconds before dividing.
2. Use only the end values of the graph for the average speed.

Answer: (b) 72

Explanation:
1. 1 m s⁻¹ = \( \frac{18}{5} \) km h⁻¹.
2. 20 m s⁻¹ = 20 × \( \frac{18}{5} \) = 4 × 18 = 72 km h⁻¹.

Teacher's Note:
1. Multiply by 18/5 to go from m s⁻¹ to km h⁻¹.
2. Check using 10 m s⁻¹ = 36 km h⁻¹, which is easy to remember.

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. Uniform velocity means no change in velocity, so acceleration is zero and (c) is true.
4. So (d) All the above.

Teacher's Note:
1. Acceleration is the rate of change of velocity.
2. If velocity does not change, the acceleration must be zero.

Answer: (c) DBCAE

Explanation:
1. First note the initial velocity u (D) and the final velocity v (B).
2. Then find the time t taken for the change (C).
3. Write the formula for acceleration (A) and substitute the values (E).
4. The order is D, B, C, A, E: (c) DBCAE.

Teacher's Note:
1. Collect the data (u, v, t) first, then write the formula, then substitute.
2. Substituting always comes at the end.

Answer: (b) DACEB

Explanation:
1. First take the same pendulum to the moon (D) and measure its time period T_M there (A).
2. Then use the formula T = 2π√(ℓ/g) (C).
3. Substituting the values gives T_E/T_M = √(g_M/g_E) and the value of g_M (E).
4. The result is checked with the known fact g_M = g_E/6 (B).
5. The order is D, A, C, E, B: (b) DACEB.

Teacher's Note:
1. The experiment (D, A) comes first, then the calculation (C, E).
2. The known value in (B) is used to verify the result.

Answer: (a) 0.5

Explanation:
1. From the graph, the time period of the pendulum is 2 s (one complete wave takes 2 s).
2. Frequency = \( \frac{1}{T} \) = \( \frac{1}{2} \) = 0.5 Hz.

Teacher's Note:
1. Frequency is the number of oscillations per second.
2. If T = 2 s, then only half an oscillation is done in one second.

Answer: (b) 5

Explanation:
1. The bob completes 1 oscillation in 2 s.
2. In 10 s the number of oscillations = \( \frac{10}{2} \) = 5.

Teacher's Note:
1. Count the complete waves in the graph, or divide the total time by the time period.
2. Both methods must give the same answer.

Answer: (b) 200

Explanation:
1. From the graph, the time period = 2 s.
2. Time for 100 oscillations = 100 × 2 = 200 s.

Teacher's Note:
1. Total time = number of oscillations × time period.
2. This is the same idea that is used in the lab to find T by timing 20 oscillations.

Answer: (b) \( 13.\overline{3} \)

Explanation:
1. 72 km h⁻¹ = 72 × \( \frac{5}{18} \) = 20 m s⁻¹ and 36 km h⁻¹ = 10 m s⁻¹.
2. Let the total distance be x. Time for the first half = \( \frac{x/2}{20} \) = \( \frac{x}{40} \) s and for the second half = \( \frac{x/2}{10} \) = \( \frac{x}{20} \) s.
3. Total time = \( \frac{x}{40} + \frac{x}{20} = \frac{3x}{40} \) s.
4. Average velocity = \( \frac{x}{3x/40} = \frac{40}{3} \) = \( 13.\overline{3} \) m s⁻¹.

Teacher's Note:
1. Equal distances at two speeds do not give the simple average of the speeds.
2. Always find the total time first, then divide the total distance by it.