CBSE Class 7 Science Advanced Physics Kinematics Short Answer Questions Set 02

Short Answer Questions

Question 1. A simple pendulum makes 20 oscillations in one second on the surface of the earth. Determine the time period of the simple pendulum on the surface of a planet where the acceleration due to gravity is one fourth of the acceleration due to gravity on the surface of the earth.
Answer:
On the earth, T₁ = \( \frac{1}{20} \) s = 2π√(ℓ/g₁).
On the planet, g₂ = \( \frac{g_1}{4} \) and T₂ = 2π√(ℓ/g₂).
\( \frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}} = \sqrt{4} = 2 \).
T₂ = 2 × \( \frac{1}{20} \) = 0.1 s.
In simple words: With one-fourth of the gravity, the pendulum swings twice as slowly, so its time period is doubled to 0.1 s.

Teacher's Note:
1. T is inversely proportional to √g.
2. First find T on the earth as 1 ÷ 20 s.

Question 2. Which of the following distance-time graphs is not possible? Explain why?

Answer:
Graphs (2) and (3) are not possible.
Graph (1): the distance stays the same as time passes, so the body is at rest. It is possible.
Graph (2): the distance decreases with time. The distance travelled can never decrease, so this is not possible.
Graph (3): a vertical line means the body has different distances at the same instant of time, which is not possible.
Graph (4): the distance increases steadily with time, so the body moves with uniform speed. It is possible.
In simple words: Distance can stay the same or grow, but it cannot shrink, and one instant of time cannot have two distances.

Teacher's Note:
1. Check the graph against two rules: distance never decreases, and each time has one distance.
2. A horizontal line means rest; a sloping straight line means uniform speed.

Question 3. The distance between two railway stations A and B is 440 km. Two trains T₁ and T₂ start simultaneously from the stations A and B, respectively, and move towards each other. The train T₁ moves with a constant speed of 60 km h⁻¹ and the train T₂ moves with a speed of 50 km h⁻¹. Determine the point where they meet and the time taken for this.
Answer:
Both trains travel for the same time before they meet. Let the trains meet at x km from A; then T₂ travels (440 − x) km.
\( \frac{x}{60} = \frac{440 - x}{50} \) ⇒ 5x = 6(440 − x) = 2640 − 6x ⇒ 11x = 2640 ⇒ x = 240 km.
The trains meet at 240 km from A.
Time = \( \frac{240}{60} \) = 4 h.
In simple words: Together the trains close 110 km every hour, so they meet after 440 ÷ 110 = 4 h, 240 km from A.

Teacher's Note:
1. The times taken by both trains are equal.
2. Check using the relative speed: 60 + 50 = 110 km h⁻¹.

Question 4. A body travels a distance of 500 m in 10 minute. Find out the speed of the body in km h⁻¹.
Answer:
Speed = \( \frac{500\ \text{m}}{10 \times 60\ \text{s}} = \frac{5}{6} \) m s⁻¹.
In km h⁻¹ = \( \frac{5}{6} \times \frac{18}{5} \) = 3 km h⁻¹.
(Check: 0.5 km in \( \frac{1}{6} \) h = 3 km h⁻¹.)
In simple words: The body covers half a kilometre in a sixth of an hour, which is 3 km h⁻¹.

Teacher's Note:
1. Convert 500 m to 0.5 km and 10 min to 1/6 h to get the answer in one step.
2. Or find the speed in m s⁻¹ and multiply by 18/5.

Question 5. A student goes to school with a uniform speed of 3 km h⁻¹ and returns to his home with a speed of 2 km h⁻¹. If the student takes 1 hour in all, then find out the distance between the school and the house.
Answer:
Let the distance be x km.
Time to school = \( \frac{x}{3} \) h; time back = \( \frac{x}{2} \) h.
\( \frac{x}{3} + \frac{x}{2} = 1 \) ⇒ \( \frac{2x + 3x}{6} = 1 \) ⇒ 5x = 6.
x = 1.2 km.
In simple words: The school is 1.2 km from the house.

Teacher's Note:
1. The distance is the same in both directions.
2. Add the two times and equate them to the total time.

Question 6. Kishore walking with a uniform speed covers 20 km in 5 h. Find the distance he covers in 7 hours.
Answer:
Speed = \( \frac{20}{5} \) = 4 km h⁻¹.
Distance in 7 h = 4 × 7 = 28 km.
In simple words: At 4 km every hour, he covers 28 km in 7 hours.

Teacher's Note:
1. First find the speed, then use distance = speed × time.
2. Check: 20 ÷ 5 = 28 ÷ 7.

Question 7. Determine the speed of the tip of a minute hand of length 10 cm of a wall clock.
Answer:
The tip makes one round (circumference of a circle of radius 10 cm = 0.1 m) in 1 hour = 3600 s.
Distance = 2πr = 2π × 0.1 = 0.2π m ≈ 0.628 m.
Speed = \( \frac{0.628}{3600} \) = 1.75 × 10⁻⁴ m s⁻¹.
In simple words: The tip of the minute hand creeps at less than two ten-thousandths of a metre per second.

Teacher's Note:
1. The minute hand takes 1 h for one full rotation.
2. Use π = 3.14 and keep the length in metres.

Question 8. The distance-time graph for the motion of three vehicles A, B and C are as shown in the figure. S_A, S_B and S_C are the speeds of A, B and C, respectively. Compare the speeds of A, B and C.

Answer:
The speed of a body is given by the slope of its distance-time graph; the steeper the line, the greater the speed.
The line of A is the steepest and the line of B is the least steep, with C in between.
So S_A > S_C > S_B.
In simple words: The steeper the line on a distance-time graph, the faster the vehicle.

Teacher's Note:
1. Compare slopes at the same time: the higher line covers more distance.
2. All three lines start at the origin, so they can be compared directly.

Question 9. Usha participated in a marathon walk and completes the journey in 10 h. If she travels half the distance with a speed of 10 km h⁻¹ and the rest of the distance with a speed of 15 km h⁻¹, then calculate the total distance of the marathon walk.
Answer:
Let the total distance be x km. Each half is \( \frac{x}{2} \) km.
t₁ = \( \frac{x/2}{10} = \frac{x}{20} \) h and t₂ = \( \frac{x/2}{15} = \frac{x}{30} \) h.
\( \frac{x}{20} + \frac{x}{30} = 10 \) ⇒ 3x + 2x = 600 ⇒ 5x = 600.
x = 120 km.
In simple words: The marathon walk is 120 km long.

Teacher's Note:
1. Take the common denominator 60 to add the two fractions.
2. Check: 60 km at 10 km h⁻¹ is 6 h, and 60 km at 15 km h⁻¹ is 4 h; total 10 h.

Question 10. Venkat went to his college from his house on his car with a uniform speed of 60 km h⁻¹ and returns to his house with a different speed. If the average speed of the car for the whole journey is 50 km h⁻¹, then find the speed of his car when it travelled from college to house.
Answer:
Let the distance between house and college be d and the speed on return be x km h⁻¹.
Total time = \( \frac{d}{60} + \frac{d}{x} \); total distance = 2d.
\( \frac{2d}{\frac{d}{60} + \frac{d}{x}} = 50 \) ⇒ \( \frac{2}{\frac{1}{60} + \frac{1}{x}} = 50 \) ⇒ \( \frac{1}{60} + \frac{1}{x} = \frac{1}{25} \).
\( \frac{1}{x} = \frac{1}{25} - \frac{1}{60} = \frac{7}{300} \) ⇒ x = \( \frac{300}{7} \) = 42.86 km h⁻¹.
In simple words: To average 50 km h⁻¹ the car must return at about 42.9 km h⁻¹.

Teacher's Note:
1. The total distance is 2d, the round trip.
2. d cancels out, so the distance is not needed.

Question 11. A boy goes to a cycle shop by walk at 5 km h⁻¹ and comes back home by cycle at a speed of 15 km h⁻¹, find out the average speed of the boy.
Answer:
Let the distance to the shop be x km.
Time going = \( \frac{x}{5} \) h, time returning = \( \frac{x}{15} \) h; total time = \( \frac{3x + x}{15} = \frac{4x}{15} \) h.
Average speed = \( \frac{2x}{4x/15} = \frac{30}{4} \) = 7.5 km h⁻¹.
In simple words: The boy's average speed is 7.5 km h⁻¹, not the simple average of 5 and 15.

Teacher's Note:
1. For equal distances, average speed = 2v₁v₂/(v₁ + v₂).
2. The slower speed counts for more because he spends more time on it.

Question 12. The distance between two stations A and B is 200 km. A train travels from A to B at a uniform speed of 60 km h⁻¹. After half an hour another train leaves B towards A with a uniform speed of 80 km h⁻¹. Determine the distance of the point from A where the two trains meet each other.
Answer:
Let the trains meet at x km from A. The second train covers (200 − x) km.
The first train travels half an hour longer: \( \frac{x}{60} = \frac{200 - x}{80} + \frac{1}{2} \).
\( \frac{x}{60} - \frac{200 - x}{80} = \frac{1}{2} \) ⇒ \( \frac{80x - 60(200 - x)}{4800} = \frac{1}{2} \) ⇒ 140x − 12000 = 2400.
140x = 14400 ⇒ x = 102.86 km from A.
In simple words: The trains meet about 102.9 km from A.

Teacher's Note:
1. The train from A has been running for 0.5 h more than the train from B.
2. Check: 102.86 ÷ 60 = 1.714 h and 97.14 ÷ 80 = 1.214 h, which is 0.5 h less.

Question 13. Ram starts his bike from rest and moves along a straight path. If he attains a speed of 72 km h⁻¹ in 10 s, determine his acceleration.
Answer:
u = 0 and v = 72 km h⁻¹ = 72 × \( \frac{5}{18} \) = 20 m s⁻¹; t = 10 s.
a = \( \frac{v - u}{t} = \frac{20 - 0}{10} \) = 2 m s⁻².
In simple words: His speed grows by 2 m s⁻¹ each second.

Teacher's Note:
1. Starting from rest means u = 0.
2. Change km h⁻¹ to m s⁻¹ before substituting.

Question 14. Raj dropped a cricket ball of mass 700 g from the top of a tower and the ball takes 5 s to reach the ground. He then determined the maximum velocity of the body by taking the acceleration of the ball as 10 m s⁻². Find out the value of the maximum velocity.
Answer:
u = 0, a = 10 m s⁻², t = 5 s.
a = \( \frac{v - u}{t} \) ⇒ 10 = \( \frac{v - 0}{5} \).
v = 10 × 5 = 50 m s⁻¹.
In simple words: The ball hits the ground at 50 m s⁻¹, its greatest velocity.

Teacher's Note:
1. The mass of the ball (700 g) is not needed.
2. Use v = u + at with u = 0.

Question 15. An ant moves along a rectangular path as shown in the figure. It starts from A and moves with a uniform speed of 0.01 m s⁻¹. Determine the average speed after 1 hour.

Answer:
The ant moves with a uniform speed all the time, so its average speed is the same as its speed.
Average speed = 0.01 m s⁻¹.
(In 1 hour = 3600 s it covers 0.01 × 3600 = 36 m, i.e. 36 ÷ 0.24 = 150 rounds of the rectangle, and 36 m ÷ 3600 s = 0.01 m s⁻¹.)
In simple words: Steady speed means the average speed is that same speed, 0.01 m s⁻¹.

Teacher's Note:
1. Average speed does not depend on the path shape when the speed is uniform.
2. Average velocity after complete rounds would be zero, but the question asks for speed.

Question 16. The distance-time graph of a body moving in a straight line is given below. Draw the speed-time graph of the motion of the body.

Answer:
On a distance-time graph, the slope gives the speed.
From O to t₁: the graph is a straight sloping line, so the speed is constant (uniform).
From t₁ to t₂: the line is horizontal, so the distance does not change and the speed is zero.
From t₂ to t₃: the line slopes up again (steeper than before), so the body moves with a constant but higher speed.
After t₃: the line is horizontal again, so the speed is zero.
The speed-time graph is therefore:

In simple words: Each straight part of the distance graph becomes a flat piece at the height of its speed, and a rest period gives zero speed.

Teacher's Note:
1. Slope of distance-time graph = height of the speed-time graph.
2. A flat distance-time line becomes a speed-time line lying on the time axis.

Question 17. Asish took a challenge to cover a distance of 1 km in 1 minute on his racing bicycle. If he covers half the distance in \( \frac{3}{5} \)th of a minute, then with what speed should he ride the remaining journey on his bicycle?
Answer:
Distance left = \( \frac{1000}{2} \) = 500 m.
Time left = \( \left(1 - \frac{3}{5}\right) \times 60 \) = 24 s.
Speed required = \( \frac{500}{24} \) = 20.83 m s⁻¹ (= 75 km h⁻¹).
In simple words: He must cover the last 500 m in 24 s, riding at about 20.8 m s⁻¹.

Teacher's Note:
1. Find the remaining time first: 1 min − 36 s = 24 s.
2. Divide the remaining distance by the remaining time.

Question 18. A body moves along a straight line. In every t seconds the speed of the body decreases uniformly to half of the original speed. Draw the speed-time graph of the body.
Answer:
Let the initial speed be S. After t seconds the speed is \( \frac{S}{2} \), after 2t seconds it is \( \frac{S}{4} \), after 3t seconds it is \( \frac{S}{8} \), and so on.
Plot time on the X-axis and speed on the Y-axis through the points (0, S), (t, S/2), (2t, S/4) and (3t, S/8). The speed falls continuously, so we get a downward-curving graph that gets closer and closer to the time axis:

In simple words: The speed is halved again and again, so the graph is a curve that falls quickly at first and then flattens out.

Teacher's Note:
1. Mark the points for t, 2t and 3t first, then join them with a smooth curve.
2. The speed gets smaller each time but never becomes exactly zero in this pattern.