NCERT Solutions for Class 7 Science Chapter 8 Reproduction in Plants

1. Calculate the speed of a car that travels 150 metres in 10 seconds. Express your answer in km/h.

Answer:

Distance = 150 m
Time taken $=10 \mathrm{~s}$

$
\begin{aligned}
& \text { SPeed }=\frac{\text { Distance covered }}{\text { Time taken }}=\frac{150 \mathrm{~m}}{10 \mathrm{~s}} \\
& \text { SPeed in } \mathrm{km} / \mathrm{hr}=15 \times \frac{18}{5}=54 \mathrm{~km} / \mathrm{h}
\end{aligned}
$

2. A runner completes 400 metres in 50 seconds. Another runner completes the same distance in 45 seconds. Who has a greater speed and by how much?

Answer:

Runner 1:
Distance $=400 \mathrm{~m}$,
Time $=50 \mathrm{~s}$

$
\text { SPeed }=\frac{\text { Distance covered }}{\text { Time taken }}=\frac{400 \mathrm{~m}}{50 \mathrm{~s}}=8 \mathrm{~m} / \mathrm{s}
$

Runner 2:
Distance $=400 \mathrm{~m}$,
Time $=50 \mathrm{~s}$

$
\text { SPeed }=\frac{\text { Distance covered }}{\text { Time taken }}=\frac{400 \mathrm{~m}}{45 \mathrm{~s}}=8.89 \mathrm{~m} / \mathrm{s}
$

Difference $=8.89-8=0.89 \mathrm{~m} / \mathrm{s}$
Hence, speed of runner 2 is greater by approximately $0.89 \mathrm{~m} / \mathrm{s}$.

3. A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time does it take?

Answer:

$
\begin{aligned}
& \text { Speed }=25 \mathrm{~m} / \mathrm{s} \\
& \text { Distance }=360 \mathrm{~km}=3,60,000 \mathrm{~m} \\
& \text { Time taken }=\frac{\text { Distance }}{\text { Speed }}=\frac{3,60,0000 \mathrm{~m}}{25 \mathrm{~m} / \mathrm{s}} \\
& =14,400 \mathrm{~s} \\
& \Rightarrow 240 \mathrm{mins} \\
& \Rightarrow 4 \text { hours }
\end{aligned}
$

4. A train travels 180 km in 3 h. Find its speed in:
(i) km/h
(ii) m/s
(iii) What distance will it travel in 4 h if it maintains the same speed throughout the journey?

Answer:

Distance $=180 \mathrm{~km}$, time $=3 \mathrm{~h}$
(i) Speed $=\frac{\text { Distance }}{\text { Time }}=\frac{180 \mathrm{~km}}{3 \mathrm{~h}}=60 \mathrm{Km} / \mathrm{h}$
(ii) Speed in $\mathrm{m} / \mathrm{s}=60 \times \frac{5}{18}=16.677 \mathrm{~m} / \mathrm{s}$
(iii) Time $=4 \mathrm{~h}$, Speed $=60 \mathrm{~km} / \mathrm{h}$

Distance $=$ Speed $\times$ Time

$
=60 \times 4=240 \mathrm{~km}
$

5. The fastest galloping horse can reach the speed of approximately 18 m/s. How does this compare to the speed of a train moving at 72 km/h?

Answer:

Speed of horse $=18 \mathrm{~m} / \mathrm{s}$
Speed of train $=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s}$
The train is faster by $2 \mathrm{rn} / \mathrm{s}$ than the fastest galloping horse.

6. Distinguish between uniform and non-uniform motion using the example of a car moving on a straight highway with no traffic and a car moving in city traffic.
Answer:

Uniform motion: If an objects covers equal distances in equal distance in equal intervals of time, its motion is said to be uniform. A car moving on a straight highway with no traffic is an example of uniform motion.

Non-uniform motion: If an object covers unequal distances in equal interval of time, its motion is said to be non-uniform. A car moving in a traffic is an example of nonuniform motion.

7. Data for an object covering distances in diff erent intervals of time are given in the following table. If the object is in uniform motion, fi ll in the gaps in the table.

Answer:

In uniform motion, object covers equal distances in equal intervals of time. Hence, the speed of an object remains constant throughout the motion.

The object covers 8 m in every 10 seconds. Hence, the speed of an object remains constant at 0.8 m/s throughout the motion.

8. A car covers 60 km in the fi rst hour, 70 km in the second hour, and 50 km in the third hour. Is the motion uniform? Justify your answer. Find the average speed of the car.

Answer:

The car covers different distances in each hour. Hence, the motion of the car is nonuniform.
Total distance $=60 \mathrm{~km}+70 \mathrm{~km}+50 \mathrm{~km}=180 \mathrm{~km}$
Total time $=3$ hours

$
\begin{aligned}
& \text { Average speed }=\frac{\text { Total distance travelled }}{\text { Total time taken }} \\
& =\frac{180 \mathrm{~km}}{3 \mathrm{~h}}=60 \mathrm{~km} / \mathrm{h} .
\end{aligned}
$
Hence, the average speed of the car is $60 \mathrm{~km} / \mathrm{h}$.

9. Which type of motion is more common in daily life — uniform or non-uniform? Provide three examples from your experience to support your answer.

Answer:

In our daily life, most motions are nonuniform because object do not move at the same speed all the time. Their speed changes due to factors like traffic, rough or uneven roads and other obstacles.

Examples:
• Travelling in a bus on an uneven road
• Playing cricket
• Walking through a crowded market

10. Data for the motion of an object are given in the following table. State whether the speed of the object is uniform or non-uniform. Find the average speed.

Answer:

$
\begin{aligned}
&\text { The object exhibits non-uniform motion, because it covers unequal distances in equal time intervals. }\\
&\begin{aligned}
& \text { Total distance travelled }=60 \mathrm{~m} \\
& \text { Total time taken }=100 \mathrm{~s} \\
& \text { Average speed }=\frac{\text { Total distance }}{\text { Total time taken }} \\
& =\frac{60}{100} \\
& =0.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
$

11. A vehicle moves along a straight line and covers a distance of 2 km. In the fi rst 500 m, it moves with a speed of 10 m/s and in the next 500 m, it moves with a speed of 5 m/s. With what speed should it move the remaining distance so that the journey is complete in 200 s? What is the average speed of the vehicle for the entire journey?

Answer:

Given
Total distance $=2 \mathrm{~km}=2000 \mathrm{~m}$,
Total time $=200$
Step 1: Time taken to cover the first 500 m

$
\text { Time }=\frac{\text { Distance }}{\text { Speed }}=\frac{500}{10}=50 \mathrm{~s}
$

Step 2: Time taken to cover the next 500 m

$
\text { Time }=\frac{500}{10}=100 \mathrm{~s}
$

Step 3: Remaining distance $=2000-1000=1000 \mathrm{~m}$
Remaining time $=200-150=50 \mathrm{~s}$

Step 4: Speed required to cover the remaining 1000 m
Speed $=\frac{1000}{50}=20 \mathrm{~m} / \mathrm{s}$

Step 5: Average speed $=\frac{\text { Totaldistance }}{\text { Totaltimetaken }}=\frac{2000}{200}=10 \mathrm{~m} / \mathrm{s}$

Exploratory Projects

1. Construct a fl oating bowl-type water clock. Experiment by using bowls of diff erent sizes and making holes of diff erent sizes in them so that the sinking time of the bowl can be close to 24 minutes.

Answer:

2. Design an activity for measuring the pulse rate (number of times the pulse of a person beats in 1 minute) of your friends. Think of an activity where you can use your pulse to measure time and develop a story over that idea.

Answer:

Story:
You and your friends are on a very special mission assignment called “Team Pulse Trackers.” One day, your science teacher gives you an undercover assignment. You will assess your teammates' health using a unique piece of equipment - your pulse!

There is a catch - you cannot time your demonstration using a stopwatch. You will be timing using your own pulse beats!

What you will need:

A notebook
A pen
A quiet place
A few friends
Steps:

Locate Your Pulse:
Place two fingers (not your thumb) on either your wrist or your neck side to locate your own pulse.

Count Your Own Pulse for 1 Minute:
Initially, use clock and record the number of beats of your own pulse in 1 minute. This will serve as your reference pulse rate.

Example: If your pulse beats 72 times in one minute, then 72 pulse beats = 1 minute.

Now use your own pulse as a timer:
You will now replace the clock with your pulse beats. Count the 72 beats silently.

Now measure your friend's pulse:
While you are counting your own 72 pulse beats, ask your friend to let you to feel their wrist pulse

Conclusion (What You Learned):
Your pulse rate rises after exercising.

You can use your body (pulse) as a time measuring device — just like a stopwatch.

Measuring and comparing pulse rates is a fun way to learn about heartbeats and well-being. Count how many times your friend's pulse beats in the same time.

Write It Down:
Write your friend's name down and their pulse count while you used your 1 minute pulse timer.

Compare Pulse Rates:
Try with different friends after running, walking, or resting, and compare how their pulse rates increased!

3. What might be the reasons for the slight diff erences in the time periods of a pendulum of a given length in diff erent readings taken in Activity 8.2. Think of ways to control those and repeat the activity to check if the diff erence in readings is reduced.

Answer:

Small inconsistencies may occurr in the time period readings although the length of the pendulum is the same if slight mistakes were made in carrying out the activity.

Some Possible Reasons for the Difference:
1. Not letting the pendulum go cleanly - If you pushed on the pendulum instead of letting it swing out freely.
2. Counting incorrectly - You might count one extra, or one less oscillation.
3. Timing incorrectly - If you start or stop the stopwatch too early or late.
4. Air disturbance - If you have a fan or wind blowing in the room.
5. From an uneven surface or shaky stand - If the pendulum is not hanging from a secure point.

Some Ways to Control these Differences:
1. Let the pendulum go slow and from the same height each time - do not push it.
2. Use a steady hand and practice starting and stopping the time with the stopwatch.
3. Get help from a friend (one counts and one controls the time).
4. Use the same conditions by closing doors and windows to prevent air disturbance.
5. Use a stable and secure support to hang the pendulum.

4. Visit a playground with a few swings. Measure the time taken by a swing for 10 oscillations and calculate its time period. Repeat it a few times with children of different weights to find out if its time period is almost the same. Repeat this with swings of different lengths. Find out how the time period changes with increasing length of the swings. Is the swing also an example of a pendulum?

Answer:

Step 1: Visit a Playground
Find a local playground with swings that will allow you to monitor time and swinging. Swings act similarly to pendulums, as they swing back and forth just like a simple pendulum swings.

What you will need:
Stopwatch or phone with a timer

measuring tape or a scale (to measure length of swing)

notebook and pen

Friends (or young children) that can use the swings

Step 2: Measure the Time Period of One Swing
Choose one swing.

Ask a child (or you) to sit as still as possible in the swing.

Pull it back a little, and then let is go slowly.

Count 10 complete back-and-forth swings (oscillations), and use your stopwatch to time:

From this, you can now calculate the time period with:

\text { Time Period }=\frac{\text { Total Time }}{10}

Step 3: Try With Different Weight Children
Repeat the same experiment with:

A small child

A larger child

You will find that the weight of the child does not really matter!

Step 4: Try Swings with Different Lengths
Now take the shorter swing and longer swing. Repeat the same process.

You will find that swings with a longer length will take longer for a complete oscillation, while the shorter swing's oscillation will go by quickly.

Conclusion:

• A swing is a real-world example of a pendulum.
• When the length of the swing increased, the time period increased.
• The mass of the person does not make a noticeable difference in the time period.

5. Gather the timings of the winners of the races — 100 m, 200 m, and 400 m for men and women in the last two Olympic games. Calculate and compare their speeds. In which event is the speed the fastest?

Answer:

By comparing speeds:

Men’s 100 m: ≈ 10.20 m/s (fastest among men)

Women’s 100 m: ≈ 9.42 m/s (fastest among women)

Even though 200 m is longer, athletes maintain high speed, but the 100 m remains the fastest event overall, for both men and women—because the sprint allows athletes to run at near-maximum speed for the shortest duration.