Class 09 : Matter In Our Surroundings : NCERT Exercise Solution

 Question 1: Convert the following temperature to the Celsius scale.

(a) 293 K (b) 470 K

Answer: (a) Temperature in Celsius scale = Temperature in Kelvin scale - 273

⇒ 293 K = 293 K – 273 = 20⁰C

(b) Temperature in Celsius scale = Temperature in Kelvin scale - 273

⇒ 470 K = 470 K – 273 = 197⁰C

Question 2: Convert the following temperature to the Kelvin scale.
(a) 25⁰C (b) 373⁰C

Answer: (a) Temperature in Kelvin scale = Temperature in Celsius scale + 273
= 25⁰C + 273 = 298 K

Answer: (b) Temperature in Kelvin scale = Temperature in Celsius scale + 273
= 373⁰C + 273 = 646 K

Question 3: Give reason for the following observations.

(a) Naphthalene balls disappear with time without leaving any solid.

Answer: Naphthalene ball is a sublimate and a sublimate turns into vapour without changing into liquid. Thus, naphthalene balls disappear with time without leaving any solid.

(b) We can get the smell of perfume sitting several meters away.

Answer: Perfume turns into gas at room temperature. The vapour of perfume travels up to several meters because of diffusion. That’s why we can get the smell of perfume sitting several meters away.

Question 4: Arrange the following substances in increasing order of forces of attraction between the particles – Water, Sugar, Oxygen.

Answer: Oxygen < Water < Sugar

Explanation: Oxygen is a gas, thus force of attraction is negligible between particles. Water is a liquid, thus force of attraction between particles is more than liquid and less than solid. Sugar is a solid, thus force of attraction between particles is greatest.

Question 5: What is the physical state of water at (a) 25⁰C (b) 0⁰C (c) 100⁰C

Answer: (a) At 25⁰C – water is in liquid state.
(b) At 0⁰C – water is in solid state.
At 100⁰C – water is in transition state, i.e. in liquid and gas both.

Question 6: Give two reasons to justify;

(a) Water at room temperature is a liquid.

Answer: At room temperature:

• Water has definite volume, but not definite shape as it takes the shape of the container in which it is kept.
• Water flows at room temperature.

(b) An iron almirah is a solid at room temperature.

Answer: An iron almirah is a solid at room temperature because:

• It has definite shape.
• It has definite volume.

Question 7: Why is ice at 273K more effective in cooling than water at the same temperature.

Answer: At 273K ice requires more latent heat to melt into water, while water at 273K requires less latent heat; to come to the room temperature. So, ice at 273 K is more effective in cooling than water at the same temperature.

Question 8: What produces more severe burns, boiling water or steam?

Answer: Steam produces more severe burns than boiling water, because steam has more latent heat than boiling water.

Question 9: Name A,B,C,D,E and F in the following diagram showing change in its state.

Answer:
A – Heating - Melting
B – Heating - Vapourisation
C – Cooling – Condensation - Liquefaction
D – Cooling – Freezing
E – Sublimation
F – Solidification

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Class 09 : Motion : NCERT In Text Solution

 Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance.

Suppose a ball starts moving from point A and it returns back at same point A, then the distance will be equal to 20 meters while displacement will be zero.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Answer:

Given, side of the square field = 10m

Therefore, perimeter = 10 m x 4 = 40 m

Farmer moves along the boundary in 40s.

Displacement after 2 m 20 s = 2 x 60 s + 20 s = 140 s =?

Since in 40 s farmer moves 40 m

∴ in 1s distance covered by farmer $=\frac{40}{40}m=1m$

∴ in 140s distance covered by farmer $=1\times 140m=140m$

Now, number of rotation to cover 140m along the boundry$=\frac{\text{Total distance}}{\text{Perimeter}}$

$\frac{140m}{40m}=3.5$ round

Thus, after 3.5 round farmer will at point C of the field.

∴ Displacement $AC=\sqrt{{\left(10m\right)}^2+{\left(10m\right)}^2}$

$=\sqrt{100m^2+100m^2}$

$=\sqrt{200m^2}$

$=\sqrt{2\times 100m^2}$

$=10\sqrt{2}m$

${\displaystyle \mathrm{<br>}}$

Question 3: Which of the following is true for displacement?

(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.

Answer: None

Question 4: Distinguish between speed and velocity.

Answer: Speed has only magnitude while velocity has both magnitude and direction.

Question 5: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: When distance is equal to displacement.

Question 6: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 7: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion the path of an object will look like a straight line.

Question 8: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 ms-1.

Answer: Here, we have, speed $=3\times {10}^{8}m{s}^{-1}$

Time = 5 minute

$=5\times 60s=300$ second

We know that, Distance = Speed x Time

⇒ Distance $=300\times 3\times {10}^{8}m{s}^{-1}$

$=900\times {10}^{8}m$

$=9\times {10}^{10}m$

Question 9: When will you say a body is in

(i) uniform acceleration?

Answer: When rate of change of motion is same in equal intervals of time.

(ii) non-uniform acceleration?

Answer: When rate of change of motion is not same in equal intervals of time.

Question 10: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Answer: Here we have, u=80 km/h $=80\times \frac{5}{18}=\frac{200}{9}$ m/s

v=60 km/h $=60\times \frac{5}{18}=\frac{50}{3}$ m/s and $t=5s$

∴ Acceleration (a) =?

We know that $v=u+at$

Or, $a=\frac{v-u}{t}$

$=\frac{\frac{50}{3}-\frac{200}{9}}{5}=-\frac{50}{9\times 5}$

$=-\frac{10}{9}$ m s-2

Question 11: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

Answer: Here we have,
Initial velocity, u = 0,
Final velocity, v = 40km/h $=40\times \frac{5}{18}=\frac{100}{9}$ m/s
Time (t) = 10 minute = 60 × 10 = 600s
Acceleration (a) =?

We know that $v=u+at$

Or, $a=\frac{v-u}{t}$

$=\frac{100}{9\times 600}=\frac{1}{54}$ m s-2

Question 12: What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer:

(a) The slope of the distance-time graph for an object in uniform motion is straight line.

(b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 13: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: When the slope of distance-time graph is a straight line parallel to time axis, the object is moving with uniform motion.

Question 14: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: When the slope of a speed time graph is a straight line parallel to the time axis, the object is moving with uniform speed.

Question 15: What is the quantity which is measured by the area occupied below the velocity-time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity time graph.

Question 16: A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled.

Answer: Here we have, Initial velocity (u) = 0
Acceleration (a) = 0.1ms-2
Time (t) = 2 minute = 120 second

(a) The speed acquired:

We know that, v = u + at

$\Rightarrow v=0+0.1m/s^2\times 120s$

$\Rightarrow v=120m/s$

Thus, the bus will acquire a speed of 120 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

We know that, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=0\times 120s+\frac{1}{2}\times 0.1m/s^2\times {\left(120s\right)}^2$

$=\frac{1}{2}\times 1440m=720m$

Thus, bus will travel a distance of 720 m in the given time of 2 minute.

Question 17: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.

Answer: Here,we have,

Initial velocity, $u=90km/h$

$=\frac{90\times 1000m}{60\times 60s}=25m/s$

Final velocity $v=0$

Acceleration, $a=-0.5m/s^2$

Thus, distance travelled =?

We know that, $v^2=u^2+2as$

$\Rightarrow 0={(25m/s)}^2+2\times -0.5m/s^2\times s$

$\Rightarrow 0=625m^2{s}^{-2}-1m{s}^{-2}s$

$\Rightarrow 1m{s}^{-2}s=625m^2{s}^{-2}$

$\Rightarrow s=\frac{625m^2{s}^{-2}}{1m{s}^{-2}}=625m$

Therefore, train will go 625 m before it brought to rest.

Question 18: A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?

Answer: Here we have,
Initial velocity, u = 0
Acceleration (a) = 2cm/s2 = 0.02m/s2
Time (t) = 3s
Therefore, Final velocity, v =?

We know that, $v=u+at$

$\therefore v=0+0.02m/s^2\times 3s$

$\Rightarrow v=0.06m/s$

Therefore, the final velocity of trolley will be 0.06m/s after start

Question 19: A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?

Answer: Here we have,
Acceleration, a = 4m/s2
Initial velocity, u =0
Time, t = 10s
Therefore, Distance (s) covered =?

We know that, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow s=0\times 10s+\frac{1}{2}\times 4m/s^2\times {\left(10s\right)}^2$

$\Rightarrow s=\frac{1}{2}\times 4m/s^2\times 100s^2$

$\Rightarrow s=2\times 100m=200m$

Thus, racing car will cover a distance of 200m after start in 10 s with given acceleration.

Question 20: A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Answer: Here we have,

Initial velocity (u) = 5m/s

Final velocity (v) =0 (Since from where stone starts falling its velocity will become zero)

Acceleration (a) = -10m/s2

(Since given acceleration is in downward direction, i.e. the velocity of the stone is decreasing, thus acceleration is taken as negative)

Height, i.e. Distance, s =?

Time (t) taken to reach the height =?

We know that, $v^2=u^2+2as$

$\Rightarrow 0={(5m/s)}^2+2\times -10m/s^2\times s$

$\Rightarrow 0=25m^2{s}^2-20m/s^2\times s$

$\Rightarrow 20m/s^2\times s=25m^2{s}^2$

$\Rightarrow s=\frac{25m^2{s}^2}{20m/s^2}$

$\Rightarrow s=1.25m$

Now, we know that, $v=u+at$

$\Rightarrow 0=5m{s}^{-1}+(-10m{s}^{-2})\times t$

$\Rightarrow 0=5m{s}^{-1}-10m{s}^{-2}\times t$

$\Rightarrow 10m{s}^{-2}\times t=5m{s}^{-1}$

$\Rightarrow t=\frac{5m{s}^{-1}}{10m{s}^{-2}}$

$\Rightarrow t=\frac{1}{2}s=0.5s$

Thus, stone will attain a height of 1.25m. And time taken to attain this height is 0.5s

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