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Science Que-Ans (Each of 4 Mark )

Motion · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 24 questions.

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Que-Ans (Each of 4 Mark )

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24 questions · timed · auto-graded

Question 14 Marks
An electron moving with a velocity of 5 × 104 m s-1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s–2 in the direction of its initial motion. Calculate the time in which the electron would acquire a velocity double of its initial velocity. How much distance the electron would cover in this time?
Answer
Given initial velocity, u = 5 × 104ms-1 and acceleration, a = 104ms-2
  1. Fina; velocity = v = 2u = 2 × 5 × 104ms-1 = 10 × 104ms-1
    To find t, use v = u + at
    or  $\text{t}=\frac{\text{v-u}}{\text{a}}$
    $\Big(\frac{10\times10^2-5\times10^4}{10^4}\Big)=\frac{5\times10^4}{10^4}=5\text{s}$
  1. Using $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
    $=(5\times10^4)\times5+\frac{1}{2}(10^4)\times(5)^2=25\times10^4+\frac{25}{2}\times10^4=37.5\times10^4\text{m}$
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Question 24 Marks
Show by using the graphical method that:

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

where the symbols have their usual meanings.

Answer
Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC.

Suppose the body travels a distance s in time t. In the figure, the distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABC. Thus:

Distance travelled = Area of figure OABC

= Area of rectangle OABC + area of triangle ABD

Now, we will find out the area of rectangle OABC and area of triangle ABD.

  1. Area of rectangle OADC = OA × OC

= u × t

= ut

  1. Area of triangle ABD $=\Big(\frac{1}{2}\Big)\times\text{Area of rectangle AEBD}$

$=\Big(\frac{1}{2}\Big)\times\text{AD}\times\text{BD}$

$=\Big(\frac{1}{2}\Big)\text{at}^2$

Distance travelled, s = Area of rectangle OADC + Area of triangle ABD

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

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Question 34 Marks
The figure shows an x-t graph of a particle moving along a straight line. What is the sign of the acceleration during the intervals OA, AB, BC, and CD?

Answer
  1. From the graph we can infer that in the interval OA, it’s almost a straight line with a positive slope. Therefore velocity is positive that is with increase in time distance also increases and thus acceleration is also positive as well as uniform.
  2. In the interval AB, the line has a negative slope as with increase in time distance decreases. Here velocity is negative and it shows retardation.
  3. BC represents more downfalls in the slope of the line which means velocity will become more negative and hence it shows retardation with greater magnitude than AB.
  4. CD shows that with increase in time distance is increasing again with a positive slope. Therefore velocity is positive which implies acceleration is also positive.
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Question 44 Marks
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Plot a velocity–time graph for the same.

Answer

From the graph,

  1. Initial velocity, u = 0 [since displacement and time is zero ]
  1.  Velocity after 100s, $\text{v}=\frac{\text{Displacement}}{\text{Time}}$ [Here, displacement = zero and time = 100s]

 $\Rightarrow\text{v}=\frac{0}{100}=0$

Therefore,
 
v
0
2
0
t
0
50
100

 Velocity-time graph plotted from the above data is shown  below 

 

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Question 54 Marks
A boy walks from his classroom to the bookshop along a straight corridor towards North. He covers a distance of 20m in 25 seconds to reach the bookshop. After buying a book, he travels the same distance in the same time to reach back in the classroom. Find (a) average speed, and (b) average velocity, of the boy.
Answer
  1. Total distance covered in going to the bookshop and coming back to the classroom = 20 + 20 = 40m

Total time taken = 25 + 25 = 50 sec

Average speed $=\frac{\text{Total distance}}{\text{Total time}}=\frac{40}{50}=0.8\text{m/s}$

  1. Average velocity $=\frac{\text{Total displacement}}{\text{Total time}}=\frac{0}{50}=0\text{m/s}$
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Question 64 Marks
The graph given alongside shows how the speed of a car changes with time:

  1. What is the initial speed of the car?
  2. What is the maximum speed attained by the car?
  3. Which part of the graph shows zero acceleration
  4. Which part of the graph shows varying retardation?
  5. Find the distance travelled in first 8 hours.
Answer
  1. Initial speed of the car is 10 km/h
  2. Maximum speed attained by the car is 35km/h
  3. BC represents zero acceleration.
  4. CD represents varying retardation.
  5. Distance travelled is given by the area enclosed within the curve. So,

Distance travelled = Area or trapezium + Area of rectangle

So distance travelled,

$= \Big(\frac{1}{2}\Big)(8+5)(25) + (8)(10)$

$= 242.5\text{km}$

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Question 74 Marks
Describe an experiment to show that sound cannot pass through vacuum.
Answer

Sound cannot travel through vacuum. This can be shown by the following experiment:

  1. A ringing electric bell is placed inside an air tight glass jar containing air. We can hear the sound of ringing bell clearly. Thus, when air is present as medium in the bell jar, sound can travel through it and reach our ears.
  2. The bell jar containing ringing bell is placed over the plate of a vacuum pump. Air is gradually removed from the bell jar by switching on the vacuum pump. As more and more air is removed from the bell jar, the sound of ringing bell becomes fainter and fainter. And when all the air is removed from the bell jar, no sound can be heard at all. Thus, when vacuum is created in the bell jar, then the sound of ringing bell placed inside it cannot be heard.

This shows that sound cannot travel through vacuum.


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Question 94 Marks
Study the speed-time graph of a car given alongside and answer the following questions:

  1. What type of motion is represented by OA?
  2. What type of motion is represented by AB?
  3. What type of motion is represented by BC?
  4. What is the acceleration of car from O to A?
  5. What is the acceleration of car from A to B?
  6. What is the retardation of car from B to C?
Answer
  1. OA represents uniform acceleration.
  2. AB represents constant speed.
  3. BC represents uniform retardation.
  4. Acceleration of car from O to A = slope of line OA

$\text{a}=\frac{(40-0)}{(10-0)}\text{m/s}^2$

$=4\text{m/s}^2$

  1. Acceleration of car from A to B is zero as it has uniform speed.
  1. Retardation of car from B to C = slope of line BC.

$\text{a}=\frac{(40-0)}{(50-30)}\text{m/s}^2$

$=2\text{m/s}^2$

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Question 104 Marks
Using following data, draw time-displacement graph for a moving object:
Time (s)
0
2
4
6
8
10
12
14
16
Displacement (m)
0
2
4
4
4
6
4
2
0
 Use this graph to find average velocity for first 4s, for next 4s and for last 6s.
Answer
Therefore, displacement-time graph is shown in figure.

Average velocity for first $4\text{s}=\frac{\text{Change in displacement}}{\text{Total time takan}}$

$\text{v}=\frac{4-0}{4-0}=\frac{4\text{m}}{4\text{s}}=1\text{ms}^{-1}$

Average velocity for next 4s (i.e., in the interval of 4s to 8s), $\text{v}=\frac{4-4}{8-4}=\frac{0}{4}=0$

Average velocity for last $6\text{s}=\frac{(0-6)\text{m}}{(16-10)\text{s}}=\Big(\frac{-6}{6}\Big)=1\text{ms}^{-1}$

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Question 114 Marks
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Answer
From second equation of motion,

Distance travelled in t see $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

Distance travelled in 4s $\text{s}_4=\text{u}\times4+\frac{1}{2}\text{a}(4)^2$

$=4\text{u}+\frac{1}{2}\times\text{a}\times16=4\text{u}+8\text{a}$ ($\text{s}_4=$ distance travelled in 4th sec)

Again, distance travelled in 5s $\text{s}_5=\text{ut}+\frac{1}{2}\text{at}^2$

$=\text{u}\times5+\frac{1}{2}\text{a}(5)^2=5\text{u}+\frac{25}{2}\text{a}$ ($\text{s}_5=$ distance travelled in 5th sec)

So, distance travelled in the interval between 4th and 5th second.

$\text{s}=\text{s}_5-\text{s}_4=\Big(5\text{u}+\frac{25}{2}\text{a}\Big)-4\text{u}+8\text{a}$

$=5\text{u}+\frac{25}{2}\text{a}-4\text{u}-8\text{a}$

$=5\text{u}-4\text{u}\frac{25}{2}\text{a}=8\text{a}$

$=\text{u}+\frac{25\text{a}-16\text{a}}{2}=\text{u}+\frac{9}{2}\text{a}$

so, the relation will be $(\text{u}+\frac{9}{2}\text{a})$

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Question 124 Marks
A body starting from rest travels with uniform acceleration. If it travels 100m in 5s, what is the value of acceleration?
Answer
Initial velocity, u = 0m/s

Time, t = 5s

Distance, s = 100m

Acceleration, a = ?

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

$100=0\times5+\frac{1}{2}\times\text{a}\times5\times5$

$100=0+\frac{25\text{a}}{2}$

$\text{a}=\frac{200}{25}=8\text{m/s}^2$

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Question 134 Marks
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th second.
Answer
Using the equation of motion $\text{s}=\text{ut}+\frac{1}{2}\text{at}^{2}$

Distance travelled in 5 seconds, $\text{s}=\text{u}\times5+\frac{1}{2}\text{a}\times5^{2}$

or $\text{s}=5\text{u}+\frac{25}{2}\text{a}$

Similary, distance travelled in 4seconds, $\text{s}=4\text{u}+\frac{16}{2}\text{a}$

Distance travelled in the interval between 4th and 5th seconds

$=(\text{s-s'})=(\text{u}+\frac{9}{2}\text{a})\text{m}$

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Question 144 Marks
A car is travelling along the road at 8ms-1. It accelerates at 1ms-2 for a distance of 18m. How fast is it then travelling?
Answer
We have to find the final velocity of the moving object. And we have the following information:

Initial velocity, (u) = 8m/s

Acceleration, (a) = 1m/s2

Distance, (s) = 18m

So applying 3rd equation of motion to calculate the final velocity,

$\text{v}=\sqrt{\text{u}^2+2\text{as}}$

Where,

(a) - Acceleration

(ν) - Final velocity

(u) - Initial velocity

(s) - Distance

Put the values in the above equation to get the value of final velocity,

$\text{v}=\sqrt{64+36}\text{m/s}$

$=\sqrt{100}\text{m/s}$

$=10\text{m/s}$

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Question 154 Marks
A car is travelling at 20m/s along a road. A child runs out into the road 50m ahead and the car driver steps on the brake pedal. What must the car's deceleration be if the car is to stop just before it reaches the child?
Answer
We have to find the deceleration. We have the following information given,

Initial velocity, (u) = 20m/s

Final velocity, (v) = 0m/s

Distance travelled, (s) = 50m

Let the deceleration for the entire journey be (a)

We can calculate acceleration by using the 3rd equation of motion,

$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$

Where,

(s) - Displacement

(u) - Initial velocity

(a) - Acceleration

(v) - Final velocity

Put the values in above equation to find the deceleration,

$\text{a}=\Big[\frac{0-400}{2(50)}\Big]\text{m/s}^2$

$=\Big(-\frac{400}{100}\Big)\text{m/s}^2$

$=-4\text{m/s}^2$

Hence, deceleration is 4m/s2.

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Question 164 Marks
Show by means of graphical method that:
v = u + at
where the symbols have their usual meanings.
Answer
Consider the velocity-time graph of a body shown in figure.

The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.

Now, Initial velocity of the body, u = OA ...(1)

And, Final velocity of the body, v = BC ...(2)

But from the graph BC = BD + DC

Therefore, v = BD + DC ...(3)

Again DC = OA

So, v = BD + OA

Now, from equation (1), OA = u

So, v = BD + u ...(4)

We should find out the value of BD now. We know the slope of a velocity-time graph is equal to the acceleration, a.

Thus, Acceleration, a = slope of line AB

or $\text{a} =\frac{\text{BD}}{\text{AD}}$

But AD = OC = t, so putting t in place of AD in the above relation, we get:

$\text{a}=\frac{\text{BD}}{\text{t}}$

Now, putting this value of BD in equation (4), we get:

v = u + at

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Question 174 Marks
Derive the following equation of motion by the graphical method:
v2 = u2 + 2as
where the symbols have their usual meanings.
Answer

Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration a from A to B, and after time t its final velocity becomes v which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.

The distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

Distance travelled, s = Area of trapezium OABC

$\text{s}=\frac{(\text{Sum of parallel sides})\times\text{Height}}{2}$

$\text{s}=\frac{(\text{OA+CB})\times\text{OC}}{2}$

Now, OA + CB = u + v and OC = t

Putting these values in the above relation, we get:

$\text{s}=\Big(\frac{\text{u+v}}{2}\Big)\times\text{t}\ ...(1)$

Eliminate from the above equation.

This can be done by obtaining the value of t from the first equation of motion.

Thus, v = u + at (first equation of motion)

And, at = v.

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Question 184 Marks
A train travels the first 15km at a uniform speed of 30 km/h; the next 75km at a uniform speed of 50km/h; and the last 10km at a uniform speed of 20km/h.
Calculate the average speed for the entire train journey.
Answer
  1. In the first part, train travels at a speed of 30 km/h for distance of 15km.

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

$\text{t}_1=\frac{15}{30}=\frac{1}{2}\text{h}$

  1. In the second part, train travels at a speed of 50km/h for a distance of 75km.

$\text{t}_2=\frac{75}{50}=\frac{3}{2}\text{h}$

  1. In the third part, train travels at a speed of 20km/h for a distance of 10km.

$\text{t}_3=\frac{10}{20}=\frac{1}{2}\text{h}$

Total distance covered = 15 + 75 + 10 = 100km

Average speed $=\frac{\text{Total distance covered}}{\text{Total time taken}}$

$=\frac{100}{\frac{5}{2}}=40\text{km/h}$

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Question 194 Marks
Define motion. What do you understand by the terms ‘uniform motion’ and ‘non-uniform motion’? Explain with examples.
Answer
A body is said to be in motion when its position changes continuously with respect to a stationary point taken as the reference point.

  • Uniform motion: A body is said to be in uniform motion if it travels equal distances in equal intervals of time in a particular direction, no matter how small these time intervals are.

For example: A car running at a constant speed of 10m/s towards east will cover the equal distance of 10m every second towards east, so its motion will be uniform.

  • Non-uniform motion: A body is said to be in non-uniform motion if it travels unequal distances in equal intervals of time.

For example: Motion of a freely falling ball from the roof of a tall building.

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Question 204 Marks

The graph given alogside shows the positions of a body at different times. Calculate the speed of the body as it moves from:

  1. A to B.
  2. B to C.
  3. C to D.
Answer
  1. The distance covered from A to B,

= 3 - 0

= 3cm

Time taken to cover the distance from A to B

= 5 - 2

= 3s

Hence speed,

$=\frac{\text{Distance}}{\text{Time}}$

$=\frac{3}{3}\text{cm/s}$

$=1\text{cm/s}$

  1. The speed of the body as it moves from B to C is zero because the distance travelled is zero.
  2. The distance covered from C to D,

= 7 - 3

= 4cm

Time taken to cover the distance from C to D,

= 9 - 7

= 2s

Hence speed,

$=\frac{\text{Distance}}{\text{Time}}$

$=\frac{4}{2}\text{cm/s}$

$=21\text{cm/s}$

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Question 214 Marks
An object is dropped from rest at a height of 150m and simultaneously another object is dropped from rest at a height 100m. What is the difference in their heights after 2s if both the objects drop with same accelerations? How does the difference in heights vary with time?
Answer
For first object given, u = 0 (because object dropped from rest) and time (t) = 2s

from second equation of motion, the distance covered by first object in 2s is

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2$

$\text{h}=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^{2}]$

$\text{h}=0+\frac{1}{2}\times10\times4=20\text{m}$

Height of first object from the ground after 2s (h1) = 150m - 20m = 130m for second object guven, u = 0 and time (t) = 2s

From second equation of motion, the distance covered by second object in 2s is 

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^2]$

$=0+\frac{1}{2}\times1=\times4=20\text{m}$

Height of second object from the ground after 2s then h2 = 100m – 20m = 80m Now, difference in the height after 2s = h1 – h2 = 130 – 80 = 50 m

The difference in hights of the objects will remain same with time as both the objects have been dropped from rest and are falling with same acceleration i.e (acceleration due to gravity).

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Question 224 Marks
How will the equations of motion for an object moving with a uniform velocity change?
Answer
We know that, the equations of uniformly accelerated motion are

  1. $\text{v}=\text{u}+\text{at}$

  2. $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

  3. $\text{v}^2=\text{u}^2+2\text{as}$

Where,  u = Lnitial velocity

v = Final velocuty

a = Acceleration

t = Time

s = Distance

For an object moving with uniform velocity (velocity which is not changing with time), then acceleration a = 0.

So, equations of motion will become (putting a = 0 in above equations)

  1. v = u
  2. s = ut
  3. v2 = u2
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Question 234 Marks
A car travels 100km at a speed of 60km/h and returns with a speed of 40km/h. Calculate the average speed for the whole journey.
Answer
In the first case, car travels at a speed of 60km/h for a distance of 100km.

In the second case, car travels at a speed of 40km/h for a distance of 100km.

Total distance travelled = 200km.

In the first case, car travels at a speed of 60km/h for a distance of 100km.

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$

$\text{t}_1=\frac{100}{60}\text{h}$

In the second case, car travels at a speed of 40km/h for a distance of 100km.

$\text{t}_2=\frac{100}{40}\text{h}$

Total distance travelled = 200km

Total time taken $=\frac{100}{60}+\frac{100}{40}$

Average speed $=\frac{\text{Total distance travelled}}{\text{Total time taken}}$

$=\frac{200}{\frac{100}{60}+\frac{100}{40}}=\frac{2}{\frac{1}{60}+\frac{1}{40}}$

$=\frac{240}{5}=48\text{km/h}$

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Question 244 Marks
The distance between Delhi and Agra is 200km. A train travels the first 100km at a speed of 50km/h. How fast must the train travel the next 100km, so as to average 70km/h for the whole journey?
Answer
Total distancce = 200km

Average speed = 70km/h

Total time taken $=\frac{\text{Total distance}}{\text{Average speed}}=\frac{200}{70}=\frac{20}{7}\text{h}$

For first part of the journey,

Distance = 100km

Speed = 50km/h

Time taken, $\text{t}_1=\frac{100}{50}=2\text{h}$

Speed $=\text{x}\ \text{km/h}$

Time taken, $\text{t}_2=\frac{100}{\text{x}}=\text{h}$

$\text{t}_1+\text{t}_2=\frac{20}{7}$

$2+\frac{100}{\text{x}}=\frac{20}{7}$

$\frac{100}{\text{x}}=\frac{6}{7}$

$700=6\text{x}$

$\Rightarrow\text{x}=116.6\text{km/h}$

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Que-Ans (Each of 4 Mark ) - Science STD 9 Questions - Vidyadip