Goyal Brothers Prakashan Class 9 ICSE Physics Solutions

A New Approach to ICSE Physics Part 1 Class 9 Solutions Motion in One Dimension

UNIT 1

EXERCISE 1

Question 1.
What do you understand by the terms

1. 1. rest

1. motion? Support your answer by giving two examples each

Answer:
(i) Rest : A body is said to be at rest if it does not change its position with respect to its immediate surroundings.
For example :

(a) A chair lying in a room.
(b) A stone lying on the ground.

(ii) Motion : A body is said to be in motion if it changes its position with respect to its immediate surroundings.
For example :

(a) A car running on the road.
(b) A football moving on the ground.

Question 2.
By giving an example, prove that rest and motion are relative terms.
Answer:
Rest and motion are relative terms. It means a body at rest in one situation at certain time can be in motion in another situation at the same time.
A person sitting in the compartment of a moving train is in the state of rest with respect to the surrounding of the compartment yet he is in state of motion, if he compare himself with the surroundings outside the compartment.

Question 3.
Define :

1. Scalar quantities.
2. Vector quantities. Give two differences between scalar and vector quantities.

Answer :
(i) Scalar quantities : The physical quantities which are expressed in magnitude only are called scalar quantities.
For example: Mass, length, time, distance, density, energy etc.
(ii) Vector quantities : The physical quantities which are ‘ expressed in magnitude as well as direction are called vector quantities.
For example : Displacement, velocity, acceleration, momentum, force etc.
Differences between scalar and vector quantities.
SCALARS :

1. Scalars are specified by one quantity only i.e. magnitude.
2. Scalars change by change in magnitude along.
3. Scalars are written or represented by ordinary letters.
4. Scalars are added by just algebraic addition or subtraction.
5. Mass, length, time, speed, etc. are some example of scalars.

VECTORS :

1. Vectors are specified by two quantities (i) magnitude and (ii) direction.
2. Vectors change when there is change in either magnitude or direction or both.
3. Vectors are written (shown) in bold face letters or letters having arrows heads on them.
4. Vectors are added or subtracted by using triangle law, parallelogram law or polygon law.
5. Displacement, velocity acceleration are some examples of vectors.

Question 4.
Pick out the scalar and vector quantities from the following list:

1. mass
2. density
3. displacement
4. distance
5. momentum
6. acceleration
7. temperature
8. time

Answer:

1. Scalar quantities : Mass, density, distance, temperature, time.
2. Vector quantities : Displacement, momentum, acceleration.

Question 5.
Define :
(i) Speed (ii) Velocity. Give two differences between speed and velocity.
Answer:
(i) Speed : The distance covered by a body in a unit time is called its speed. It is also defined as the rate of change position of a body in any direction.
speed = distance / time
(ii) Velocity : “Rate of change of displacement with time” is called velocity, or “The time rate of change of displacement of an object” is called the velocity.
V = distance/time
Differences between speed and velocity
Speed :

1. The rate of change of position of a body in any direction is known as its speed.
2. It is a scalar quantity.
3. It can be positive or zero.

Velocity :

1. The rate of change of position of a body in a particular direction is known velocity.
2. It is a vector quantity.
3. It can be positive, negative or zero.

Question 6.
Define (i) Distance (ii) Displacement. Give two differences between displacement and distance.
Answer:
(i) Distance : The length of path travelled by a body in certain interval of time is called distance.
(ii) Displacement : of an object between two points is the shortest distance between these two points.
“It is the unique path which can take the body from its initial to final position.”
Differences between displacement and distance.
Distance :

1. It is a scalar quantity.
2. Distance travelled is always positive.
3. The distance travelled by a moving body is the actual length of path.
4. Distance travelled is always greater than or equal to the displacement.
5. 

Displacement :

1. It is a vector quantity.
2. Displacement may be positive negative or zero.
3. The displacement of a body is the shortest distance between the initial and final positions of the body.
4. Displacement is always less than or equal to the distance travelled.
5. 

Question 7.
By giving one example each, define

1. variable velocity.
2. average velocity.
3. uniform velocity.

Answer:
(i) Variable velocity : When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity.
Example : A rotating fan at a constant speed has variable velocity, because of continuous change in direction.
(ii) Average velocity: The ratio of the total distance travelled in a specified direction to the total time taken by the body to travel the distance is called average velocity.
Example : If you walk to a campsite 1 km away and then back to your starting point with in 1 hour, then your average velocity will be zero because your initial and final position is same.
(iii) Uniform velocity : When a body covers equal distances in equal intervals of time (however small may be the time interval) in a specified direction, the body is laid to be moving with uniform velocity.
Example : A car moving on a straight road with constant speed has uniform velocity.

Question 8.
What do you understand by the term acceleration? When is the acceleration

1. positive
2. negative?

Answer:

• Acceleration : “Rate of change of velocity with respect to time” is called acceleration. S.I. unit of acceleration is m s-2.
• Positive acceleration : When velocity of a body increases with time, then acceleration of the body is said to be positive acceleration.
• Negative acceleration : When the velocity of a body decreases with time, then acceleration of the body is said to be negative acceleration.

Question 9.
Define the term acceleration due to gravity. State its value in C.GS. as well as in S.I. system. When is acceleration due to gravity

1. positive
2. negative?

Answer:
Acceleration due to gravity : The acceleration of a freely falling body, under the action of gravity of earth, is called acceleration due to gravity. It is represented by ‘g’.
In S.I. system, g = 9.8 ms^-2 ; In C.GS. system, g = 980 cms^-2 If a body falls towards the earth, then value of acceleration due to gravity is positive.
If the body rises vertically upwards, then value of acceleration due to gravity is negative.

Question 10.
Give an example of a body which covers a certain distance, but its displacement is zero.
Answer:
Yes, displacement of a body can be zero even if the distance covered by it is not zero. For example, if a body moves in a circle, then displacement of the body in one rotation is zero but the distance covered by it in one roation = 2πr, where r is the radius of the circle in which the body is moving.

Question 11.
Give an example of an accelerated body, moving with a uniform speed.
Answer:
When a body moves over a circular path with constant speed, then body is said to be accelerated. Acceleration is produced in the body due to continuous change in its direction.

Question 12.
What is the relation between distance and time when

1. body is moving with uniform velocity
2. body is moving with variable velocity?

Answer:

1. When body moves with uniform velocity, then body covers equal distance in equal interval of time in a specified direction.
2. When a body moves with variable velocity, then body covers unequal distance in equal interval of time in a specified direction.

Question 13.
(a) Distinguish between scalar and vector quantities,
(b) State whether following are scalar or vector quantities

1. speed
2. force
3. acceleration
4. energy

Answer:
(a) See Q. No. 3 of Exercise-1
(b)

1. Speed is a scalar quantity because it has magnitude and no direction.
2. Force is a vector quantity because it has both magnitude as well as direction.
3. Acceleration is a vector quantity.
4. Energy is a scalar quantity because it has magnitude and no direction.

Question 14.
Copy the following table and fill in the blank spaces.

Answer:

Question 15.
Draw a diagram to show the motion of a body whose speed remains constant, but velocity changes continuously.
Answer:
A body moving in a circle with constant speed but variable velocity as the velocity of the body at any point is along the tangent to the circle at that point as shown in the figure.

Practice Problems 1

Question 1.
A car covers 90 km in 1 1/2 hours towards east. Calculate
(i) displacement of car,
(ii) its velocity in (a) kmh^-1
(b) ms^-1.
Answer:

Question 2.
A race horse runs straight towards north and covers 540 m in one minute. Calculate (i) displacement of horse, (ii) its velocity in (a) ms^-1 (b) kmh^-1.
Answer:
A race horse runs straight towards north and covers 540 m in one minute.

1. Displacement = 540 m – north
2. Time = 1 minute = 60 s
   
   

Practice Problems 2

Question 1.
The change in velocity of a motor bike is 54 kmlr1 in one minute. Calculate its acceleration in (a) ms^-2 (b) kmh^-2.
Answer:

Question 2.
A speeding car changes its velocity from 108 kmh^-1 to 36 kmh^-1 in 4 s. Calculate its deceleration in

1. ms^-2
2. kmh^-2.

Answer:

UNIT II
EXERCISE 2

(A) Objective Questions

Multiple choice Questions. Select the correct option.

Question 1.
A graph is a straight line parallel to the time axis in a distance – time graph. From the graph, it implies :
(a) body is stationary
(b) body is moving with a uniform speed
(c) body is moving with a variable speed
(d) none of these
Answer:
(a) body is stationary

Question 2.
The slope of a displacement – time graph represents :
(a) uniform speed
(b) non-uniform speed
(c) uniform velocity
(d) uniform a acceleration
Answer:
(c) uniform velocity

Question 3.
A body dropped from the top of a tower reaches the ground in 4s. Height of the tower is
(a) 39.2 m
(b) 44.1 m
(c) 58.8 m
(d) 78.4 m
Answer:
(d) 78.4 m
Explanation :
Initial velocity = u = 0
Time = t = 4s
As body falls downards,
so acceleration = a = + g= a = + 9.8 ms^-2
(Distance) S = ut + 1/2 at
S = (0) (4) + 1/2 × 9.8 × (4)³
s = 0 + 1/2 × 9.8 × 16
S = 9.8 × 8 = 78.4 m

Question 4.
The speed of a car reduces from 15 m/s to 5 m/s over a displacement of 10 m.
The uniform acceleration of the car is :
(a) -10 m/s²
(b) +10 m/s²
(c) 2 m/s²
(d) 0.5 m/s²
Answer:
(a) -10 m/s²
Explanation :
Initial velocity = u = 15 m/s
Final velocity = v = 5 m/s
Displacement = S = 10 m
Acceleration = a = ?
v² – u² = 2aS ; (5)² – (15)² = 2a (10)
25 – 225 = 20a
a = -220/20 = -10 ms-²

Question 5.
A body projected vertically up with a velocity 10 m/s reaches a height of 20 m. If it is projected with a velocity of 20 m/s, then the maximum height reached by the body is :
(a) 20 m
(b) 10 m
(c) 80 m
(d) 40 m
Answer:
(b) 10 m
Explanation :

Question 6.
What does the area of an acceleration – time graph represent?
(a) Uniform velocity
(b) Displacement
(c) Distance
(d) Variable velocity
Answer:
(a) Uniform velocity

Question 7.
A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km/h to 18 km/h in 5 s. What is the distance travelled by the train during this interval of time?
(a) 52 m
(b) 50 m
(c) 25 m
(d) 80 m
Answer:
(b) 50 m

Question 8.
In velocity time graph, the acceleration is
(a) – 4 m/s²
(b) 4 m/s²
(c) 10 m/s²
(d) zero

Answer:
(a) -4 m/s²

Question 9.
The distance covered in adjoining velocity – time graph is :
(a) 25 m
(b) 40 m
(c) 50 m
(d) 45 m

Answer:
(c) 50 m
Explanation :
Distance covered = Area under velocity – time graph
=1/2 × base × height
a= 1/2 × 5 × 20 = 50 m

Question 10.
At the maximum height, a body thrown vertically upwards has :
(a) velocity not zero but acceleration zero.
(b) acceleration not zero but velocity zero.
(c) both acceleration and velocity are zero.
(d) both acceleration and velocity are not zero.
Answer:
(b) acceleration not zero but velocity zero.

(B) Subjective Questions

Question 1.
Draw displacement – time graphs f.or the following situations :

1. When a body is stationary.
2. When a body is moving with uniform velocity.
3. When a body is moving with variable velocity.

Answer:
(i) Displacement – time graph when body is stationary.

Question 2.
Draw velocity – time graphs for the following situations :

1. When a body, is moving with uniform velocity.
2. When a body is moving with variable velocity, but uniform acceleration.
3. When a body is moving with variable velocity, but uniform retardation.
4. When a body is moving with variable velocity and variable acceleration.

Answer:
Following are the velocity – time graph :
(i) When a body is moving with uniform velocity : Then velocity time graph is AB is a straight line parallel to time axis.

Question 3.
How can you find the following?

1. Velocity from a displacement – time graph.
2. Acceleration from velocity – time graph.
3. Displacement from velocity – time graph.
4. Velocity from acceleration – time graph.

Answer:
(i) Velocity from a displacement – time graph : Displacement time graph for uniform motion is a straight line (OP) inclined to time axis. Take any two points A and B on this graph OP. From A and B, draw perpendiculars on time axis as well as displacement axis.
So, by knowing the slope of displacement – time graph. We can find the velocity of the body.

(ii) Acceleration from velocity – time graph : Velocity – time graph for uniform motion is a straight line (OP) inclined to time axis.

Take any two points A and B on this graph. From A and B draw perpendicular on time axis as well as velocity axis in such a way that

So, by knowing the slope of velocity -time graph for uniform motion, we can find the acceleration of the body.
(iii) Displacement from velocity-time graph: Displacement covered by a body is equal to the area under velocity – time graph. When object moves with a uniform velocity, then velocity – time graph is a straight line (PQ) parallel to time axis.

(iv) Velocity from acceleration – time graph : Area under the acceleration – time graph gives the velocity of the body. When the body moves with variable velocity but uniform acceleration, then acceleration – time graph is a straight line (PQ) parallel to time axis.

If initial velocity of the body = u = 0 Then area under acceleration – time graph = v – 0 = v = velocity of the body.

Question 4.
What do you understand by the term acceleration due to gravity? What is its value in C.G.S. and S.I. systems?
Answer:
Acceleration due to gravity : The acceleration of a freely falling body, under the action of gravity of earth, is called acceleration due to gravity. It is represented by ‘g’.
In S.I. system, g = 9.8 ms^-2; In C.G.S. system, g = 980 cms^-2
If a body falls towards the earth, then value of acceleration due to gravity is positive.
If the body rises vertically upwards, then value of acceleration due to gravity is negative.

Question 5.
Can you suggest about the kind of motion of a body from the following distance – time graphs?

Answer:
(a) Distance – time graph in figure (a) shows that body is stationary.
(b) Distance – time graph figure (b) shows that body is in uniform motion.
(c) Distance – time graph in figure (c) shows that initially body is in uniform motion i.e. it covers equal distance in equal intervals of time.
From O to A, body is in uniform motion and from A to B, body is at rest.

Question 6.
Can you suggest real life examples about the motion of a body from the following velocity – time graphs?

Answer:
(a) A car running on a road at constant velocity. Then v-f graph will be same as shown by (a).
(b) Figure (b) shows that car is moving with uniform acceleration. i.e. when velocity of car increases equally in equal intervals of time.
(c) Suppose a train starts from rest and accelerates uniformly for some time and then moves with a constant velocity for certain time interval. Then on applying the brakes, train is uniformly retarded and comes to rest after sometime in this situation, velocity – time graph of the train will be same as showin in figure (c).
(d) When you come home at activa from market and stops in front of your home. Then velocity of active decreases with time and its v-f graph will be same as shown on figure (d).
(e) A ball is thrown vertically upward and it returns back to earth after sometime. In this situation, graph will same as that shown in figure (e).

Question 7.
Diagram shows a velocity – time graph for a car starting from rest. The graph has three section AB, BC and CD.

1. From a study of this graph, state how the distance travelled in any section is determined.
2. Compare the distance travelled in section BC with the distance travelled in section AB.
3. In which section, car has a zero acceleration?
4. Is the magnitude of acceleration higher or lower than that of retardation ? Give a reason.

Answer:

Question 8.
Write down the type of motion of a body along the A – O – B in each of the following distance – time graphs.

Answer:

1. Body is stationary.
2. Body is in uniform motion i.e. it covers equal distance in equal intervals of time.
3. From A to O, body is in uniform motion having positive slope and from O to B, body is in uniform motion having negative slope.

Practice Problems 1

Question 1.
From the displacement – time graph shown given below calculate : –

1. Average velocity in first three seconds.
2. Displacement from initial position at the end of 13 s.
3. Time after which the body is at the initial position,
4. Average velocity after 8 s.
   

Answer:
(i) In first three seconds
Total displacement = 8m
Total time = 3 s
Average velocity = Total displacement / Totaltime = 8/3
= 2.67 ms^-1
(ii) Displacement from initial position at the end of 13s=-8m.
(iii) Body is at the initial position after 8s and 17s.
(iv) Average velocity after 8s is zero because after 8s, displacement is zero.

Question 2.
From the displacement – time graph shown given below calculate :

1. Velocity between 0 – 2 s.
2. Velocity between 8 s – 12 s.
3. Average velocity between 5 s – 12 s.
   

Answer:

Practice Problems 2

Question 1.
A train starting from rest, picks up a speed of 20 ms-1 in 200 s. It continues to move at the same rate for next 500s, and is then brought to rest in another 100 s.
(i) Plot a speed-time graph.
(ii) From graph calculate
(a) uniform rate of acceleration
(b) uniform rate of retardation
(c) total distance covered before stopping
(d) average speed.
Answer:
(i)

Question 2.
A ball is thrown up vertically, and returns back to thrower in 6 s. Assuming there is no air friction, plot a graph between velocity and time. From the graph calculate

1. deceleration
2. acceleration
3. total distance covered by ball
4. average velocity.

Answer:
A ball is thrown up vertically, and returns to thrower in 6s. It means ball takes 3s to reach the highest point and 3s to reach the earth from highest point.

Question 3.
A racing car is moving with a velocity of 50 m/s. On
applying brakes, it is uniformly retarded and comes to rest in 20 seconds. Calculate its acceleration.
Answer:

Question 4.
A body falls freely downward from a certain height. Show graphically the relation between the distance fallen and square of time. How will you determine ‘g’ from the graph?
Answer:
For a freely falling body, the displacement is directly proportional to the square of time. By knowing the slope of displacement Vs. square of time graph, we can find the acceleration due to gravity.

Question 5.
A body at rest is thrown downward from the top of tower. Draw a distance – time graph of its free fall under gravity during first 3 seconds. Show your table of values starting t = 0 with an interval of 1 second, (g = 10 ms^-2).
Answer:
Initial velocity = M = 0
Acceleration = a = +g = 10 ms^-2
when t = Is, then distance travelled (S,) is given by

Practice Problems 3

Question 1.
From the diagram given below, calculate

1. acceleration
2. deceleration
3. distance covered by body.
   

Answer:

= 5 × 15 + 3 × 15
= 75 + 45 = 120 m

Question 2.
From the velocity – time graph given below, calculate :

1. Acceleration in the region AB.
2. Deceleration in region BC.
3. Distance covered in the region ABCE.
4. Average velocity in region CED.

Answer:

Practice Problem 4

Question 1.
Diagram given below shows velocity – time graphs of car P and Q, starting from same place and in same direction. Calculate :

1. Acceleration of car P.
2. Acceleration of car Q between 2 s – 5 s.
3. At what time intervals both cars have same velocity?
4. Which car is ahead after 10 s and by how much?
   

Answer:

UNIT III

Practice Problems 1

Question 1.
A motor bike, initially at rest, picks up a velocity of 72 kmh^-1 over a distance of 40 m. Calculate

1. acceleration
2. time in which it picks up above velocity.

Answer:
Initial velocity = u = 0

Question 2.
A cyclist driving at 5 mr^-1, picks a velocity of 10 ms^-1, over a distance of 50 m. Calculate

1. acceleration
2. time in which the cyclist picks up above velocity.

Answer:
Initial velocity = u = 5 ms^-1
Final velocity = v = 10 ms^-1
Distance = S =50 m
(i) v² – u = 2aS
(10)² – (5)² = 2a (50)
100a = 100—25=75

Practice Problems 2

Question 1.
An aeroplane lands at 216 kmh^-1 and stops after covering a runway of 2 m. Calculate the acceleration and the time, in which it comes to rest.
Answer:
Initial velocity = u = 216 kmh^-1

Question 2.
A truck running at 90 kmh^-1, is brought to rest over a distance of 25 m. Calculate the retardation and time for which brakes are applied.
Answer:

Practice Problems 3

Question 1.
A racing car, initially at rest, picks up a velocity of 180 kmh^-1 in 4.5 s. Calculate

1. acceleration
2. distance covered by car.

Answer:

Question 2.
A motor bike running at 5 ms^-1, picks up a velocity of 30 ms^-1 in 5s. Calculate

1. acceleration
2. distance covered during acceleration.

Answer:

Practice Problems 4

Question 1.
A motor bike running at 90 kmh^-1 is slowed down to 18 kmh^-1 in 2.5 s. Calculate

1. acceleration
2. distance covered during slow down.

Answer:
Initial velocity of motor bike = u = 90 kmh^-1
= u = 90 x 5/18 ms^-1 = 25 ms^-1
Final velocity of motor bike = v = 18 kmh^-1

Question 2.
A cyclist driving at 36 kmh^-1 stops his motion in 2 s, by the application of brakes. Calculate

1. retardation
2. distance covered during the application of brakes.

Answer:

Practice Problems 5

Question 1.
A motor bike running at 90 kmh^-1, is slowed down to 54 kmh^-1 by the application of brakes, over a distance of 40 m. If the brakes are applied with the same force, calculate

1. total time in which bike comes to rest
2. total distance travelled by bike.

Answer:

Total distance covered from A to C = 40 + 22.5 = 62.5
Total time taken from A to C = t₁ + t₂ + = + 3 = 5s

Question 2.
A motor car slows down from 72 kmh^-1 to 36 kmh^-1 over at distance of 25 m. If the brakes are applied with the same force
calculate

1. total time in which car comes to rest
2. distance travelled by it.

Answer:

Practice Problems 6

Question 1.
A packet is dropped from a stationary helicopter, hovering at a height ‘h’ from ground level, reaches ground in 12s. Calculate

1. value of  h
2. final velocity of packet on reaching ground. (Take g = 9.8 ms 2)

Answer:

Question 2.
A boy drops a stone from a cliff, reaches the ground in 8 seconds. Calculate

1. final velocity of stone
2. height of cliff. (Take g = 9.8 ms’2)

Answer:
Initial velocity = u = 0
Time = t = 8s
Final velocity = v = ?
Height of cliff = h = ?
A cceleration = a = + g = + 9.8 ms^-1

Practice Problems 7

Question 1.
A stone thrown vertically upwards, takes 3 s to attain maximum height. Calcualte

1. initial velocity of the stone
2. maximum height attained by the stone. (Take g = 9.8 ms^-2)

Answer:

Question 2.
A stone thrown vertically upwards, takes 4 s to return to thrower. Calculate
(i) initial velocity of the stone
(ii) maximum height attained by stone. (Take g = 10 ms^-2)  
Answer:
A stone thrown vertically upwards, takes 4s to return to thrower,

Practice Problems 8

Question 1.
A spaceship is moving in space with a velocity of 50 kms^-1. Its engine fires for 10 s, such that its velocity increases to 60 kms^-1. Calculate the total distance travelled by
spaceship in 1/2 minute, from the time of firing its engine
Answer:
Initial velocity of spaceship = u = 50 kms^-1
u- 50 x 1000 ms^-1
u = 50000 ms^-1
Time = t= 10s
Final velocity of spaceship = v = 60 kms^-1
v = 60 x 1000 ms^-1 = 6000 ms^-1
Acceleration = a = ?

Question 2.
A spaceship is moving in space with a velocity of 60 kms^-1. It fires its retro engines for 20 second and velocity is reduced to 55 kms^-1. Calculate the distance travelled by the spaceship in 40 s, from the time of firing of the retro- rockets.
Answer:
Initial velocity of spaceship = u = 60 kms^-1
Final velocity of spaceship = v = 55 kms^-1
Case -1:
It decelerates for 20 s
⇒ t = 20s
v = u + at
55 = 60 +a (20)
20a = 55 – 60 = -5

A New Approach to ICSE Physics Part 1 Class 9 Solutions Upthrust and Archimedes’ principle.

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Upthrust and Archimedes’ principle.

Practice Problem 1:

Question 1.
A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate :

1. Weight of solid in SI system.
2. Upthrust on solid in SI system.
3. Apparent weight of solid in alcohol.
4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [g = 10 ms^-2]

Answer:
Density of solid = ρ= 2700 kgm³
Volume of solid = V = 0.0015 m³
Density of alcohol = ρ’ = 800 kgm³

1. Mass of solid = m = V x p
   m = 0.0015 x 2700 = 4.05 kg
   Weight of solid = mg = 4.05 x 10 = 40.5 N
2. Volume of alcohol displaced = Volume of solid
   V = 0.0015 m³
   Mass of alcohol displaced = m’ = V x p’
   m’ = 0.0015 x 800 = m’ = 1.2 kg
   Upthrust = Weight of alcohol displaced
   = m’g= 1.2 x 10= 12N
3. Apparent weight of solid in alcohol
   = Actual weight of solid – Upthrust
   = 40.5 -12 = 28.5 N
4. When a solid is immersed completely in brine solution, then upthrust acts on it in upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than actual weight of solid.

Question 2.
A stone of density 3000 kgm³ is lying submerged in water of density 1000 kgm³. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms²]
Answer:
Density of stone =ρ = 3000 kgm³
Density of water = ρ’ = 1000 kgm³
Mass of stone = m = 150 kg
Acceleration due to gravity = g = 10 ms^-2
Volume of stone = \(\mathrm{V}=\frac{m}{\rho}\)
V= \(\frac{150}{3000}=\frac{1}{20}=0.05 \mathrm{m}^{3}\)
Actual weight of stone = mg = 150 x 10 = 1500 N
Volume of water displaced = Volume of stone
V = 0.05 m³
Mass of water displaced = m’ = V x p’
m’ = 0.05 x 1000 = 50 kg
Upthrust = m’g = 50 x 10 = 500 N
Force required to lift the stone
= Actual weight of stone – upthrust
= 1500 -500 = 1000 N

Question 3.
A solid of area of cross-section 0.004 m² and length 0.60 m is completely immersed in water of density 1000 kgm³. Calculate :

1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density 1050 kgm³.
   [Take g = 10 N/kg; Density of solid = 7200 kgm³]

Answer:
Area of cross-section of solid = A = 0.004 m²
Length of the solid = l = 0.60 m
Density of water = p’ = 1000 kgm³
Acceleration due to gravity = g = 10 ms^-2
Density of solid = p = 7200 kgm³
(1) Volume of solid = V = A x l
V = 0.004 x 0.60 = 0.0024 m³
Mass of solid = m = V x p
m = 0.0024 x 7200 = 17.28 kg
Weight of the solid = mg = 17.28 x 10 = 172.8 N

(2) Volume of water displaced = Volume of solid
= V = 0.0024 m³
Mass of water displaced = m’ = V x p’
m’ = 0.0024 x 1000 = 2.4 kg
Upthrust = Weight of water displaced
= m’g = 2.4 x 10 = 24 N

(3) Apparent weight of solid=Actual weight of solid – upthrust
= 172.8-24= 148.8 N

(4) Density of brine solution =ρ_b= 1050 kgm³
Volume of brine solution displaced = Volume of solid = V
V = 0.0024 m³
Mass of brine solution displaced
= m_b = V x ρ_b = 0.0024 x 1050
m_b = 2.52 kg
Upthrust acting on solid in brine solution = Weight of brine solution displaced -m_bg
= 2.52 x 10 = 25.2 N
Apparent weight of solid in brine solution
= Actual weight – Upthurst
= 172.8-25.2= 147.6 N

Practice Problem 2:

Question 1.
A solid of density 7600 kgm³ is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm³, find the apparent weight of solid in liquid.
Answer:
Weight of solid in air = 0.950 kgf
∴ Mass of solid in air = m = 0.950 kg
Density of solid =ρ = 7600 kgm³

Apparent weight of solid in liquid
= Actual weight – Upthrust
= 0.950-0.09 = 0.860 kgf

Question 2.
A glass cylinder of length 12 x 10^-2 m and area of cross­section 5 x 10^-4 m² has a density of 2500 kgm^-3. It is immersed in a liquid of density 1500 kgm^-3, such that 3/8. of its length is above liquid. Find the apparent weight of glass cylinder in newtons.
Answer:
Length of glass cylinder = l = 12 x 10^-2 m
Area of cross-section = A = 5 x 10^-4 m²
Volume of glass cylinder = V = A x l
V= 5 x 10^-4 x 12 x 10-²
V= 0.00006 m³
Acceleration due to gravity = g = 9.8 m/s²
Density of glass cylinder = ρ = 2500 kgm³
Mass of glass cylinder = m = V x ρ

m = 0.00006 x 2500 m
= 0.15 kg
Weight of glass cylinder = mg = 0.15 x 10=1.5 N
Mass of liquid displaced by glass cylinder = V’ x ρ’
m’= 0.0000375 x 1500
m’ = 0.05625 kg
Upthrust = Weight of liquid displaced by the glass cylinder
= m’g = 0.05625 x 10 = 0.5625 N
Apparent weight of glass cylinder in liquid = Actual weight of glass cylinder – Upthrust
= 1.5 – 0.5625 = 0.9375 N

Practice Problems 3:

Question 1.
A solid weighs 0.08 kgf in air and 0.065 kgf in water.Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water = 1000 kgm³]
Answer:
Weight of solid in air = 0.08 kgf
Weight of solid in water = 0.065 kgf
Density of water = 1000 kgm³
(1) Relative density (RD) of solid
= Weight of solid in air
wt. of solid in air-wt. of solid in water

Question 2.
A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
Answer:
Relative density of solid = R.D. = 2.5
Weight of solid in water = W’ = 0.120 kgf
Weight of solid in air = W = ?

Question 3.
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
Answer:
Relative density of solid = R.D. = 4.2
Weight of solid in air = W = 0.200 kgf

Practice Problems 4:

Question 1.
A sinker is found to weigh 56.7 gf in water. When the sinker is tied to a cork of weight 6 gf, the combination is found to weigh 40.5 gf in water. Calculate R.D. of cork.
Answer:
Weight of sinker in water = 56.7 gf
Weight of cork = 6 gf
Wt. of sinker in water + Wt. of cork in air
= 56.7 + 6 = 62.7 gf                                                 …(1)
Wt. of cork in water + Wt. of sinker in water = 40.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of cork in air – Wt. of cork in water = 62.7 – 40.5 = 22.2 gf

Question 2.
A solid lighter than water is found to weigh 7.5 gf in air. When tied to a sinker the combination is found to weigh .If the sinker alone weighs 72.5 gf in water, find R.D. of solid.
Answer:
Weight of solid in air = 7.5 gf
Weight of sinker in water = 72.5 gf
Wt. of sinker in water + Wt. of solid in air
= 72.5 + 7.5 = 80.0 gf                                              …(1)
Wt. of solid in water + Wt. of sinker in water = 62.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of solid in air – Wt. of solid in water = 80- 62.5 = 17.5 gf.

Practice Problems 5:

Question 1.
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.
Answer:
Side of an aluminium cube = l = 5 cm
Volume of aluminium cube = V = Z³ = (5)³ =125 cm³
Relative density of aluminium = R.D. = 2.7
Relative density of alcohol = R.D. = 0.80
Density of water = 1 g cm^-3

Density of aluminium =ρ = 2.7 g cm^-3
Mass of aluminium = V x ρ
m = 125 x 2.7 = 337.5 g
Wt. of aluminium cube acting downwards = 337.5 gf
Volume of alcohol displaced = Volume of cube = V = 125 cm³
Upthrust due to alcohol = V x ρ_alcohol x g

Question 2.
A cube of lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.
Answer:
Length of side of cube = l = 8 cm
Volume of cube =β= (8)³ = 512 cm³
V = 512 cm³
Relative density of lead cube = R.D. = 10.6 Relative density of sugar solution = R.D. 1.4
Density of water = 1 g cm^-3

 Practice Problems 1:

Question 1.
A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm², floats in water with 3/5 of its length inside water. Calculate :
(1) apparent density of hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.
Answer:
(1) Length of hollow cylinder of copper = h_cu= l= 25 m
Length of hollow cylinder of copper inside water

Question 2.
A cork cut in the form of a cylinder floats in alcohol of density 0.8 gcm^-3, such that 3/4 of its length is outside alcohol. If the total length of cylinder is 35 cm and area of cross-section 25 cm², calculate :
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol
Answer:

Practice Problems 2:

Question 1.
A cylinder made of copper and aluminium floats in mercury of density 13.6 gem^-3, such that 0.26th part of it is below mercury. Find the density of solid.
Answer:
Density of mercury = ρ_Hg =13.6 g.cm^-3
Density of solid cylinder = ρ_solid = ?
0.26th part of the cylinder is below mercury
Let V_solid = Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder

Question 2.
An iceberg floats in sea water of density 1.17 g cm ³, such that 2/9 of its volume is above sea water. Find the density of iceberg.
Answer:
Density of sea water =ρ_w = 1.17 g cm^-3
Density of solid ice berg =ρ_i = ?

Practice Problems 3:

Question 1.
A wooden block floats in alcohol with 3/8 of its length above alcohol. If it is made to float in water, what fraction of its length is above water? Density of alcohol is 0.80 g cm^-3.
Answer:
Let length of wooden block = x

Question 2.
A hollow metal cylinder of length 10 cm floats in alcohol of density 0.80 g cm^-3, with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density 1.25 g cm^-3?
Answer:
Length of hollow metal cylinder = x= 10 cm
1 cm length of cylinder is above the alcohol
∴Length of the cylinder below alcohol = (10-1) = 9cm
Density of alcohol =ρ_alcohol = 0.80 g cm^-3
Density of copper sulphate solution = ρ_CuS04 = 1.25 g cm^-3
When block floats in alcohol By the law of floatation :

Length of metal block below copper sulphate solution = 5.76 cm
So, length of metal block above copper sulphate solution
= 10 – 5.76 = 4.24 cm

Practice Problems 4:
Question 1.
What fraction of an iceberg of density 910 kgm^-3 will be above the surface of sea water of density 1170 kgm^-3 ?
Ans.
Let volume of iceberg = V_i = x
Volume of iceberg inside sea water = V_w
Density of iceberg = ρ_i = 910 kmg^-3
Density of sea waer = ρ_w = 1170 kgm^-3
By law of floatation:
Weight of icebeig = Weight of sea water displaced by the iceberg

Question 2.
What fraction of metal of density 3400 kgm^-3 will be above the surface of mercury of
density 13600 kgm^-3, while floating in mercury?
Answer:
Density of metal =ρ_m = 3400 kgm^-3
Density of mercury = p_Hg = 13600 kgm^-3
Let volume of metal = x
and volume of metal inside mercury = y
By law of floatation:
Weight of mercury displaced by metal = wt. of metal

Practice Problems 5:

Question 1.
A balloon of volume 1000 m³ is filled with a mixture of hydrogen and helium of density 0.32 kgm^-3. If the fabric of balloon weighs 40 kgf and the density of cold air is 1.32 kgm^-3, find the tension in the tope, which is holding the balloon to ground.
Answer:
Volume of balloon = V = 1000 m³
Density of mixture of hydrogen and helium =ρ = 0.32 kgm³
Density weight of empty balloon = 40 kgf
Density of cold air = p’ = 1.32 kgm³
Volume of balloon=Volume of mixture of hydrogen and helium gas
= Volume of cold air displaced by balloon
= V= 1000 m³
Weight of mixture of hydrogen and helium gas in balloon = Vρg
= 1000 x 0.32 x g = 320 kgf
Down thrust = Weight of empty balloon + Weight of mixture of hydrogen and helium gas
= 40+ 320 = 360 kgf
Upthrust = Weight of cold air displaced by balloon = Vρ’g
= 1000 x 1.32 x g= 1320kgf
Tension in the rope = Upthrust – downthrust
= 1320-360 = 960 kgf

Question 2.
A balloon of volume 800 cm³ is filled with hydrogen gas of density 9 x 10^-5 gem^-3. If the empty balloon weighs 0.3 gf and density of air is 1.3 x 10^-3 gem^-3, calculate the lifting power of balloon.
Answer:
Volume of balloon = V = 800 cm³
Density of hydrogen gas = p_H = 9 x 10^-5 g cm^-3
Weight of empty balloon = 0.3 gf
Density of air = ρ_a = 1.3 x 10^-3 g cm^-3
Weight of hydrogen gas in balloon = Vρ_Hg
= 800 x 9 x 10^-5 x g
= 72 x 10^-3gf
= 0.072 gf
Volume of balloon = Volume of hydrogen gas in the balloon = Volume of air displaced by balloon.
Downthrust = Wt. of empty balloon + wt. of hydrogen gas in balloon
= 0.3 + 0.072 = 0.372 gf
Upthrust = Wt. of air displaced by balloon
= Vρ_ag
= 800 x 1.3 x 1o^-3 x  g
= 8 x 1.3 x 1o^-1 x g
= 10.4 x 10^-1 x g = 1.04 gf
Lifting power of balloon = Upthrust – Down thrust
=1.04-0.372 = 0.668 gf

Question 3.
A balloon of volume 120 m³ is filled with hot air, of density 38 kg^-3. If the fabric of balloon weighs 12 kg, such that an additional equipment of wt. x is attached to it, calculate the magnitude of Density of cold air is 1.30 kgm^-3.
Answer:
Volume of balloon = V = 120 m³
Density of hot air = ρ_{hot air} = 0.38 kgm^-3
Mass of empty balloon = 12 kg
Weight of the empty balloon = 12 kgf
Weight of the additional equipment attached with the balloon
=x kgf
Density of cold air = ρ_coldair = 1.30 Kgm³
Volume of balloon=Volume of hot air inside the balloon=Volume
of cold air displaced by balloon = V = 120 m³ Weight of hot air = Vp_hotair g
= 120 x 0.38 x g = 45.6kgf
Weight of empty balloon + Weight of hot air inside the balloon +
Weight of equipment = Downthrust
12 + 45.6 + x = Downthrust
Downthrust = 57.6 + x
Upthrust = Weight of cold air diplaced by balloon
=Vρ_coldair g
= 120 x 1.30 xg= 156kgf
By law of floatation :
Downthrust = Upthrust
57.6+x= 156
x= 156-57.6
x =98.4 kgf

Practice Problems 6:

Question 1.
A test tube weighing 17 gf, floats in alcohol to the level P. When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.
Answer:
When tube floats in alcohol :
Weight of test tube = 17 gf
By law of floatation :

Weight of alcohol displaced by test tube = Weight of test tube = 17 gf
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3g lead stones are added in it.
.’. Weight of test tube = 17 gf + 3 gf = 20 gf
Weight of water displaced by test tube = 20 gf
Volume of alcohol displaced = Volume of water displaced

Question 2.
A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.
Answer:
When tube floats in water :
Weight of test tube = 150 gf
By law of floatation:
Weight of water displaced = Weight of test tube = 150 gf
When test tube floats in alcohol :
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
∴Weight of test tube in alcohol = 150 – 27 = 123 gf
By law of floatation:
Weight of alcohol displaced by test tube = Weight of test tube in alcohol = 123 gf
As volume of alcohol displaced = Volume of water displaced
∴ R.D. of alcohol

QUESTIONS BASED ON ICSE EXAMINATIONS
(A) Objective Questions

Multiple Choice Questions.
Select the correct option.

1.The force experienced by a body when partially or fully immersed in water is called :
(a) aparent weight
(b) upthrust
(c) down thrust
(d) none of these

2. When a body is floating in a liquid :
(a) The weight of the body is less than the upthrust due to immersed part of the body
(b) The weight of body is more than the upthrust due to the immersed part of the body
(c) The weight of body is equal to the upthrust due to the immersed part of the body
(d) none of the above

3. With the increase in the density of the fluid, the upthrust experienced by a body immersed in it :
(a) decreases
(b) increases
(c) remains same
(d) none of these

4. The apparent weight of a body in a fluid is :
(a) equal to weight of fluid displaced
(b) volume of fluid displaced
(c) difference between its weight in air and weight of fluid displaced
(d) none of the above

5. The phenomenon due to which a solid experiences upward force when immersed in water is called :
(a) floatation
(b)    buoyancy
(c)    density
(d)    none of these

6. When an object sinks in a liquid, its :
(a) buoyant force is more than the weight of object
(b) buoyant force is less than the weight of object
(c) buoyant force is equal to the weight of the object
(d) none of the above

7. The SI unit of density is :
(a)   gem^-3
(b)    kgem^-3
(c)    kgm-³     
(d)   gm-³

8. When a body is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to :
(a) volume of liquid displaced by it
(b) weight of liquid displaced by it
(c) both (a) and (b)
(d) none of the above

9. The ratio between the mass of a substance and the mass of an equal volume of water at 4°C is called :
(a)   relative density  
(b)   density
(c)    weight
(d)   pressure

10. A body has density 9.6 gcm ^-3. Its density in SI system is :
(a) 96 kgm^-3
(b) 960 kgm^-3
(c)  9600 kgm^-3  
(d) 96,000 kgm^-3
Ans:
Explanation :
Density = 9.6 g cm^-3
\(=\frac{9.6}{10^{3}} \times 10^{6} \mathrm{kgm}^{-3}=9600 \mathrm{kgm}^{-3}\)

(B) Subjective Questions

Question 1.
A wooden block floats in water with two third of its volume submerged.
(1) Calculate density of wood.
(2) When the same block is placed in oil, three quarter of its volume is immersed in oil. Calculate the density of oil.
Answer:

Question 2.
A.metal cube of 5cm edge and relative density 9 is suspended by a thread so as to be completely immersed in a liquid of relative density 1.2. Find the tension in the thread.
Answer:
Volume of metal cube = (side)³ = 5³ = 125cm³
Density of cube = 9 g cm^-3
Weight of cube acting downward = mg = v × d
F₁ ↓= 125 x 9 = 1125 gf
Density of liquid d_L– 1.2 g cm^-3
.’. Upthrust due to liquid in the upward direction.
F₂↑ = v d_L= 125 x 1.2 = 150.0 gf Tension in the string = Net downward force = F₁  – F₂
= 1125- 150 = 975 gf= 9.75 N

Question 3.
A weather forecasting plastic balloon of volume 15 m³ contains hydrogen of density 0.09 kgm^-3. The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon is 7.15 . kg. The balloon is floating in air of density 1.3 kgm^-3.
(1) Calculate the mass of hydrogen in balloon.
(2)    Calculate the mass of hydrogen and the balloon.
(3) If the mass of equipment is x kg, write down the total mass of hydrogen, the balloon and the equipment,
(4) Calculate the mass of air displaced by balloon.
(5) Using the law of floatation, calculate the mass of equipment.
Answer:
Volume of Hydrogen V = 15 m³
Density of hydrogen = d= 0.09 kg m^-3
(1)  Mass of hydrogen in balloon = Vd= 15 × 0.09
= 1.35 kg

The mass of empty balloon alone = 7.15 kg
(2) The mass of hydrogen and balloon = 1.35 + 7.15
= 8.50 kg
Mass of equipment = x kg

(3) Total mass of hydrogen + Balloon + Equipment
= (8.50 + :c) kg
Density of air = 1.3 kg m^-3

(4) Mass of air displaced by balloon = v x d=15 x 1.3
= 19.5 kg

(5) According to law of floatation
Total downward wt. = UPTHRUST
8.5+x= 19.5
Mass of equipment x = 11 kg

Question 4.
(a) State the principle of floatation.
(b) The mass of a block made of certain material is 1.35 kg and its volume is 1.5 x 10^-3 m³.

1. Find the density of block.
2. Will this block float or sink? Give reasons for your answer.

Answer:
(a) PRINCIPLE OF FLOATATION : “When a solid is floating in a fluid, the weight of whole solid acting vertically downward at its CENTRE OF GRAVITY, is equal to the weight of fluid displaced by the IMMERSED part of solid acting upward, at its CENTRE OF BUOYANCY or at the centre of the BULK OF LIQUID displaced.”

OR

“The weight of a floating body is equal to the weight of the liquid displaced by its SUBMERGED part.”

(b) Mass of block = m = 1.35 kg
Volume of block = V = 1.5 x 10^-3 m³
(1) Density of block = \(\frac{m}{\mathrm{V}}=\frac{1.35}{1.5 \times 10^{-3}}=900 \mathrm{kgm}^{-3}\)
(2)   Density of block (900 kgm^-3) is less than density of water (1000 kgm^-3)
.’. Block will float in water

Question 5.
(a) State Archimedes’ Principle.
(b) A block of mass 7 kg and volume 0.07 m³ floats in a liquid of density 140 kg/m³. Calculate :

1. Volume of block above the surface of liquid.
2. Density of block.

Answer:
(a) ARCHIMEDES’ PRINCIPLE : “Whenever a body is immersed in a liquid (fluid), wholly or partially, it loses weight equal to the weight of liquid displaced by it.”

(b) Mass of block = m = 7 kg
Volume of block = V = 0.07 m³
Density of liquid =ρ_l = 140 kgm^-3
Let V’ = Volume of block immersed in the liquid
By law of floatation:

Weight of block = Weight of liquid displaced by the immersed pa

Question 6.
(a) A body whose volume is 100 cm³ weighs 1 kgf in air. Find its weight in water.
(b) Why is it easier to swim in sea water than in river water?
Answer:
(a) Volume of body = V = 100 cm³ = 10^-4 m³
Weight of body in air = 1 kgf
Density of water = p_w = 1000 kgm³
We know volume of water displaced = Volume of body
= V = 10^-4 m³
Upthrust = Weight of water displaced by body = Vp_wg
= 10-⁴ x 1000 x g .
= 10-¹ kgf =0.1 kgf
Weight of body in water=Weight of body in air – Upthrust
= 1-0.1 =0.90 kgf

(b) With smaller portion of man’s body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body.
While to displace the same weight of river water, a larger portion of the body will have to be submerged in water. It is easier for man to swim in sea water.

Question 7.
Why does a ship made of iron not sink in water, while an iron nail sinks in it?
Answer:
Density of iron is more than density of water, therefore weight of iron nail is more than wt. of water displaced by it and nail SINKS. While shape of iron ship is made in such a way that it displaces MORE WEIGHT OF WATER than its own weight. Secondly the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF SHIP LESS THAN THAT OF WATER and hence ship floats on water.

Question 8.
A solid of density 5000 kgm^-3 weighs 0.5 kgf in air. It is completely immersed in a liquid of density 800 kgm^-3. Calculate the apparent weight of the solid in liquid.
Answer:
Density of solid, d_s = 5000 kg m^-3
Weight of body in air = 0.5 kgf
mg = 0.5 kgf
∴  m – 0.5 kg

(1) Apparent weight of the solid in water = 0.5 – 0.08
= 0.42 kgf

(2) Apparent weight of body in liquid of density 800 kg m^-3 is zero.
Density of solid is less than density of liquid i.e. upthrust is more than weight of body.

Question 9.
(a) A body dipped in a liquid experiences an upthrust. State the factors on which the upthrust depends
(b) While floating, is the weight of body greater than, equal to or less than upthrust?
Answer:
(a) Factors on which upthrust depends are :

1. Volume of body immersed in fluid.
   Upthrust is maximum when body completely immersed in the fluid.
2. Density of the fluid.
   Upthrust α density of fluid
   Larger the density of the fluid, large will be the upthrust acting on the body.

(b) When the body floats then weight of the body is equal to the upthrust acting on the body.

Question 10.
A sinker is first weighed alone under water. It is then tied to a cork and again weighed under water. In which of the two cases weight under water is less and why?
Answer:
Weight of sinker, when tied to a cork, under water is less than that when it is alone weighed under water. Because cork displaces more water than its own weight and hence large upthrust acts on the sinker.

Question 11.
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm³ of water, calculate relative density of solid.
Answer:
Weight of solid in air =105 kgf
Volume of solid = Volume of water displaced = 30000 cm³
= 30000 x 10^-6 m³ = 0.03 m³
p_w = Density of water = 1000 kgm^-3
Wt. of water displaced by solid .

Question 12.
A test tube loaded with lead shots weighs 25 gf and floats upto the mark X in water. When the test tube is made to float in brine solution, it needs 5 gf more of lead shots to float upto level X. Find the relative density of brine solution.
Answer:
When test tube floats is water :
Weight of test tube = 25 gf
By law of floatation
Weight of water displaced = Weight of test tube = 25 gf

When test tube floats in brine solution, it needs 5 gf more of lead shots to float upto same level as in water.
Weight of test tube = 25 + 5 = 30 gf

By law of floatation :
Weight of brine solution displaced = Weight of test tube = 30 gf As, volume of brine solution displaced = Volume of water displaced

Question 13.
A wooden block is weighed with iron, such that combination just floats in water at room temperature. State your observations when :
(1) water is heated above room temperature
(2) water is cooled below 4°C. Give reasons to your answers in (1) and (2).
Answer:
(1) We know density of water decreases with rise in temperature and hence upthrust decreases.
(2) Density of water is maximum at 4°C. When water cooled below 4°C, then its density and hence upthrust acting on it decreases. So, wooden block weighed with iron, sinks more than earlier.

Question 14
A rubber ball floats in water with 2/7 of its volume above the surface of water. Calculate the average relative density of rubber ball.
Answer:
Let volume of rubber ball = V

Question 15.
A cube of ice whose side is 4.0 cm is allowed to melt. The volume of water formed is found to be 58.24 cm³. Find the density of ice.
Answer:
Side of ice cube = l = 4 cm
Volume of ice cube = V = β = (4)³ = 64 cm³
Volume of water = V = 5824 cm³
Density of ice cube = ρ_i= ?
Density of water = ρ_w = 1 gem^-3
By law of floatation:
Volume of ice cube x Density of ice=Volume of water x Density of water
64 x ρ_i = 58.24 x 1
\(\rho_{i}=\frac{58.24}{64}\)
ρ_i =0.91 gcm^-3

Question 16.
A jeweller claims to make ornaments of pure gold of relative density 19.3. A customer buys from him a bangle of weight 25.25 gf. The customer then weighs the bangle under water and finds its weight 23.075 g with the help of suitable calculations explain whether the bangle is of pure gold or not.
[R.D. of gold is 11.61 and bangle is not of pure gold] Ans. R.D of pure gold = 19.3 …given
Answer:

Question 17.
(a) When a piece of ice floating in water melts, the level of water inside the glass remains same. Explain.
(b) An inflated balloon is placed inside a big glass jar which is connected to an evacuating pump. What will you observe when the evacuating pump starts working? Give a reason for your answer.
Answer:
(a) A piece of ice displaces an amount of water equal to its own weight. Volume of water displaced is equal to volume of submerged part of ice cube. When ice cube melts, its volume decreases and gets occupied in that volume of water which is displaced by it. As a result, level of water inside the glass remains same when piece of ice (ice cube) melts.

(b) When evacuating pump starts working, pressure inside the glass jar reduces. As the pressure inside the balloon is more than pressure outside the balloon inside the glass jar, so balloon with burst.

Question 18.
(a) A trawler is fully loaded in sea water to maximum capacity. What will happen to this trawler, if moved to river water? Explain your answer.
(b) A body of mass 50 g is floting in water. What is the apparent weight of body in water? Explain your answer.
Answer:
(a) Density of sea water is more than the density of river water.So river water offers less upthrust to the trawler as compared to sea water.So, when a trawler is fully loaded sea water to maximum capacity, is moved to river water, it will sink.

(b) Mass of body = 50 g
Apparent weight of body = Weight of body in air – Weight of water displaced by body.
But when a body floats, then weight of body in air is equal to the weight of water displaced by the body.
=> Apparent weight of the body = 0

Question 19.
A body of mass ‘m’ is floating in a liquid of density ‘p’
(1) what is the apparent weight of body?
(2) what is the loss of weight of body?
Answer:
Mass of body = m
Density of liquid = ρ
(1) Apparent weight of body = Weight of body in air – Weight of liquid displaces by body.
When a body floats in the liquid, then weight of the body in a liquid is equal to weight of liquid displaced by the body.
=> Apparent weight of body = 0

(2) Loss in weight of body is equal to the weight of liquid displaced by the body.

Question 20.
A block of wood of volume 25 cm³ floats in water with 20 cm³ of its volume immersed in water.
Calculate :
(1) density of wood
(2) the weight of block of wood.
Answer:
Volume of wooden block = V = 25 cm³
Volume of wooden block immersed in water = 20 cm³
Volume of water displaced by wooden block=volume of wooden
block immersed in water = 20 cm³
Density of water = ρ_water = 1 gcm^-3
Density of wooden block =ρ_wood = ?
By law of floatation:
Volume of wooden block x Density of wood=Volume of water displaced x Density of water

Question 21.
A solid body weighs 2.10 N. in air. Its relative density is 8.4. How much will the body weigh if placed
(1) in water,
(2) in liquid of relative density 1.2?
Answer:
Weight of solid body in air = 2.10 N
R.D. of solid = 8.4

Weight of body in water = Weight of body in air – Weight of water displaced by body
= 2.10 -0.25 = 1.85 N

(2) Upthrust due to water = Weight of water displaced by body
= 0.25 N
Upthrust due to liquid = Upthrust due to water ×R.D. of liquid
= 0.25 x 1.2 = 0.30 N
Weight of body in liquid = Weight of body in air – Upthrust due to liquid
= 2.10-0.30= 1.80 N

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given
A New Approach to ICSE Physics Part 1 Class 9 Solutions Upthrust and Archimedes’Principle. are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Measurements and Experimentation

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Measurements and Experimentation

Unit 1
Exercise 1

(A) Objective Questions

I. Multiple choice Questions.
Select the correct option:

1. Which of the following is not a fundamental unit?
(a) Second
(b) Ampere
(c) Candela
(d) Newton
Ans. (d) Newton
Explanation : Second, Ampere and Candela are the fundamental units while Newton is a derived unit.

2. Which of the following is a fundamental unit?
(a) m/s²
(b) Joule
(c) Newton
(d) metre
Ans. (d) metre
Explanation : m/s², Joule and Newton are derived units while metre is a fundamental unit.

3. Which is not a unit of distance?
(a) metre
(b) millimetre
(c) Leap year
(d) kilometre
Ans. (c) Leap year
Explanation : Leap year is a unit of time while the metre, millimetre and kilometre are the units of distance.

II Fill in the blanks

1. The unit is which we measure the quantity is called constant quantity.
2. One light year is equal to 9.46 × 10¹⁵ m.
3. One mean solar day = 86400 sec
4. One year = 3.1536 × 10⁷ sec
5. One micrometre = 10^-6 m.

(B) Subjective Questions

Question 1.
What do you understand by the term measurement?
Answer:
“Measurement implies comparison of a physical quantity with a standard unit to find out how many times the given standard is contained in the physical quantity.”
Physics, like other branches of science requires experimental study which involves measurement.

Question 2.
What do you understand by the terms

1. unit
2. magnitude, as applied to a physical quantity?

Answer:
(i) Unit : Unit “is a standard quantity of the same kind with which a physical quantity is compared for measuring it. ” In order to measure a physical quantity, a standard is needed (which is acceptable internationally). The standard should be some convenient, definite and easily reproducible quantity of the same kind in terms of which the physical quantity as a whole is expressed. This standard is called a unit
(ii) Magnitude of a physical quantity : The number of times a standard quantity is present in a given physical quantity is called magnitude of physical quantity.
Physical quantity = Magnitude × Unit

Question 3.
A body measures 25 m. State the unit and the magnitude of unit in the statement.
Answer:
Here S.I. unit of length i.e. metre (m) has been used. Magnitude of the given quantity = 25
Metre : It is defined as 1,650,763,73 times the wavelength of specified orange red spectral line a emission spectrum of Krypton-86 or 1,553,164.1 times the wavelength of the red line in emission spectrum of cadmium.
or one metre is defined as the distance travelled by the light in 1/299,792,458 of a second in air/vacuum.

Question 4.
State four characteristics of a standard unit.
Answer:
Characteristics of standard unit :

1. It should be of convenient size.
2. It should not change with respect to place and time.
3. It should be well defined.
4. It should be easily reproduced.

Question 5.
Define the term fundamental unit. Name fundamental units of mass; length; time; current and temperature.
Answer:
Fundamental unit : A fundamental or basic unit is that which is independent of any other unit or which can neither be changed nor can be related to any other fundamental unit. e.g. units of mass, length, time and temperature.

Question 6.
What do you understand by the term derived unit? Give three examples.
Answer:
Derived units. “Derived units are those which can be expressed in terms of fundamental units.”
Example.


2. S.I. unit of area i.e. m² is a derived unit.
Area = length × breadth
Now metre is unit of length and breadth, so S.I. unit of area is obtained by multiplying the fundamental unit ‘m’ with itself. So, m² is the derived unit of area.
3. Density = Mass/volume
S.I. unit of density i.e. kg/m³ is the derived unit of density because it can be obtained by combining two fundamental units kilogram and metre.

Question 7.

(a) Define metre according to old definition.
(b) Define metre in terms of wavelength of light.
(c) Why is the metre length in terms of wavelength of light considered more accurate?

Answer:
(a) Metre : One metre is defined as the one ten millionth part of distance from the pole to the equator.
(b) Metre : One metre is defined as 1,650, 763.73 times the wavelength of specified orange red spectral line in emission spectrum of Krypton = 86.
OR
One metre is defined as 1,553,164.1 times the wavelength of the red line in emission spectrum of cadmium.
(c) Metre length in terms of wavelength of light is considered more accurate because

1. The wavelength of light does not change with time, temperature, pressure etc.
2. It can be reproduced anywhere at any time because Krypton is available every where.

Question 8.
Name the convenient unit you will use to measure :

(a) length of a hall
(b) width of a book
(c) diameter of hair
(d) distance between two cities.

Answer:

(a) Foot (Ft)
(b) Centimetre (cm)
(c) Micrometre (µm)
(d) Kilometre (km)

Question 9.
(a) Define mass.
(b) State the units in which mass is measured in (1) C.GS. system (2) S.I. system.
(c) Name the most convenient unit of mass you will use to measure :

1. Mass of small amount of a medicine.
2. The grain output of a state
3. The bag of sugar
4. Mass of a cricket ball.

Answer:
(a) Mass: The quantity of matter contained in a body is known as its mass.
(b) In C.GS. system, mass is measured in gram. In S.I. system, mass is measured in Kilogram.

Question 10.
(a) Define time.
(b) State or define the following terms :

1. Solar day
2. Mean solar day
3. An hour
4. Minute
5. Second
6. Year.

Answer:
(a) Time : It is defined as the time interval between two events
(b)
(i) Solar day : The time taken by the earth to complete one rotation about its own axis is called solar day.
(ii) Mean solar day : The average of the varying solar days, when the earth completes one revolution around the sun, is called mean solar day.
(iii) An hour : It is defined as the 1/24 th part of the mean solar day.
(iv) Minute : It is defined as the 1/1440 part of the mean solar day.
(v) Second : “A second is defined as 1/86400 th part of a mean solar day.”
OR
Second may also be defined “as to be equal to the duration of9,192,631,770 vibrations corresponding to the transition between two hyperfme levels of caesium – 133 atom in the ground state.”
(vi) Year : One year is defined as the time in which earth completes one complete revolution around the sun.

Unit II

Practice Problems 1

Question 1.
A student calculates experimentally the value of density of iron as 7.4 gem^-3. If the actual density of iron is 7.6 gcm^-3, calculate the percentage error in experiment.
Answer:

Question 2.
A student finds that boiling point of water in a particular experiment is 97.8°C. If the actual boiling point of water is 99.4°C, calculate the percentage error.
Answer:
Experimental value of boiling point of water = B.P₁ = 97.8°C
Actual value of boiling point of water = B.P₂ = 99.4°C
Absolute error = B.P.₂ – B.P₁ = 99.4 – 97.8 = 1.6°C

Question 3.
A pupil determines velocity of sound as 320 ms^-1. If actual velocity of sound is 332 ms^-1, calculate the percentage error.
Answer:
Velocity of sound determined by pupil = V₁ = 320 ms^-1
Actual value of velocity of sound = V₂ = 332 ms^-1
Absolute error = V₂ – V₁ = 332 – 320 = 12 ms^-1

Exercise 2

Question 1.
(a) What do you understand by the term order of magnitude of a quantity?
(b) Why are physical quantities expressed in the order of magnitude? Support your answer by an example.
Answer:
(a) Order of a magnitude of a quantity : The exponent part of a particular measurement is called order of magnitude of a quantity.
OR
The order of magnitude of a given numerical quantity is the nearest power of ten to which its value can be written, (b) Measurement of certain physical quantities are either too large or too small that these cannot be expressed conveniently. It is difficult to write or remember them. So such quantities can be expressed in the order of magnitude.
For example : The diameter of the sun is 1,390,000,000 m. It is difficult to write or remember such a measurement. So it is expressed as 1.39 × 109 m.
Here power of ten i.e. 9 (i.e. exponent part of the measurement) gives the order of magnitude of the given quantity.
So order of magnitude of diameter of the sun is 10⁹ m.

Question 2.
Express the order of magnitude of the following quantities :

1. 12578935 m
2. 222444888 kg
3. 0.000,000,127 s
4. 0.000,000,000,00027 m

Answer:

Question 3.

(a) What do you understand by the term degree of accuracy?
(b) Amongst the various physical measurements recorded in an experiment, which physical measurement determines the degree of accuracy?

Answer:

(a) Degree of accuracy : It means that we can measure a quantity, without any error of estimation.
In any experiment, all observations should be taken with same degree of accuracy.
(b) Amongst the various physical measurements recorded in an experiment, least accurate observation determines the degree of accuracy.

Question 4.

(a) State the formula for calculating percentage error
(b) Is it possible to increase the degree of accuracy by mathematical manipulations? Support your answer by an example.

Answer:
(a) The percentage error can be calculated by the formula :

(b) It is not possible to increase the degree of accuracy by mathematical manipulations.
For examples : When a number of values are added or subtracted, the result cannot be more accurate than the least accurate value.

In the above addition 72.5 has least accuracy. When we . say 72.5, it implies that value lies between 72.45 and 72.55 and 72.5 is the most probable value. Thus the error in 72.5 is +0.05. As the final result cannot be more accurate than least accurate observation, so the correct and most reliable answer in the above addition is 72.9.

Question 5.
State the factors which determine number of significant figures for the calculation of final result of an experiment.
Answer:
Factors which determine number of significant figures for the calculation of final result of an experiment are :

1. The nature of experiment.
2. The accuracy with which various measurements are made.

Question 6.
The final result of calculations in an experiment is 125,347,200. Express the number in terms of significant places when

1. accuracy is between 1 and 10
2. accuracy is between 1 and 100
3. accuracy is between 1 and 1000

Answer:
Final result of calculations in an experiment = 125,347,200

1. When accuracy lies between 1 and 10, then final result may be written as 1.2 × 10⁸.
2. When accuracy lies between 1 and 100, then final result may be written as 1.25 × 10⁸.
3. When accuracy lies between 1 and 1000 than final result may be written as 1.253 × 10⁸.

Unit 3

Practice Problems 1

Question 1.
The main scale of vernier callipers has 10 divisions in a centimetre and 10 vernier scale divisions coincide with 9 main scale divisions. Calculate

1. pitch
2. L.C. of vernier callipers.

Answer:
Main scale divisions of vernier callipers in one centimetre = 10

Question 2.
In a vernier callipers 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate

1. pitch
2. L.C. of vernier callipers.

Answer:
Main scale divisions of vernier callipers in one centimetre = 20 Unit

Practice Problems 2

Question 1.
Figure shows the position of vernier scale, while measuring the external length of a wooden cylinder.

1. What is the length recorded by main scale?
   
2. Which reading of vernier scale coincides with main scale?
3. Calculate the length.

Answer:
Main scale divisions of vernier callipers in one centimetre = 10

Question 2.
In figure for vernier callipers, calculate the length recorded.

Answer:
Main scale divisions of vernier callipers in one centimetre = 10

Practice Problems 3

Question 1.
(a) A vernier scale has 10 divisions. It slides over a main scale, whose pitch is 1.0 mm. If the number of divisions on the left hand of zero of the vernier scale on the main scale is 56 and the 8th vernier scale division coincides with the main scale, calculate the length in centimetres.
(b) If the above instrument has a negative error of 0.07 cm, calculate corrected length.
Answer:
No. of divisions on vernier scale = 10
Pitch = 1.0 mm

Question 2.
(a) A vernier scale has 20 divisions. It slides over a main scale, whose pitch is 0.5 mm. If the number of divisions on the left hand of the zero of vernier on the main scale is 38 and the 18th vernier scale division coincides with main scale, calculate the diameter of the sphere, held in the jaws of vernier callipers.
(b) If the vernier has a negative error of 0.04 cm, calculate the corrected radius of sphere.
Answer:
No. of divisions on vernier scale = 20
Pitch 0.5 mm

Practice Problems 4

Question 1.
The least count of a vernier callipers is 0.0025 cm and it has an error of + 0.0125 cm. While measuring the length of a cylinder, the reading on main scale is 7.55 cm, and 12th vernier scale division coincides with main scale. Calculate the corrected length.
Answer:
Least count (L.C.) = 0.0025 cm
Error = +0.0125 cm
Correction = – (Error) = – (+0.0125) = – 0.0125 cm
Main scale reading = 7.55 cm
Vernier scale division (V.S.D.) coinciding with main scale = 12th
Length recorded = Main scale reading + L.C. × V.S.D.
= 7.55 + 0.0025 × 12
= 7.55+ 0.0300 = 7.58 cm
Correct length = Length recorded + Correction
= 7.58+ (-0.0125)
= 7.58 – 0.0125 = 7.5675 cm = 7.567 cm

Question 2.
The least count of a vernier callipers is 0.01 cm and it has an error of + 0.07 cm. While measuring the radius of a sphere, the main scale reading is 2.90 cm and the 5th vernier scale division coincides with main scale. Calculate the correct radius.
Answer:
Least count (L.C.) = 0.01 cm
Error = + 0.07 cm
Correction = (Error) = – (+ 0.07) = – 0.07 cm
Main scale reading = 2.90 cm
Vernier scale division (V.S.D.) coinciding with main scale = 5th Observed diameter of sphere = Main scale reading + L.C. × V.S.D.
= 2.90 + 0.01 x 5 = 2.90 + 0.0 = 2.95 cm
Corrected diameter = Observed diameter + Correction
= 2.95 + (-0.07) = 2.95 – 0.07 = 2.88 cm
∴ Corrected radius = 2.88/2 = 1.44 cm

Exercise 3

Question 1.
Who invented vernier callipers?
Answer:
Vernier callipers was invented by Pierre Vernier.

Question 2.
What is the need for measuring length with vernier callipers?
Answer:
For measuring the exact length with greater accuracy, especially when we are measuring a very small length, we use an appliance vernier calliper. A vernier calliper can measure accurately upto 1/100 th part of a centimetre.

Question 3.
Up to how many decimal places can a common vernier callipers measure the length in cm?
Answer:
A common vernier calliper can measure the length accurately upto two places of decimal when length is measured in centimetre
i.e. upto 1/100 th part of a centimetre.

Question 4.
Define the terms :

1. pitch
2. least count as applied to a vernier callipers.

Answer:

1. Pitch : “The pitch of a screw is the distance moved by the screw in one complete rotation of its head.”
   OR
   Pitch may also be defined as “the distance between two consecutive threads of screw measured along the axis of screw.”
   
2. Least Count of Vernier Calliper : Least count of a vernier callipers is the difference between one main scale division (M.S.D.) and one vernier scale division (V.S.D.)

Question 5.
State the formula for determining :

1. pitch
2. least count for a vernier callipers.

Answer:

Question 6.
State the formula for calculating length if :

1. Number of vernier scale division coinciding with main scale and number of division of main scale on left hand side of zero of vernier scale are known.
2. The reading of main scale is known and the number of vernier scale divisions coinciding with main scale are known.

Answer:

1. If we know the number of vernier scale divisions (V.S.D.) coinciding with main scale and number of main scale divisions (M.S.D.) on left hand side of zero of vernier scale then Length recorded = Main scale reading + L.C. × V.S.D.
2. Same as in part (i).

Question 7.
(a) What do you understand by the term zero error?
(b) When does a vernier callipers has

1. positive error
2. negative error?

(c) State the correction if

1. positive error is 7 divisions
2. negative error is 7 divisions, when the least count is 0.01 cm.

Answer:
(a) Zero Error : A vernier callipers is said to have a zero error when zero of the main scale does not coincide with zero of vernier scale.
(b) Positive Error : If the zero of the vernier scale is on right hand side of zero of the main scale, then error is said to be positive and correction is said to be negative.

Negative Error : If the zero of the vernier scale is on the left hand side of zero of the main scale, the error is said be negative and the correction is said to be positive.

(c) When positive error is 7 divisions and L.C. is 0.01 cm
Then correction = – (+ 7 × L.C.)
= -7 × 0.01 cm = – 0.07 cm
When negative error is 7 divisions and count (L.C.) is 0.01 cm
Then correction = – (- 7 × L.C.)
= – ( – 7 × 0.01) cm = + 0.07 cm

Question 8.
Which part of vernier callipers is used to measure

(a) external diameter of a cylinder
(b) internal diameter of a hollow cylinder
(c) internal length of a hollow cylinder?

Answer:

(a) External Jaws of a vernier callipers are used to measure the external diameter of cylinder.
(b) Internal Jaws are used to measure internal diameter of a hollow cylinder.
(c) Tail of vernier callipers is used to measure the internal length of a hollow cylinder.

Unit 4

Practice Problems 1

Question 1.
The circular scale of a screw gauge has 50 divisions. Its spindle moves by 2 mm on sleeve, when given four complete rotations calculate

1. pitch
2. least count.

Answer:
Number of circular scale divisions (C.S.D.) = 50
Distance moved by screw (spindle) on sleeve = 2 mm
Number of complete rotations given = 4

Question 2.
The circular scale of a screw gauge has 100 divisions. Its spindle moves forward by 2.5 mm when given five complete turns. Calculate

1. pitch
2. least count of the screw gauge.

Answer:
Number of circular scale divisions = 100
Distance moved by spindle (screw) = 2.5 mm
No. of complete rotations given = 5

Practice Problems 2

Question 1.
Figure shows a screw gauge in which circular scale has 200 divisions. Calculate the least count and radius of wire.

Answer:
No. of circular scale divisions = 200
Pitch = 1 mm

Question 2.
Figure shows a screw gauge in which circular scale has 100 divisions. Calculate the least count and the diameter of a wire.

Answer:
No. of circular scale division = 100
Pitch =0.5 mm

Practice Problems 3

Question 1.
A micrometre screw gauge having a positive zero error of 5 divisions is used to measure diameter of wire, when reading on main scale is 3rd division and 48th circular scale division coincides with base line. If the micrometer has 10 divisions to a centimetre on main scale and 100 divisions on circular scale, calculate

1. Pitch of screw
2. Least count of screw
3. Observed diameter
4. Corrected diameter.

Answer:

Question 2.
A micrometre screw gauge has a positive zero error of 7 divisions, such that its main scale is marked in 1/2 mm and the circular scale has 100 divisions. The spindle of the screw advances by 1 division complete rotation.
If this screw gauge reading is 9 divisions on main scale and 67 divisions on circular scale for the diameter of a thin wire, calculate

1. Pitch
2. L.C.
3. Observed diameter
4. Corrected diameter.

Answer:

Question 3.
The thimble of a screw gauge has 50 divisions for one rotation. The spindle advances 1 mm when the screw is turned through two rotations.

1. What is the pitch of screw?
2. What is the least count of screw gauge?
3. When the screw gauge is used to measure the diameter of wire the reading on sleeve is found to be 0.5 mm and reading on thimble is found 27 divisions. What is the diameter of wire in centimetres?

Answer:
Pitch of screw gauge is the distance moved by spindle in one revolution = 1/2 = 0 – 5 mm

Practice Problems 4

Question 1.
A micrometre screw gauge has a negative zero error of 8 divisions. While measuring the diameter of a wire the reading on main scale is 3 divisions and 24th circular scale division coincides with base line.
If the number of divisions on the main scale are 20 to a centimetre and circular scale has 50 divisions, calculate

1. pitch
2. observed diameter.
3. least count
4. corrected diameter.

Answer:

Question 2.
A micrometre screw gauge has a negative zero error of 7 divisions. While measuring the diameter of a wire the reading on main scale is 2 divisions and 79th circular scale division coincides with base line.
If the number of divisions on main scale is 10 to a centimetre and circular scale has 100 divisions, calculate

1. pitch
2. observed diameter
3. least count
4. corrected diameter.

Answer:

Exercise 4

Question 1.
For what range of measurement is micrometre screw gauge used?
Answer:
Micrometre screw gauge is used to measure upto the accuracy of 0.001 cm.

Question 2.
What do you understand by the following terms as applied to micrometre screw gauge?

1. Sleeve cylinder
2. Sleeve scale
3. Thimble
4. Thimble scale
5. Base line.

Answer:

1. Sleeve cylinder : A hollow cylinder attached to a nut of the screw gauge is known as sleeve cylinder.
   The spindle of the screw passes through sleeve cylinder
2. Sleeve scale : It is also known as main scale. A reference line or base line graduated in mm, drawn on the sleeve cylinder, parallel to axis of nut is known as sleeve scale.
3. Thimble : A hollow circular cylinder connected to the screw, which rotates along with nut on turning, is called thimble.
4. Thimble scale : It is also known as circular scale. A scale marked on tapered end of a hollow cylinder, which can move over the sleeve cylinder, is known as thimble scale.
5. Base line : A reference line drawn on the sleeve cylinder parallel to the axis of nut is known as base line.

Question 3.
What is the function of ratchet in screw gauge?
Answer:
When the flattened end of the screw comes in contact with stud, ratchet becomes free and makes a rattling noise. It indicates that screw should not be further pushed towards the stud.

Question 4.
What do you understand by the terms

(a) pitch of screw
(b) least count of screw?

Answer:

(a) Pitch of screw : The pitch of screw is defined as the distance between two consecutive threads he screw, measured along the axis of the screw.
(b) Least count of the screw : Least distance of the screw is defined as the smallest distance moved by its tip when the screw turn through one division marked on it.

Question 5.
State the formula for calculating

1. pitch of screw
2. least count of screw.

Answer:

Question 6.
What do you understand by the following terms as applied to screw gauge?

(a) Zero error
(b) Positive zero error
(c) Negative zero error.

Answer:

(a) Zero error : If the zero of the main scale does not coincide with zero of circular scale on bringing the screw end in contact with the stud, the screw gauge is said to have zero error.
(b) Positive zero error : If the zero of the circular scale is below the reference line of the main scale, then screw gauge is said to have positive zero error and the correction is negative.

(c) Negative zero error : If the zero of the circular scale is above the reference line of the main scale, then screw gauge is said to have negative zero error and correction is positive.

Question 7.
How do you account for (a) positive zero error (b) negative zero error, for calculating correct diameter of wires?
Answer:
(a) Positive zero error : If the zero line, marked on circular scale, is below the reference line of the main scale, then there is a positive zero error and the correction is negative. In the figure 5th circular scale division is coinciding with reference line.
∴ Correction
= – Coinciding division of C.S. × L.C.
= – 5 × 0.001 cm = -0.005 cm
If the observed diameter is 0.557 cm, then:
Corrected diameter
= Observed diameter + Correction
= 0.557 cm – 0.005 cm = 0.552 cm

(b) Negative zero error : If the zero line marked on circular scale, is above the reference line of the main scale, then there is a negative error and the correction is positive.
In the figure, there is 96th division on the circualr scale which coincides with reference line.
∴ Correction – + [n- coinciding division of C.S. × L.C.]
where n is the total number of circular scale divisions.
∴ Correction = + [100 – 96] × 0.001 cm
= 0.004 cm
If observed diameter is 0.557 cm, then :
Corrected diameter
= Observed diameter + Correction
= 0. 557 cm + 0.004 cm
= 0.561 cm

Unit 5

Exercise 5

Question 1.

(a) What do you understand by the term volume of substance?
(b) State the unit of volume in SI system.

Answer:
(a) Volume : The space occupied a substance (solid, liquid or gas) is called volume.
(b) SI unit of volume is Cubic metre (m³).
One cubic metre : Is the volume occupied by a cube whose each side is equal to 1 m.

Question 2.
How is SI system of unit of volume is related to 1 litre ? Explain.
Answer:

Question 3.
In which unit, volume of liquid is measured? How is this unit is related to S.I. unit of volume?
Answer:
The volume of liquid is measured in litre of its sub-multiple millilitre (mL).

Question 4.
Explain the method in steps to find the volume of an irregular solid with the help of measuring cylinder.
Answer:
Volume of an irregular solid

1. Take a measuring cylinder and fill water up to certain level. Note down the level of water in measuring cylinder. Let it be V₁.
2. Tie the irregular solid body with a thin and strong thread and lower the body gently so that the solid body is completely immersed in the water. The level of water rises. Solid body displaces water of its own volume. Note down the new level of water. Let it be V₂.
3. Take the difference of two level of water, i.e., (V₂ – V₁). This will give the volume of irregular solid body.

Question 5.
Amongst the units of volume (i) cm³ (ii) m³ (iii) litre (iv) millilitre, which is most suitable for measuring :

(a) Volume of a swimming tank
(b) Volume of a glass filled with milk
(c) Volume of an exercise book
(d) Volume of air in the room.

Answer:

(a) litre
(b) cm³
(c) millilitre
(d) m³

Question 6.
Find the volume of a book of length 25 cm, breadth 18 cm and height 2 cm in m³.
Answer:

Question 7.
The level of water in a measuring cylinder is 12.5 ml. When a stone is lowered in it, the volume is 21.0 ml. Find the volume of the stone.
Answer:
Level of water in measuring cylinder = V₁ = 12.5 ml
When stone is lowered, then level of water in measuring cylinder

Question 8.
A measuring cylinder is filled with water upto a level of 30 ml. A solid body is immersed in it so that the level of water rises to 37 ml. Now solid body is tied with a cork and then immersed in water so that the water level rises to 40 ml. Find the volume of solid body and the cork.
Answer:
Level of water in measuring cylinder = V₁ = 30 ml
Level of water in measuring cylinder when a solid body is immersed in it V₂ = 37 ml
Level of water in measuring cylinder when a cork tied with the solid is immersed in water = V₃ = 40 ml
Volume of solid body = V₂ – V₁ = 37 – 30 = 7 ml or 7 cm³
Volume or cork = V₃ – V₂ = 40 – 37
= 3 ml or 3 cm³ [∵ 1 ml = 1 cm³]

Unit 6

Practice Problems 1

Question 1.
Calculate the time period of simple pendulum of length 0. 84 m when g = 9.8 ms^-2.
Answer:

Question 2.
Calculate the time period of simple pendulum of length 1.44 m on the surface of moon. The acceleration due to gravity on the surface of moon is 1/6 the acceleration due to gravity on earth, [g = 9.8 ms^-2]
Answer:
Length of simple pendulum = l = 1.44 m
Time period (T) = ?
Acceleration due to gravity on the surface of moon

Practice Problems 2

Question 1.
Length of second’s pendulum is 100 cm. Find the length of another pendulum whose time period is 2.4 s.
Answer:
We know time period of second’s pendulum is 2 s.

Question 2.
A pendulum of length 36 cm has time period 1.2 s. Find the time period of another pendulum, whose length is 81 cm.
Answer:

Question 3.
Calculate the length of second’s pendulum on the surface of moon when acceleration due to gravity on moon is 1.63 ms^-2.
Answer:
Length of second’s pendulum = l = ?

Practice Problems 3

Question 1.
The length of two pendulum are 110 cm and 27.5 cm. Calculate the ratio of their time periods.
Answer:

Question 2.
A pendulum 100 cm and another pendulum 4 cm long are oscillating at the same time. Calculate the ratio of their time periods.
Answer:

Practice Problems 4

Question 1.
The time periods of two pendulums are 1.44 s and 0.36 s respectively. Calculate the ratio of their lengths.
Answer:

Question 2.
The time period of two pendulums are 2 s and 3 s respectively. Find the ratio of their lengths.
Answer:

Exercise 6

Question 1.

(a) Define simple pendulum.
(b) State two factors which determine time period of a simple pendulum.
(c) Write an expression for the time period of a simple pendulum.

Answer:
(a) Simple Pendulum : A simple pendulum consists of a heavy point mass (called bob) suspended from a rigid support by a massless, inextensible string.
(b) Factors on which time period of a simple pendulum depends :

1. i.e., if length increases, time period increases. That is why in summer pendulum of clock goes slow.
   
2. That is why when clock is taken to a mountain where ‘g’ decreases with altitude, time period increases and pendulum takes more time to complete an oscillation and hence the clock goes slow.
3. Mass or material of bob : Time period of simple pendulum is independent of mass.
4. Amplitude : Time period of simple pendulum is independent of amplitude. So long as swing is not too large.

(c) Expression for time period of a simple pendulum is given as :

Question 2.
Define the following in connection with a simple pendulum.

(a) Time period
(b) Oscillation
(c) Amplitude
(d) Effective length.

Answer:
(a) Time period (T) : “is the time taken to complete one oscillation.” Its unit is second (s) and time period is denoted by ‘T”
(b) Oscillation : “One complete to and fro motion of the pendulum” is called an oscillation.
i.e., motion of bob from B to C and then C to B is one oscillation.

(c) Amplitude : The maximum displacement of bob from mean position on either side is called amplitude
Amplitude = AB or AC. It is denoted by ‘a’.
(d) Effective length : The length between the point of suspension and centre of gravity of bob of a pendulum is called effective length.

Question 3.
(a) What is a second’s pendulum?
(b) A second’s pendulum is taken on the surface of moon where acceleration due to gravity is l/6th of that of earth. Will the time period of pendulum remain same or increase or decrease? Give a reason.
Answer:
(a) Seconds’ pendulum : “A pendulum which has time period of two seconds” is called seconds’ pendulum.
OR
Seconds’ pendulum may also be defined as “a pendulum which completes one oscillation in two seconds.”
(b) We know time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity

Question 4.
Which of the following do not affect the time period of a simple pendulum?

(a) mass of bob
(b) size of bob
(c) effective length of pendulum
(d) acceleration due to gravity
(e) amplitude.

Answer:
(a) Mass of the bob, and (b) Size of the bob, do not affect the time period of a pendulum. Also time period of pendulum is independent of the amplitude provided this is not too great.

Question 5.
A simple pendulum is hollow from within and its time period is T. How is the time period of pendulum affected when :

(a) 1/4 of bob is filled with mercury
(b) 3/4 of bob is filled with mercury
(c) The bob is completely filled with mercury?

Answer:
We know that time period of a simple pendulum is independent of its mass. So in all the above said cases, time period of simple pendulumn remains same.

Question 6.
Two simple pendulums, A and B have equal lengths but their bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? What is the reason for your answer?
Answer:
We know that time period of simple pendulum at a place is given by

and this expression does not contain weight of bob i.e. is independent of the weight of bob.
∴ Time period of both pendulums will be same.
∴ Ratio of their time periods =1 : 1

Question 7.
State the numerical value of the frequency of oscillation of a second’s pendulum. Does it depend on the amplitude of oscillation?
Answer:

Question 8.
(a) Name the two factors on which time period of a simple pendulum depends.
(b) Name the devices commonly used to measure
(i) mass and
(ii) weight of a body.
Answer:
(a) Factors on which time period of a simple pendulum depends :

1. 
   i.e., if length increases, time period increases. That is why in summer pendulum of clock goes slow.
2. That is why when clock is taken to a mountain where ‘g’ decreases with altitude, time period increases and pendulum takes more time to complete an oscillation and hence the clock goes slow.
   
3. Mass or material of bob : Time period of simple pendulum is independent of mass.
4. Amplitude : Time period of simple pendulum is independent of amplitude. So long as swing is not too large.

(b)

1. Mass of measured by physical balance.
2. Weight of a body is measured by spring balance.

Question 9.
Draw a graph of l, the length of simple pendulum against T², the square of its time period.
Answer:
Nature : The graph of length (l) of simple pendulum against square of its time period (T²) is a straight line inclined to time axis.

Question 10.
What do you understand by (a) amplitude and (b) frequency of oscillations of simple pendulum?
Answer:

(a) Amplitude : The maximum displacement of bob from mean position on either side is called amplitude.
Amplitude = AB or AC. It is denoted by ‘a’.
(b) Frequency: “It is the number of vibrations or oscillations made in one second.”It is denoted by f or n and its unit is Hertz (Hz) or per second (s-1).

Unit 7

Exercise 7

Question 1.

(a) What do you understand by the term graph?
(b) What do you understand by the terms (i) independent variable, (ii) dependent variable?
(c) Amongst the independent variable and dependent variable, which is plotted on X-axis?

Answer:
(a) Graph : A pictorial representation of two physical variables, recorded by ah experimenter is called graph.
(b)

1. Independent variable : A variable whose variation does not depend on that of another is known as independent variable.
2. ependent variable : A variable whose variation depends upon another variable is known as dependent variable.

(c) The independent variable is always plotted on x – axis.

Question 2.

(a) State how will you choose a scale for the graph.
(b) State the two ratios of a scale, which are suitable for plotting points.
(c) State the two ratios of a scale, which are not suitable for plotting points.

Answer:
(a) We can choose any convenient scale to represent a given variable on a given axis, such that the whole range of variations are well spread out on the whole graph paper, to give the graph line a suitable size.
For this a round number, nearest to or slightly less than minimum value should be taken as origin and a round number nearest to or slightly more than the maximum value should be taken at the far end of the respective axis for a given variable.
(b) Two ratios of a scale suitable for ploting points are 1 : 2 and 1 : 4.
(c) Two ratios of a scale not suitable for plotting points are 1 : 3 and 1 : 7. Because such scales are impractical and pose difficulty in plotting intermediate points.

Question 3.
State three important precautions which must be followed while plotting points on a graph.
Answer:
Precautions for plotting points on a graph :

1. The points marked on graph paper should be sharp, but not thick.
2. Ordinates of points should be written close to the plotted point.
3. It is not necessary that graph line should pass through all points. A best fit line should be drawn.

Question 4.
State two important precautions for drawing a graph line.
Answer:
Precautions for drawing a graph line :

1. The graph line should be thin, single straight line and sharp.
2. It is not necessary that graph line should pass through all the points. A best fit graph line should be drawn.

Question 5.

(a) What is a best fit line for a graph?
(b) What does best fit line show regarding the variables plotted and the work of experimenter?

Answer:

(a) A best bit line for a graph means a line which either passes through maximum number of points or passes closest to the maximum number of points, which appear on either side of the line.
(b) A best fit line shows that two variable quantities are directly proportional to each other. With its help, experimenter can easily understand nature of proportional relations between two variable quantities.

Question 6.

(a) What do you understand by the term constant of proportionality?
(b) How can proportionality constant be determined from the best fit straight line graph?

Answer:

(a) Constant of proportionality : If a quantity say X is directly proportional to another quantity Y, then X is written as X = KY, where K is called constant of proportionality.
(b) Constant of proportionality : can be determined from the best fit straight line by calculating the slope of graph by using the formula. Slope of graph

Question 7.
State three uses of graph.
Answer:
Uses of a graph :

(a) One can determine constant of proportionality by calculating slope of graph.
(b) It can be used to calculate mean average value of large number of observations.
(c) It can be used for verifying already known physical laws.
(d) It can also show the weakness of the experimenter at some particular instant during the course of experiment.

Question 8.
How does a graph help in determining the proportional relationship between two quantities?
Answer:
It has been found that if a graph is plotted between pressure of an enclosed gas at constant temperature, against its volume, the graph line is a smooth curve, which does not meet X-axis or Y- axis on extending as shown in figure.
From the figure, it is clear that pressure of gas is not directly proportional to volume of gas.

However, if a graph is plotted between pressure and inverse of volume, the graph line is a straight line as illustrated in figure. From the straight line graph we can say:
Pressure is inversely proportional to volume.


Similarly, if a graph is plotted between length and time period of a simple pendulum, the graph line is a curve, which has a tendency to meet X-axis or Y-axis when produced towards origin, as shown in the figure.

From the figure, it is clear that length of a simple pendulum is not proportional to its time period.
However, if a graph is plotted between length and (Time)², the graph line is a straight line. Thus, we can say :
From the above discussion it is very clear that graph line helps to determine the nature of proportional relationship between two variable quantities.

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Measurements and Experimentation are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1

Unit I
Exercise 1

Question 1.
There is a positively charged sphere A and negatively charged sphere B, such that they are brought in electrical contact by a copper wire. Answer the following questions :

(a) Which sphere is at higher potential before electrical contact on the basis of convention?
(b) Which sphere is at lower potential before electrical contact on the basis of convention?
(c) In which direction conventional current flows?
(d) In which direction electronic current flows?
(e) What is potential of the spheres after electrical contact?

Answer:
Sphere A is positively charged and sphere B is negatively charged.
Both the sphere A and B are brought in electrical contact by a copper wire.

(a) On the basis of convention, positively charged sphere A is at higher potential before electrical contact.
(b) On the basis of convention, negatively charged sphere B is lower potential before electrical contact.
(c) Conventional current flows from sphere A to sphere B i. e. from a body at higher potential to the body at lower potential.
(d) Electronic current flower from sphere B to sphere A i.e. from a body at lower potential to the body at higher potential.
(e) After electrical contact, both the spheres will be at same potential.

Question 2.

(a) What do you understand by the term electric potential?
(b) Define electric potential in terms of energy spent.
(c) State the unit of electric potential and define it.

Answer:
(a) Electric potential : The amount of work done in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(b) Electric potential : The amount of energy spent in moving a unit positive charge from infinity to a given point in an electric field is called electric potential.
(c) Volt is the SI unit of electric potential.
One volt : When one coulomb of charge is brought from infinity to a given point in an electric field, such that work done is one joule, then electric potential is said to be one volt.
OR
Electric potential is said to be one volt if one Joule of work is done in moving one coulomb of charge from infinity to a given point in an electric field.

Question 3.

(a) What do you understand by the term quantity of electric charge?
(b) State SI unit of electric charge and define it.
(c) How many electrons constitute one unit electric charge in SI system?

Answer:

(a) Quantity of electric charge : The number of charge (electrons) which drift from lower to higher potential is called quantity of charge.
(b) SI unit of electric charge is coulomb (C).
One coulomb : The quantity of electric charge which will deposit 0.00118 g of silver on the cathode, when passed through silver nitrate is called one coulomb.
(c) 6.25 × 10¹⁸ electrons constitute one unit (IC) electric charge in SI system.

Question 4.

(a) What do you understand by the term electric current?
(b) State and define the SI unit of electric current.
(c) State the relation between electric current; number of electrons moving in a circuit and time in seconds.

Answer:
(a) Electric current : The rate of flow of electric charge in an electric circuit is called electric current.
(b) Ampere (A) is the SI unit of electric current.
One ampere : When one coulomb charge flows through an electric circuit in one second, then the electric current flowing the circuit is said to be ampere.
(c) If Q is the charge (in coulombs) flowing through conductor in time t (in seconds) such that current I flows through the conductor then
Rate of flow of charge = Q/t
We know rate of flow of charge = I = Electric current.
⇒ I = Q/t

Question 5.
How electric current flows in (i) solids, (ii) liquids?
Answer:
(i) Flow of electric current in solids : In solids, the positive charges are associated with atomic nuclei. As the nuclei are firmly packed and closely held by inter-atomic forces, therefore, positive charges cannot drift.
On the other hand, negative charges (electrons) are not held firmly. Thus, when a potential difference, however small, is applied they start drifting from lower to higher potential.
The continuous drift of electrons, through the body of a solid conductor constitutes the current.
(ii) Flow of electric current in liquids : Within a liquid no electrons move. However, when a negatively charged and a positively charged electrodes are placed in a liquid, it sets up an electric field.
Under the influence of the electric field, the positively charged ions migrate towards the negatively charged electrode and vice versa.
At the cathode the positively charged ions gain electrons. At the anode the negatively charged ions lose same number of electrons.
Thus, in a way number of electrons given by the cathode is equal to the number of electrons accepted by the anode. The sum up, we can say that simultaneous movement and discharge of positive and negative ions in the opposite directions constitutes the current in the liquids.

Question 6.

(a) Define the term potential difference.
(b) How is potential difference related to work done and quantity of charge?

Answer:
(a) Potential difference : The amount of work done in moving a unit positive charge from one point to another point in an electric field is called potential difference.
(b) If Q = Charge moving from one point to another in an electric field.
W = Work done to move the charge Q from one point to another.
V = Potential difference between two points.
Then work done in moving Q units of charge = W
Work done in moving one unit of charge = W/Q
But work done in moving one unit of charge = Potential difference = V
⇒ V = W/Q

Practice Problems 1

Question 1.
A charge of 5000 C flows through an electric circuit in 2 hours and 30 minutes. Calculate the magnitude of current in circuit.
Answer:

Question 2.
A charge of 8860 C flows through an electric circuit in 2 min and 40 s. Calculate the magnitude of current in circuit.
Answer:

Practice Problems 2

Question 1.
A battery can supply a charge of 25 × 10⁴ C. If the current is drawn from battery at the rate of 2.5 A, calculate the time in which battery will discharge completely.
Answer:

Question 2.
A dry cell can supply a charge of 800 C. If continuous current of 8.0 mA is drawn, calculate the time in which cell will discharge completely.
Answer:

Practice Problems 3

Question 1.
Calculate the total number of electrons flowing through a circuit in 20 mins and 40 s, if a current of 40 μA flows through the circuit.
[1 e^– = 1.6 × 10^-19 C]
Answer:

Question 2.
4 × 10²⁰ electrons flow through a circuit in 10 hours. Calculate magnitude of current. [1 e- = 1.6 × 10^-19 C]
Answer:

Practice Problems 4

Question 1.
What is the electrical potential at a point in an electric field when 24 J of work is done in moving a charge of 96 C from infinity?
Answer:

Question 2.
A charge of 75 C is brought from infinity to a given point in an electric field, when amount of work done is 3.75 J. Calculate the electrical potential at that point.
Answer:

Practice Problems 5

Question 1.
A work of 25 J and 30 J is done when 5 C charge is moved first to point A and then to point B from infinity. Calculate the potential difference between points A and B.
Answer:

Question 2.
A charge of 25 C is moved from infinity to points A and B in an electric field when the work done to do so is 10 J and 10.5 J respectively. Calculate the potential difference between the points A and B.
Answer:

Unit II
Exercise 2

(A) Objective Questions

Multiple Choice Questions.
Select the correct option :

1. SI unit potential difference is :
(a) coulomb
(b) kelvin
(c) volt
(d) ampere
Ans. (c) volt

2. Current in a circuit flows :
(a) in a direction from high potential to low potential
(b) in a direction from low potential to high potential
(c) in a direction of flow of electron
(d) in any direction
Ans. (a) in a direction from high potential to low potential

3. In a metallic conductor, electric current is thought to be due to movement of :
(a) ions
(b) amperes
(c) electrons
(d) protons
Ans. (c) electrons

4. Assuming that the charges of an electron is 1.6 × 10^-19 coulombs, the number of electrons passing through a section of wire per sec, when the wire carries a current of one ampere is :
(a) 0.625 × 10¹⁹
(b) 1.6 × 10^-19
(c) 1.6 × 10¹⁹
(d) 0.627 × 10^-17
Ans. (a) 0.625 × 10¹⁹

5. Which of the following is best conductor of electricity?
(a) copper
(b) gold
(c) platinum
(d) silver
Ans. (d) silver

(B) Subjective Questions

Question 1.
What do you understand by the term electric cell?
Answer:
Electric cell is an arrangement which maintains constant potential difference between conductors.
A cell basically consists of two conducting rods, which are called electrodes, immersed in a solution, which is called the electrolyte.

Question 2.
Draw a neat and labelled diagram of simple voltaic cell showing clearly the direction of flow of conventional current and direction of flow of electrons.
Answer:
Simple voltaic cell was invented by Alessandro Volta in year 1800. It was the first device which could creat a constant potential difference between two plates with the help of chemical energy.

Question 3.
Briefly describe the theory of simple voltaic cell.
Answer:
Theory of simple voltaic cell: Amongst the zinc and copper plates, zinc is more electro-positive (ionisation potential – 0.76 V) as compared to copper
(ionisation potential + 0.34 V) in electrochemical series. The dilute sulphuric acid is used as an electrolyte in the ionised state.

When zinc plate comes in contact with (H⁺) hydrogen ions, it being more electro-positive, ionises to form zinc ions and free electrons.

The free electrons so formed, take the passage of least resistance, and hence, move out in the external circuit through the copper wires. The zinc ions, however, enter in the dilute sulphuric acid. Since Zn²⁺ ions are positively charged, they repel H⁺ ions in the acid solution, with the result that H⁺ ions start crowding at the copper plate.
The copper plate in turn loses its free electrons to H⁺ ions, which . form nascent hydrogen.

The nascent hydrogen atoms so formed unite to form molecular hydrogen.

From the above explanation, it is clear that free electrons actually drift from zinc to copper in external circuit, and hence, current should flow from zinc to copper.
However, we still continue saying the electric charge flows from copper to zinc and this current is called conventional current, whereas the actual flow of electrons is from zinc to copper which constitutes electronic current.
The emf between the zinc copper plate in the external circuit is 0.34 V – (-0.75 V)= 1.10 V.

Question 4.
What do you understand by the following terms?

1. electric circuit
2. closed electric circuit
3. open electric circuit.

Answer:

1. Electric circuit : The path along which electric current flows is known as electric circuit.
2. Closed electric circuit : W’hen the path of an electric circuit starting from one terminal of the cell, ends at the other terminal of cell, without any break, then such a circuit is called closed circuit.
3. Open electric circuit : When the path of an electric circuit, starting from one terminal of the cell, is broken at some point, then such a circuit is called open electric circuit.

Question 5.
State two conditions necessary for a circuit, such that electric current flows through it.
Answer:
For the flow of electric current through a circuit, following are the necessary conditions :

1. Electric circuit must be closed or complete.
2. Every part of the circuit is a conductor.

Question 6.
Draw a neat diagram showing

1. closed electric circuit
2. open electric circuit.

Answer:

1. Closed circuit
   
2. Open circuit
   

Question 7.
Name four electric conductors and four electric insulators.
Answer:

• Conductors : Silver, copper, aluminium and iron.
• Insulators : Plastic, nylon, dry wood and rubber.

Question 8.

(a) What do you understand by the term electric resistance?
(b) Why does the filament of an electric bulb in an electric circuit get white hot, but not the connecting wires?

Answer:

(a) Electric resistance : The obstruction offered to the passage of electric current by a material is called resistance of the material.
(b) Filament of an electric bulb is made up of tungsten having high resistance. Due to its high resistance, on passing electric current through it, a electrical energy changes into heat energy. So much heat is produced that filament of bulb becomes white hot and gives light.
Resistance of connecting wires is very low and hence the connecting wires do not get heated.

Question 9.
Is it correct to say that a resistance wire is an insulator or a bad conductor? Explain your answer.
Answer:
It is not correct to say that a resistance wire is an insulator or bad conductor. From resistance, it is implied that a gien material will conduct electricity, but will also offer obstruction to the passage of electric current.

Question 10.

(a) What do you understand by the term series circuit?
(b) State two characteristics of resistances in the series circuit.
(c) Draw a diagram showing two bulbs connected in series to a dry cell.

Answer:
(a) Series circuit : When a number of resistances are connected in an electrical circuit in such a way that positive of one resistance acts as the negative of the other resistance, then resistances are said to be in series.
OR
A number of resistors are said to be in series if these are joined end to end and same current flows through each one of them when a potential difference is applied across the combination.

(b) Characteristics of resistances in series :

1. The sum total of resistances in series increases with increase in number of resistors.
2. The potential difference remaining constant, the current in series circuit decreases with the increase in number of resistors in series.
3. All the elements in series circuit work simultaneously. If the circuit is broken anywhere between the elements, none of the elements work.

(c) Two bulbs glowing dimly

Question 11.

(a) What do you understand by the term parallel circuit?
(b) State two characteristics of resistance in the parallel circuit.
(c) Draw a diagram showing two bulbs connected in parallel to a dry cell.

Answer:
(a) Parallel circuit : When a number of resistances (bulbs) are connected in an electrical circuit in such a way that all of them are connected to common positive and common negative terminal of a cell, then the resistance (bulbs) are said to be connected in parallel.
OR
A number of resistors are said to be connected in parallel if one end of each resistor is connected to one point and other end of each resistor is connected to another point so that the potential difference across each resistor is same.

(b) Characteristics of resistances in Parallel :

1. The sum total of resistances in parallel decreases with the increase in number of resistors.
2. The current flowing in any resistor in parallel will be inversely proportional to resistance i.e., more the resistance, less the current.
3. Each resistor in parallel functions independently with respect to the other resistors in parallel.

(c)

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 1 are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound

Exercise

(A) Objective Questions

Multiple Choice Questions.
Select the correct option :

1. In case of longitudinal waves, the particles of the medium vibrate :
(a) in the direction of wave propagation
(b) opposite to the direction of wave propagation
(c) at right angles to the direction of wave propagation
(c) none of the above
Ans. (a) in the direction of wave propagation

2. A longitudinal wave consists of :
(a) crest and trough in the medium
(b) compression and rarefaction in the medium
(c) both (a) and (b)
(d) neither (a) nor (b)
Ans. (b) compression and rarefaction in the medium

3. The longitudinal waves can propagate only in .
(a) solids
(b) liquids
(c) gases
(d) all of these
Ans. (d) all of these

4. A part of the longitudinal wave in which particles of the medium are closer than the normal particles is called :
(a) rarefaction
(b) crest
(c) trough
(d) compression
Ans. (d) compression

5. A part of longitudinal wave in which particles of the medium are farther away than the normal particles is called :
(a) rarefaction
(b) trough
(c) compression
(d) crest
Ans. (a) rarefaction

6. In the region of compression or rarefaction, in a longitudinal wave, the physical quantity which does not change is :
(a) pressure
(b) mass
(c) density
(d) volume
Ans. (b) mass

7. The wavelength is the linear distance between the :
(a) two consecutive compressions
(b) two consecutive rarefactions
(c) one compression and one rarefaction
(d) both (a) and (b)
Ans. (d) both (a) and (b)

8. The number of oscillations passing through a point in unit time is called :
(a) vibration
(b) frequency
(c) wavelength
(d) amplitude
Ans. (b) frequency .

9. The SI unit of frequency is
(a) hertz
(b) gauss
(c) decibel
(d) none of these
Ans. (a) hertz

10. If the frequency of a wave is 25 Hz. the total number of compressions and rarefactions passing through a point in 1 second is :
(a) 25
(b) 50
(c) 100
(d) none of these
Ans. (b) 50

11. Which of the following is an elastic wave?
(a) light wave
(b) ratio wave
(c) sound wave
(d) microwave
Ans. (c) sound wave

(B) Subjective Questions

Question 1.

(a) What do you understand by the term sound energy?
(b) State three conditions necessary for hearing sound.

Answer:
(a) Sound energy : “It is a form of energy that produces the sensation of hearing in our ears.” Sound is produced when a body vibrates.
(b) Necessary conditions for hearing sound :

1. There must be a vibrating body, capable of transferring its energy to its surroundings.
2. These must be a material medium to pick the energy and then propagate it in forward direction.
3. There must be a receiver, so as to receive the sound vibrations and then transmit them to the brain for final interpretation, such as human ear.

Question 2.
Describe briefly an experiment to prove that vibrating bodies produce sound.
Answer:
Experiment : Take a tuning fork. It is U-shaped fork made of steel provided with a handle.

Strike the tuning fork with a rubber hammer and hold it close to ear. A sound is heard. Now take a freely suspended pith ball and touch one end of the tuning fork (which is already hit with a rubber hammer) to it. It is observed that pith ball repeatedly flies outward. This experiment too proves that sound is produced by a vibrating body.

Question 3.

(a) What do you understand by the term infrasonic vibrations?
(b) What do you understand by the term sonic vibrations? State the range of sonic vibrations for the human ear.

Answer:
(a) Infrasonic vibrations : Those vibrations whose frequency is less than 20 hertz are known as infrasonic vibrations
(b) Sonic vibrations : Sonic vibrations are also known as audio vibrations.
Those vibrations whose frequency is from 20 hertz to 20000 hertz are known as sonic vibrations.
The range of sonic vibrations is from 20 vibrations per second to 20000 vibrations per second.

Question 4.

(a) What do you understand by the term ultrasonic vibrations?
(b) Name three animals which can hear ultrasonic vibrations.

Answer:

(a) Ultrasonic vibrations : Those vibrations, whose frequency is more than 20000 hertz and are not perceived by human ear, are known as ultrasonic vibrations.
(b) Dogs, bats and dolphins can hear ultrasonic vibrations.

Question 5.
How do bats locate their prey during flight?
Answer:
Bats produce ultra sound which returns after striking an obstacle in their way. By hearing the reflected round, bats can judge the distance and direction of obstacle/prey in their way and hence bats can catch their prey during flight.

Question 6.
What is Galton’s whistle? To what use is it put?
Answer:
A special whistle which can produce ultra sound, not heard by humans, is called Galton’s whistle.
It is used to train dogs because they can hear ultrasounds upto a frequency of 40000 hertz.

Question 7.
State four practical uses of ultrasonic vibrations.
Answer:
Uses of ultrasonic vibrations :

1. These are used for dissipating fogs on the runways at the airports.
2. These are used in the ultrasound scanning of internal organs of human body.
3. These are used for making dish washing machines. In these machines, water and detergents are vibrated with ultlrasonics vibrator. The vibrating particles of the dissolved detergent rub against the plates and clean them.
4. These are used in SONAR (Sound navigation and ranging) to detect and find the distance of objects under water.

Question 8.
Describe an experiment to prove that material medium is necessary for the propagation of sound.
Answer:
A material medium is necessary for the propagation of sound. It can be proved with the help of following experiment :
An electric circuit consisting of a cell, a switch and an electric bell is arranged inside a bell-jar, which stands on the platform of an evacuating pump.
The switch of the bell is pressed to close the electric circuit. Sound is heard when there is air within the bell-jar. Air is now gradually pumped out of the bell-jar. The intensity of sound goes on decreasing, and no sound is heard when the air is completely removed from the bell-jar. It is because, the air which acts as a medium for the propagation of sound energy is removed.

From above experiment, it is clear that sound can not be heard in the absence of air i.e. a material medium is necessary for the propagation of sound.

Question 9.
Why do astronauts talk to each other through radio telephone in space?
Answer:
Sound can not travel through vacuum. There is no material medium in the space for propagation of sound. Hence astronauts talk to each other through radio telephone in space. In radio telephone, message of one astronauts reach to the other with the help of radio waves, which can travel through vacuum.

Question 10.
Define the terms :

1. wavelength
2. amplitude
3. frequency.

Answer:
(i) Wavlength : The linear distance between the two consecutive particles of a vibrating medium in the same phase is called its wavelength. It is denoted by Greek letter lambda (λ).
OR
The distance travelled by the wave in one time period of vibration of the particles of medium is called its wavelength.
OR
In a longitudinal wave, distance between the two consecutive compressions or rare fractions is called wavelength.
(ii) Amplitude : The maximum displacement of a vibrating particles about its mean position is called amplitude.
(iii) Frequency : It is denoted by letter (ƒ). The number of complete vibrations executed by a vibrating particle of a medium about its mean position in one second is called its frequency.
OR
It may be defined as the number of waves passing through one particular point in one second.

Question 11.
State four differences between the sound wave and the light wave.
Answer:
Differences between the sound wave and light wave :
Light waves :

1. They are produced from the electrons in an excited state.
2. They travel at a very high speed of 3 x 108 ms-1 in air.
3. They can travel through vacuum.
4. Their velocity does not change with the change in temperature, humidity, etc.
5. They can produce the sensation of vision.

Sound waves :

1. They are produced due to vibrations of various objects in a material medium.
2. They travel in air at a very low speed of 332 ms-1 at 20°C.
3. They cannot travel through vacuum and always require some material medium for propagation.
4. Their velocity changes with the change in velocity, humidity, etc.
5. They can produce the sensation of hearing.

Question 12.
What is meant by the term wave motion?
Answer:
Wave motion : The transference of energy when the particles of a medium, move about their mean position is called wave motion.

Question 13.
State the relation between the wavelength and the frequency.
Answer:
Consider a wave is propagating through a medium,
Let,ƒ = Frequency of wave
λ = Wavelength of wave
T = Time period of the wave.
In the time ‘T’, distance covered by wave = λ
In the time 1 second, distance covered by wave = λ/T
But distance covered by wave in one second = v = Velocity of the wave

Which is the required relation between wavelength and frequency.

Question 14.
What kinds of the waves are produced in solids, liquids and gases?
Answer:
Elastic waves or material waves are produced in solids, liquids and gases.

Question 15.
The sound of an explosion on the surface of lake is heard by a boatman 100 m away and a diver 100 m below the point of explosion.

1. Of the two persons mentioned (boatman and diver), who would hear the sound first?
2. Give reason for your answer in (i).
3. If the sound takes ‘t’ seconds to reach the boatman, approximately how mcuh time it will take to reach the diver?

Answer:

1. Diver would hear the sound first.
2. Because velocity of sound in water (1450 ms^-1) is more than the velocity of sound in air (332 ms^-1).
3. Time taken by sound to reach the diver would be less than t, where t is the time taken by sound to reach the boatman.

Note : Time taken by sound to reach the diver would be 4.36 times less than the time taken by sound to reach the boatman.

Question 16.
What is approximate value of speed of sound in iron as compared to that in air? Illustrate your answer with a simple experiment.
Answer:
Speed of sound in iron is approximately sixteen times the speed of sound in air.
[Speed of sound in iron = 5100 ms^-1; Speed of sound in air=332 ms^-1]
Experiment : If we put our ears to rails, we can hear the sound of the train through metal. But at the same time, when we stand near by the railway track, we are not able to hear the sound.
This occurs because when we put our ears to rails, the sound travels through iron. But in standing position, sound travels through air and due to smaller speed of sound in air as compared to iron, we are not able to hear the sound.

Question 17.
How does a bat avoid obstacles in its way when in flight?
Answer:
Bats make a series of twittering sound, so high pitched that human ear can not hear. These sound waves strike against the obstacles less in their path of flight and send back echoes to the bat’s ear. The echoes tell the bats, how they must turn in the air to avoid colliding with the obstacles or with one another.

Question 18.
A continuous disturbance is created on the surface of water in a ripple tank with a small piece of cork floating on it. Describe the motion of the cork. What does the motion of the cork tell about the disturbance?
Answer:
Cork will move up and down about the mean position at the same position along horizontal.
It tells as that only the disturbance/energy is transferred from one particle to the other but particles of the medium do not move from one position to the other.

Question 19.
Draw a displacement-time graph for water wave.
Answer:
Displacement-time graph for water wave :

Practice Problems 1

Question 1.
200 waves pass through a point in one second. Calculate the time period of wave.
Answer:
Waves passing through a point in 1 second = 200
We know, frequency (ƒ) = Number of waves passing through point in 1 second.
∴ ƒ = 200Hz

Question 2.
A bat emits an ultrasonic sound of frequency 0.25 MHz. Calculate the time in which one vibration is completed.
Answer:
Frequency of ultrasound = ƒ = 0.25 MHz
ƒ = 0.25 × 10⁶ Hz

Practice Problems 2

Question 1.
The sonic boom of an aircraft has a time period of 0.00005s. Calculate the frequency of sound produced.
Answer:

Question 2.
An electromagnetic wave has a time period of 4 × 10^-8 s. Calculate its frequency in MHz.
Answer:
Time period = T = 4 × 10^-8 s
Frequency of sound = ƒ = ?

Practice Problems 3

Question 1.
An ultraviolet radiation has a wavelength of 300Å. If the velocity of electromagnetic wave is 3 × 10⁸ ms^-1. Calculate

1. frequency
2. time period.

Answer:

Question 2.
The wavelength of the vibrations produced on the surface of water is 2 cm. If the wave velocity is 16 ms^-1, calculate

1. no. of waves produced in one second
2. time required to produce one wave.

Answer:

Practice Problems 4

Question 1.
A continuous progressive transverse wave of frequency 8 Hz moves across the surface of a ripple tank

(a) With reference to the frequency, describe the movement of water on the surface
(b) If the wavelength of transverse wave is 32 mm, calculate the speed with which wave travels across the surface of water.

Answer:

Question 2.
A thin metal plate is placed against the teeth of cog wheel. Cog wheel is rotated at a speed of 120 rotations per minute and has 160 teeth. Calculate :

1. frequency of node produced.
2. speed of sound, if wavelength is 1.05 m.
3. what will be the effect when speed of cog wheel is doubled?

Answer:
Frequency for rotation of wheel = 120 rotations per minute

Practice Problems 5

Question 1.
A sound wave of wavelength 1/3 m has a frequency 996 Hz. Keeping the medium same, if frequency changes to 1328 Hz. Calculate

1. velcoity of sound
2. new wavelength.

Answer:

Question 2.
Two tuning forks A and B of frequencies 256 Hz and 192 Hz respectively are vibrated in air. If the wavelength of A is 1.25 m, calculate the wavelength produced by B.
Answer:

Practice Problems 6

Question 1.
The distance between one crest and one trough of a sea wave is 4.5 m. If the waves are produced at the rate of 240/min, calculate

1. time period
2. wave velocity.

Answer:
Distance between one crest and one trough of a sea wave = 4.5 m

Question 2.
The distance between three consecutive crests of wave is 60 cm. If the waves are produced at the rate of 180/ min, calculate

1. wavelength
2. time period
3. wave velocity.

Answer:

Practice Problems 7

Question 1.
The diagram given below shows a displacement distance graph of a wave. If the velocity of wave is 160 ms-1, calculate

1. wavelength
2. frequency
3. amplitude.

Answer:

Question 2.
From diagram given below calculate

1. velocity of P and Q
2. frequency of P, when frequency of Q is 512 Hz.

Assume that both wave are travelling in same medium.

Answer:

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Sound are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 2

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 2.

EXERCISE 
(A) Objective Questions

Multiple Choice Questions.
Select the correct option:

1. A bar magnet is rubbed on a bar of steel along its length 20 times. The bar of steel gets magnetised due to the process of :
(a) induction
(b) conduction
(c) friction
(d) none of the these

2.The magnetic strength of a bar magnet is :
(a) maximum at its centre
(b) same along the magnet
(c) maximum near its ends
(d) none of these

3. The surest test of  magnet   is :
(a) repulsion
(b) attraction
(c) induction
(d) none of these

4. Nickel is a :
(a) ferromagnetic substance
(b) paramagnetic substance
(c) diamagnetic substance
(d) none of these

5. The substance which form a strong temporary magnet is:
(a) steel
(b) platinum
(c) soft iron   
(d) manganese

6. The place around a magnet where its influence can be detected is called :
(a) magnetic lines of force
(b) magnetic pole
(c) magnetic field
(d) magnetic space

(B) Subjective Questions

Self Objective Questions

Question 1.
What do you meant by the term pole of a magnet? Magnetically speaking, what is the difference between a piece of brass, a piece of soft iron and a piece of lode- stone?
Answer:
Pole of a magnet: Each end of a bar magnet is called its pole. The point situated slightly inside a bar magnet, where most of its magnetic power is concentrated, is called magnetic pole or pole of a magnet.
Brass is not a magnetic substance and it is not affected by magnetic field. It does not stick to a magnet.
Soft iron is a ferromagnetic substance and gets strongly attracted towards a magnet. Soft iron can not attract other magnetic substances unless gets magnetised.
Lode stone is naturally magnetized piece of mineral magnetite. It can attract other magnetic substance.

Question 2.
(a) What are magnetic and non-magnetic substances? Give at least two examples of each.
(b) Fill the blank spaces in the table given below :

Answer:
(a) Magnetic substance : Those substances which are affected by the magnetic field are known as magnetic substances.
For example : Iron, nickel, cobalt etc. are the magnetic substances.
Non-magnetic substances : Those substance which are not affected by the magnetic field are known as non­magnetic substances.
For example : Paper, glass, wood etc.

Question 3.
Define : magnetic field, magnetic meridian, geographical meridian, declination and magnetic equator.
Answer:
Magnetic field : The space surrounding a magnet within which the magnet has its influence is called magnetic field.
Magnetic meridian: The vertical plane containing the magnetic axis of a free suspended magnet at rest, under the action of magnetic intensity of earth is called magnetic meridian. Geographical meridian : The vertical plane which contains geographical north and south poles of earth at a given place is called geographical meridian.
Decimation : The phenomenon due to which the earth’s geographical meridian is inclined to earth’s magnetic meridian is called declination.
Magnetic equator : An imaginary line right bisecting the effective length of bar magnet is called magnetic equator.

Question 4.
Why do lines of magnetic force never cross? Why do they never pass through a neutral point?
Answer:
No two lines of magnetic force cross each other because in that case there would be two directions of resultant magnetic force at a given point, which is not possible.
Magnetic lines of force never pass through neutral point because at neutral point magnetic field due to bar magnet is neutralised by earth’s magnetic field.

Question 5.
Define : Isogonic line, agonic line isoclinic line.
Answer:
Isogonic lines : A line which joins all the places on earth, having same angle of declination is called isogonic line.
Agonic line: A line which joins all the places on earth, having zero angle of declination is called agonic line.
Isoclinic line : A line joining all the places on globe, having same angle of dip or inclination is called isoclinic line.

Question 6.
How do you account for the following facts?
(a) Iron becomes magnetised when placed in a coil carrying direct current.
(b) Bar magnets lose their magnetism when heated strongly.
(c) Steel makes better permanent magnet than soft iron.
(d) Soft iron keepers help to prevent the magnets from losing their magnetic properties.
Answer:
(a) Iron is a magnetic substance and hence its each atom behaves as a tiny magnet. When iron piece is placed in a coil carrying direct current, then all the north poles of all the atoms of iron will align themselves in one direction and all the south poles of all the atoms of iron align themselves in a direction opposite to that to which their north poles point. As a result, iron piece gets magnetised.

(b) Bar magnets lose their magnetism when heated strongly. Due to heat energy, the kinetic energy of the molecules of a barg magnet increases. Thus from straight line molecular chains, they form closed molecular chains and hence, magnetism is lost.

(c) Steel makes better permanent magnet than soft iron because on magnetising steel, steel retain their magnetic behaviour for longer time even after the removal of source which is magnetising the steel.
While the soft iron retains the properties of magnetism only so long as the current is passing through the coil i.e. as long as the source which is magnetising the soft iron is present.

(d) In magnets, external fields like earth’s magnetic field can randomize the domains. Perhaps stray fields caused by flowing currents in near by electric circuits can also disturb i   the alignment of domains lying inside a magnet. Given enough time, such magnets may find their domains randomly oriented and hence their net magnetisation may get lost. A keeper for magnets is just a strong permanent magnet that keeps all the domains pointing the same way and realign those that may have gone stray and hence magnet, can retain its magnetism for a long time.

Question 7.
State briefly (a) the molecular theory of magnetism, (b) the modern views on magnetism.
Answer:
(a) Ewing suggested the molecular theory of magnetism as , follows:

1. Each molecule of a magnetic substance, whether it is magnetised or unmagnetised, is an independent magnet.

2. In a magnetised substance, the molecules are arranged in an order so as to produce an external effect. In this order, all the north poles of the molecules of the magnetised substances point to one direction and all their south poles point to a direction opposite to that to which their north poles points.

3. In an unmagnetised substance, the molecules are not arranged in any order, so they neutralise the magnetic forces of each other.

4. The molecular theory of magnetism was a considerable step forward but later there came an electrical explanation for the magnetism of atoms. Atoms consist of negatively charged particles (electrons) which revolve around the positively changed nucleus. Electrical current loops are formed in an atom due to the circulation of these electrons. Each current loop behaves a magnetic dipole and hence produce magnetic field. Also electrons are also spinning like tops and this adds further magnetism to the atom.

Question 8.
Describe various methods of magnetising a piece of iron.
Answer:
METHODS OF MAGNETISATION :

1. Single Touch Method : The specimen to be magnetised is placed flat on the table. A permanent bar magnet is taken and its one pole is placed on one end of the piece. The bar magnet is then drawn to the other end, keeping it in the inclined position as shown in figure. The permanent magnet is then lifted and the process is repeated several times. The specimen is then turned over and the other side is also magnetised in the same way.

The specimen gets magnetised. Its starting end gets the same polarity as the polarity of the magnet touching it. The polarity at the other end of the specimen is opposite to that of the magnetising pole.

2. Divided Touch Method: The specimen to be magnetised is placed flat on a table. Opposite poles of two strong bar magnets of equal strength are placed together in the middle of the specimen. The ends of bar magnets are drawn towards the opposite ends of the specimen, keeping the bar magnets inclined as shown in figure. The magnets are then lifted. The operation is repeated several times

The specimen is then turned over and the other side is also magnetised in the same way. The end of the specimen where the south pole of the bar magnet leaves, becomes north pole. Similarly, the end of the specimen where the north pole of the bar magnet leaves, becomes south pole. For strong magnetisation, the two ends of the specimen are supported on teh two poles of two other bar magnets, such that the pole of each magnet being the same as that of the stroking magnet over it.

3. Double Touch Method : This method is almost similar to the divided touch method. The only difference is that a piece of wood or cork is placed between the two opposite poles of the permanent magnets. The magnets are then moved together from the middle to the one end and then to the other end without lifting them from the specimen as shown in figure. This process is repeated several times.
The polarities on the end of the specimen are of the opposite nature to that of the nearer poles of stroking magnets figure. For strong magnetisation, the specimen is mounted on two permanent bar magnets as mentioned in the divided touch method.

4. Electrical Method of Magnetisation : The specimen to be magnetised is placed inside a long coil of insulated copper wire. A strong direct current is passed through the coil for some time, when the specimen is magnetised.
If the specimen is a steel bar, it becomes a permanent magnet. However, if the specimen is a soft iron bar, it becomes a strong magnet, but it retains the properties of magnetism only so long as the current is passed through the coil. As soon as the current is stopped, it loses its magnetism. The magnet formed by the passage of electric current by using soft iron core is called electromagnet.

Question 9.
What is magnetic induction? Explain it giving a suitable experiment.
Answer:
Magnetic induction :
The Phenomenon due to which a piece of steel or iron behaves like a magnet when placed near a strong magnet is called magnetic induction.
Experiment : Take a freely suspended magnetic needle and bring near its south pole, the south end of a bar magnet. The needle gets longer affects the south and of magnetic needle as shown in figure (a).

Place a flat piece of iron AB, in between the bar magnet and magnetic needle. It is observed that south end of needle is repelled. Remove the iron piece AB. It is observed that needle . comes back to its original position. Repeat the experiment, but remove bar magnet instead of iron piece. We will observe that needle does not get repelled and remains continuously in its original position.
From this experiment, it is clear that soft iron piece behaves as a magnet only when a bar magnet is placed near it, when a bar magnet is removed then soft iron piece loses its magnetism.

Question 10.
Repulsion is a surer test of magnetic condition of a body than attraction. Explain.
Answer:
Repulsion is the surest test of magnetism because the attraction can be caused between two unlike poles of the two magnets or between the magnet and magnetic substance such as iron, nickel etc. But repulsion is caused when two similar poles approach each other.

Question 11.
There are two knitting needles. One of them is magnetised. How will you find out which one is magnetised, if no other magnet is available?
Answer:
When an iron bar is magnetised, it slightly increases in length due to setting of molecular magnets along straight chains. So, on precisely measuring the length of knitting needle, the knitting needle which is slightly longer in length than other is magnetised.

Question 12.
Describe two methods of determining the arrangement of the lines of force in the field close to a bar magnet. Give a brief explanation of each method.
Answer:
First method : Place a card board on the top of a bar magnet and scatter some iron filings uniformly over the whole of card board. Now tap it with a pencil. The filing are magnetised by induction and arrange themselves in curved lines as shown in figure. The curved lines are called the magnetic lines of force which may be defined as the lines in a magnetic field along which free magnetic poles tend to be driven if free to do so.

Second method : Lines of force can be traced on a paper by using tracing needle or small magnetic compass needle.
The bar magnet is placed on a sheet of paper and its boundary is drawn with a sharp pencil. A point X is marked along the boundary towards the north of the magnet.

The tracing needle is then placed at point X in such a way that its one end points towards the point X. With the help of pencil the direction of other end of needle is marked on paper. Let it be point Y.
Now, shift the needle from the point X and place it in such a way that its one end points towards the point Y. The direction of other end of needle is marked by pencil. Let it be point Z. The process is continued till a closed curve is obtained.

This curve is called magnetic line of force.

If we plot a number of such curves around the magnet, starting from different points then the space so enclosed is called magnetic field.

Question 13.
Draw diagrams showing the arrangements of the lines of force for:
(a) a single magnet.
(b) two magnets in line, with unlike poles facing one another.
(c) a piece of soft iron laid in line with magnetic field.
Answer:
(a) Arrangement of the lines of force single magnet :

(b) Arrangement of lines of force for two magnets in line, with unlike poles facing one another :

(c) Arrangement of lines of force for a piece of soft iron laid in line with the magnetic field:

AB is a soft iron rod whose end A behaves as south pole and end B behaves a north pole when soft iron rod is placed in a line with magnetic field.

Question 14.
Give short account of the earth’s magnetic field.
Answer:
When a bar magnet is suspended freely, then it aligns itself along geographical north-south direction i.e. north pole of the magnet points towards the geographical north and south pole of the magnet points towards geographical south direction. William Gilbert suggest that earth itself behaves as a huge magnet. It was assumed that:

1. A huge magnet is bured at the centre of earth.
2. The south end of earth’s magnet is towards the earth’s geographic north and vice-versa.
3. The axis of earth’s magnet is not in line with the geographical axis, but makes a small angle with it.
   The diagram on next page show the earth as a magnet and the magnetic lines of force around it

Question 15.
Give the various methods for demagnetising a magnet.
Answer:
1. A magnet can be damagnetised by any of the following methods:
Electrical Method : An insulated copper coil is wound around a card board tube and inside it is placed a permanent bar magnet. The coil is placed in East-West direction and its ends are connected to a step-down transformer. The alternating current is switched on for one minute, then the bar magnet gets demagnetised.

Reason : When the current rapidly changes the direction in the insulated copper coil, the polarity set up in the coil also rapidly changes. This in turn acts inductively on the bar magnet, whose molecular magnets rapidly try to align themselves with the changing magnetic polarity. Thus, molecular magnets form closed chains and hence, the bar magnet gets demagnetised.

2. By Rough Handling : When a magnet is rough-handled (i.e. it is allowed to drop repeatedly) or hammered, it gets completely demagnetised.
Reason : In a bar magnet the molecular magnets are arranged in straight line chains. On hammering or rough handling, they gain kinetic energy and vibrate rapidly about their mean positions. In doing so they form closed magnetic chains and hence, the magnetism is lost.

3. By Heating : When a magnet is heated to red hot temperature and then allowed to cool, it loses its magnetism. Reason : Due to heat energy, the kinetic energy of the molecules increases. Thus, from straight line molecular chains, they form closed molecular chains and hence, the magnetism is lost.

4. By Induction : When a given magnet is placed in contact with another similar magnet (i.e., the other magnet should be of same strength), such that their similar poles are facing each others then both the magnets get demagnetised in a couple of days.
Reason: It is because both the magnets will induce opposite polarity in each other. In doing so the molecular magnets in each magnet form closed molecular chains and hence, they get demagnetised.

5. By Self-Induction : A single bar magnet has a tendency to lose its magnetism.
Reason : In a bar magnet the molecular magnets (dipoles) are arranged in straight line chains. However, two or more parallel chains have their north and south poles facing each other. Thus, dipoles act inductively on each other and hence, turn to form closed molecular chains. Thus, single bar magnet gets demagnetised.

Question 16.
Describe two simple experiments to support the statement that magnetism is a property of the molecules of a magnet.
Answer:
Magnetism is a property of molecules of a magnet.
Experiment-1 : Take a bar magnet and break it into as small parts as possible. We shall find that each small part retain the original polarity i.e. even the smallest part of the magnet has its own north and south pole and can attract the magnetic substances.
It shows that, if it were possible to break a magnet into its molecules then each molecule would retain the property of a magnet.

Experiment-2 : Take a soft iron bar. Place it in a magnetic field. We shall find that there is small increase in its length on getting magnetised.
A soft iron bar becomes magnetised and get lengthened only if each molecule of soft iron bar behave as a magnet. Due to setting of molecular magnets along straight chains, there is slight increase in length of soft iron bar. So we can say that magnetism is a property of molecules of a magnet.

Question 17.
Explain, why steel is used in preference to soft iron for making permanent magnets while soft iron is used in preference to steel for making electromagnets.
Answer:
Steel is used in preference to soft iron for making permanent magnets because steel acts as a magnet, even on the removal of inducing magnet and also steel has a very high retention. Soft iron is used in preference to steel for making permanent magnets because soft iron behave like magnet as long as there is an inducing magnet and also soft iron has a very poor magnetic retention.

Question 18.
Describe, how you will proceed to determine the position of the pole of a bar magnet.
Answer:
Fix a white sheet of paper on a wooden drawing board and in the middle of it draw a straight line. On the straight line place a magnetic needle. Turn the drawing board in the clockwise or anticlockwise direction, till the magnetic needle is in line with

line drawn on drawing board. At this point drawing board is in magnetic meridian. Remove the magnetic needle. Place a bar magnet such that its axis coincides with this line. Mark the outline of the magnet with a pencil. Place the compass needle near one end of the bar magnet. In this position, the action of the earth’s field is ineffective in deflecting the needle and the compass needle is acted on by the nearest pole only. Market the two ends of the needle by two dots A and B as shown in figure.

Change the position of the compass needle and repeat the process. Join the two marks A, B and A,, B, by straight lines. It will be found that the straight lines, when produced, intersect at a point near the end of the magnet. This point of intersection indicates the position of the magnetic pole. Similarly, the other pole may be ascertained in the same way as described above. The length between the two poles is called the effective length of the magnet which is found to be about 0.84 times the actual length of the magnet.

Question 19.
Draw lines of force surrounding a bar magnet when it is placed in the magnetic meridian with its
(a) north pole pointing geographic, north
(b) north pole pointing geographic south.
Answer:
(a) Bar magnet placed in magnetic meridian with its north pole pointing geographic north.

(b) Bar magnet placed in magnetic meridian with its north pole pointing geographic south.

Bar magnets are often stored in pair as shown in the figure. E and F being pieces of metal.

1. Name the metal used for E and F.
2. Why are E and F placed in contact with the poles of the magnets as shown in the diagram?
3. Mark on the diagram the poles of the second magnet,
4. What is the material of darkened part?
   

Answer:
1. Soft iron is used for E and F.

2.E and F are placed in contact with the poles of magnets to preserve the strength of magnet.

3. Darkened part is a non-magnetic material like wood.

Question 21.
The figure shows a freely suspended magnet in rest position. Copy the diagram and on it show : 
(a) Angle of declination
(b) Angle in dip

Answer:

Question 22.
(a) Since every iron atom is a tiny magnet, why are not all iron bar magnets?
(b) If a magnet is carefully broken into two pieces as shown in figure (i), how does the magnetic strength of each piece compare with that of original magnet? If another magnet is carefully broken in half along its long axis shown in figure (ii), how would the strength of each piece compare with that of original magnet?
Answer:
(a) Each molecule of a magnetic substance is an independent magnet. But in an unmagnetised iron piece, molecular/ atoms (tiny magnets) are not arranged in any order and hence they neturalise the magnetic forces of each other As a result, any unmagnetised iron piece can not behave as a magnet.
(b)
1. When a magnet is cut into two equal parts as shown in figure (i), then pole strength of each piece remains same as that of original magnet.

2. When a magnet is cut into two equal parts as shown in figure (ii), then pole strength of each pole is half as that of the pole strength of original magnet.

Question 23.
Draw the magnetic flux pattern near a bar magnet placed with its axis in the magnetic meridian and the south pole pointing towards geographic north.
Answer:

Question 24.
Draw a clearly labelled diagram, to show how a steel bar is magnetised by a divided touch method. A written description is not required.
Answer:

Question 25.
Define the terms magnetic declination and dip with reference to freely suspended magnet.
(a) What do you understand by the terms magnetic meridian and geographic meridian?
(b) At what places on the earth will the angle of dip be (1) maximum and (2) minimum?
Answer:
Magnetic declination : The angle through which freely suspended magnetic needle is inclined to the geographic axis is known as magnetic declination.
OR
The angle between the geographic meridian and magnetic meridian it is given place is called declination.

Magnetic dip : The angle between the horizontal axis passing through freely suspended magnet and the direction of earth’s magnetic field is called magnetic dip.
(a) Magnetic meridian : The vertical plane containing the magnetic axis of a freely suspended magnet at rest, under the action of magnetic intensity of earth is called magnetic meridian.
Geographic meridian : The vertical plane which contains geographic north and south pole of earth at a given place is called geographic meridian.
(b) The angle of dip is maximum e. 90° at the magnetic poles. The angle of dip is minimum i.e. 0° at the magnetic equator.

Question 26.
(a) What are magnetic keepers? What are they used for?
(b) Explain the ‘molecular theory’ of magnetism with the help of a diagram.
Answer:
(a) Magnetic keeper : A magnetic keeper is a ferromagnetic bar made from soft iron or steel, which is placed across the poles of a permanent magnet.Magnetic keepers are used to preserve the strength of the magnet by completing the magnetic circuit.
(b) Ewing suggested the molecular theory of magnetism as follows:
1. Each molecule of a magnetic substance, whether it is magnetised or unmagnetised, is an independent magnet.

2. In a magnetised substance, the molecules are arranged in an order so as to produce an external effect. In this order, all the north poles of the molecules of the magnetised substances point to one direction and all their south poles point to a direction opposite to that to which their north poles points.

3. In an unmagnetised substance, the molecules are not arranged in any order, so they neutralise the magnetic forces of each other.

Question 27.
What do you understand by the term magnetic declination?
Answer:
Magnetic declination : The angle through which freely suspended magnetic needle is inclined to the geographic axis is known as magnetic declination.
OR
The angle between the geographic meridian and magnetic meridian it is given place is called declination.

Magnetic dip : The angle between the horizontal axis passing through freely suspended magnet and the direction of earth’s magnetic field is called magnetic dip.

(a) Magnetic meridian : The vertical plane containing the magnetic axis of a freely suspended magnet at rest, under the action of magnetic intensity of earth is called magnetic meridian.
Geographic meridian: The vertical plane which contains geographic north and south pole of earth at a given place is called geographic meridian.
(b) The angle of dip is maximum i. e. 90° at the magnetic poles. The angle of dip is minimum i.e. 0° at the magnetic equator.

Question 28.
(a) Explain the mechanism by which unmagnetised iron nails get attracted to a magnet when brought near it.
(b) State any two properties of magnet.
Answer:
(a) Every atom of an iron nail behaves as a tiny magnet. Due to the random orientations of these tiny magnets, iron nail does not behave as a magnet.
But when iron nail is placed near a magnet, then due to induced magnetism, all the atoms (tiny magnets) align themselves in a particular direction. As a result, the end of the iron nail nearer to magnet acquires the opposite polarity and hence get attracted towards the magnet.
(b) Properties of a magnet :

1. Freely suspended magnet always align itself in the geographic north and geographic south direction.
2. Like poles of magnets repel each other while unlike poles attract.

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given  A New Approach to ICSE Physics Part 1 Class 9 Solutions Electricity and Magnetism – 2 are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.

 

A New Approach to ICSE Physics Part 1 Class 9 Solutions Light

These Solutions are part of A New Approach to ICSE Physics Part 1 Class 9 Solutions. Here we have given A New Approach to ICSE Physics Part 1 Class 9 Solutions Light.

Unit I
Exercise 1

Question 1.
(a) What do you understand by the following terms?

1. Light
2. Diffused light

(b) By giving one example and one use explain or define

1. Regular reflection
2. Irregular reflection.

Answer:
(a)
(i) Light : Light is a form of energy which produces in us sensation of seeing.
(ii) Diffused light : Light obtained after reflection from rough surface is known as diffused light.
It is a soft light with neither the intensity nor the glare of direct light. It is scattered and comes from all directions. It does not cause harsh shadows.
(b)
(i) Regular reflection : The phenomenon due to which a parallel beam of light travelling through a certain medium, on striking some polished surface, bounces off from it, as parallel beam, in some other direction is called regular reflection.
For example : Reflection taking place from the objects like looking glass, still water, oil, highly polished metals is regular reflection.
Regular reflection is useful in the formation of images. We can see our face in a mirror only due to regular reflection.
(ii) Irregular reflection or Diffused reflection : The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called irregular reflection.
For example : Reflection taking place from ground, walls, trees, suspended particles in air is irregular reflection.
Use : It helps in the general illumination of places and helps us to see things around us.

Question 2.
By drawing a neat diagram define the following :

1. Mirror
2. Incident ray
3. Reflected ray
4. Angle of incidence
5. Angle of reflection
6. Normal

Answer:

1. Mirror is a highly polished and smooth surface which reflects almost the entire light falling on it. A plane mirror is made by silvering one side of a glass plate as shown in figure.
   
2. Incident ray : The light ray striking a reflecting surface is called the incident ray.
3. Reflected ray : The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.
4. Angle of incidence : The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter “i”.
5. Angle of reflection : The angle which the reflected ray makes with the normal at point of incidence, is called the angle of reflection. It is denoted by the letter “r”.
   
6. Normal : The perpendicular drawn at the point of incidence, to the surface of mirror is called normal.

Question 3.
State the laws of reflection.
Answer:
Laws of reflection :

1. The incident ray, the reflected ray and the normal ray at the point of incidence, lie in the same plane.
2. The angle of incidence i is equal to the angle of reflection r i.e. ∠i = ∠r.

Question 4.
A ray of light strikes a plane mirror, such that angle with the mirror is 20°. What is value of angle of reflection? What is the angle between the incident ray and the reflected ray?
Answer:
∵ Light ray makes an angle of 20° with the mirror
∴ ∠ABM = 20°
Angle of incidence = ∠i = 90° – 20°
∠i = 70°
∵ ∠i = r ∴ ∠r = 70°

Angle between incident ray and reflected ray = ∠i + ∠r = 70°+ 70°= 140°

Question 5.
Prove experimentally that images are formed as far behind in a plane mirror as the object is in front of it.
Answer:
Consider an object‘O’ situated in front of a plane mirror MM,. A ray of light which starts from point ‘O’ perpendicularly, is reflected back along the same path (See figure).
However, another ray which moves along OB is reflected along BC, obeying the laws of reflection, such that BN is the normal. Produce OA and CB backward, such that they meet at point I. Then ‘I’ is image of ‘O’.

Thus, in particular OA = IA.

Question 6.
Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
Answer:
Consider a ray of light AB, incident on plane mirror in position MM’, such that BC is the reflected ray and BN is the normal.
∠ABM = ∠CBN = ∠i
∠ABC = 2 ∠i …(i)
Let the mirror be rotated through an angle ‘0’ about point B, such that M₁M₁ is the new position of the mirror and BN₁ is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is ∠ABN₁ whose magnitude is (i + θ). Let BD be the reflected ray, such that ∠DNB₁ is the new angle of reflection.

Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.

Question 7.

(a) What do you understand by the term lateral inversion?
(b) A printed card has letters PHYSICS. By drawing the diagram show the appearance of the letters. (No ray diagram is required).

Answer:
(a) Lateral Inversion. The phenomenon due to which the image of an object turns through an angle of 180° through verticle axis rather the horizontal axis, such that right side of image appears as left or vice-versa is called Lateral Inversion. During lateral inversion the left side of object appears as right side of image and vice-versa. In a way the image turns through the angle of 180° about vertical axis.

Question 8.

(a) State the mirror formula for the formation of total number of images formed in two plane mirrors, held at an angle.
(b) Calculate the number of images formed in two plane mirrors, when they are held at the angle of (i) 72° (ii) 36°.

Answer:
(a) If θ = Angle of inclination between two mirrors
n = number of images formed

Numeral one is subtracted because of the loss of one image due to overlapping of the images.
(b) (i) When θ = 72°

Question 9.
Draw a neat two ray diagram for the formation of images in two plane mirrors, when mirrors are

1. at right angles to each other
2. facing each other.

Answer:
(a) When two mirrors are inclined at right angles
‘O’ is an object placed in between two mirrors XY and
XZ, inclined at an angle of 90°. (See figure)

Taking normal incidence, I₁ and I₂ are the images formed in the plane mirror XY and XZ respectively as far behind the mirrors, as point ‘O’ is in front of them.
However, image I₁ acts as a virtual object for image mirror XZ₁ and forms an image I₃. Similarly, image I₂ acts as a virtual object for the image mirror XY₁ and forms the image I₄. The images I₃ and I₄ overlap to form a very bright image. Thus, on the whole three images are seen. In order to draw two-ray diagrams, from the position FE of the eye, draw two rays meeting at I₃,I₄ such that these ray intersect the mirror XZ at D and C.
Now draw two rays from point I₁ to join C and D intersecting mirror XY at A and B. Join O with A and B.
Similarly, in order to show image I₂, draw two rays from I₂ to the position of eye FE, such that they intersect at H and G Join H and G to ‘O’ so as to form incident beam.
(b) When two mirrors are parallel to each other

Consider two plane mirrors XY and PQ facing each other and ‘A’ as an object situated anywhere between them (See figure).
It is clear that mirror XY forms its image in mirror PQ and vice versa. These images by themsleves will act as image mirror or virtual mirrors.
Let us consider the normal incidence towards the mirror XY for the object A. First of all an image X₁ is formed as far behind, as the object is in front of it. This image X₁ will fall in front of mirror PQ, and hence, forms an image X₂. The image X₂ falls in front of mirror XY and hence, forms an image X₃. Thus, it continues and infinite images can be formed.
Similarly, taking normal incidence for mirror PQ image P₁,P₂,P₃ etc. are formed.
In order to draw ray diagram, from point A, draw a divergent beam, meeting mirror PQ at points 1 and 2. With P₁ as reference point, draw rays 1, 3 and 2, 4 meeting mirror XY. With P₃ as reference point draw rays, such that they enter the eye.

Question 10.
Why are infinite images not seen when two plane mirrors are facing each other?
Answer:
Infinite images are not seen when two plane mirrors are facing each other because :

1. After every successive reflection, some amount of light energy is absorbed. Thus luminosity of image goes on decreasing, till they are no longer visible.
2. As the distance of images from the eye goes on increasing, it is unable to resolve far off images.

Question 11.

(a) State four characteristics of the image formed in a plane mirror.
(b) State three ways in which the image formed in a plane mirror differs from the image formed in a pin hole camera.

Answer:
(a) Characteristics of Image formed by a Plane Mirror.

1. Image is of the same size as that of object.
2. Image is laterally inverted.
3. It is upright.
4. It is virtual.
5. Image formed is as far behind the mirror as the object infront of the mirror.

(b) Image formed by plane mirror :

1. Image formed by plane mirror is virtual.
2. It can not be obtained on the screen.
3. Size of image is same as that of the object.
4. Image appears to be inverted only on the vertical axis.

Image formed by Pinhole camera :

1. Image formed by pinhole camera is real.
2. It can be obtained on the screen.
3. Size of the image is smaller than the object.
4. Image appears to be inverted on both the vertical axis and horizontal axis.

Question 12.
(a) What should be the minimum size of a plane mirror, so that a person 182 cm high can see himself completely?
(b) A boy stands 4 m away from plane mirror. If the boy moves 1/2 m towards mirror, what is now the distance between the boy and his image? Give a reason for your answer.
Answer:
(a) Size (height) of a person = 182 cm
We know to see the full length image of a person, we need a plane mirror of size as half of the size of the person.
So, size of plane mirror required = Height of person/2 = 182/2 = 91 cm
(b) Distance of boy from the plane mirror = u = 4 m When boy moves 1/2 m towards the mirror.
Then, distance between boy and mirror = 4 – 1/2 = 7/2 m
We know, in a plane mirror, the image formed is as behind the mirror as the object is in front of it.
So, distance between the mirror and image of boy = 7/2 m
Distance between the boy and his image
= Distance of boy from mirror + Distance of image of boy from mirror

Question 13.
State four uses of a plane mirror.
Answer:
Four uses of plane mirror are :

1. Plane mirrors are used in construction of reflecting periscope.
2.  They are used as looking glass.
3. They are used in solar cookers for reflecting the rays of the sun into the cooker.
4. They are used for signalling purpose.

Question 14.

(a) Draw a neat diagram of reflecting periscope.
(b) State two advantages and two disadvantages of the reflecting periscope.

Answer:
(a)

(b) Advantages of Reflecting Periscope :

1. It is used to see above the head of crowds.
2. It is used by soldiers in trench warfare.

Disadvantages of Reflecting Periscope :

1. The final images is not brightly illuminated as light energy is absorbed due to two successive reflections.
2. Any deposition of moisture of dust on the mirror reduces the reflection almost to nil, and hence, the periscope cannot be used in places where there is a lot of dust or moisture.

Question 15.
What must be the minimum length of a plane mirror in which a person can see himself full length? Draw a diagram to justify your answer. Does the distance of person from the mirror affect the above answer?
Answer:
Minimum Height Of Plane Mirror Required For A Person To See Full Length Consider a person AB, such that A represents the highest point on his head, and B the lowest point on the foot such that E is the fixed eye level (see figure). The person will be able to see every part of his body if the can see points A and B. Let MN be the minimum length of mirror fixed on the wall, such that the rays AM and BN, after reflection, reach the eye of person, thereby forming an image A₁B₁ when produced backward.

In ∆AEA₁, CM is parallel to AE and C is the mid-point of AA₁. M is the mid-point of A₁E.
Similarly, in ∆BEB₁, ND is parallel to BE and D is the mid-point of BB₁
∴ N in the mid-point of B₁E.
Now in ∆A₁B₁E, M is mid-point of A₁E and N is mid-point of B₁E.
∴ MN is paralle and half of A₁B₁
But, A₁B₁ = AB
So, MN = 1/2 AB.
Thus, in order to see full length, a person requires a plane mirror which is half his own height. This relation is true for any distance of object from plane mirror.

Question 16.
An insect is sitting in front of a plane mirror at a distance of one metre from it.

1. Where is the image of insect formed?
2. What is the distance between insect and its image?
3. State any two characteristics of image formed in a plane mirror.

Answer:
(i) Distance of insect from plane mirror = u = 1 m
We know in case of plane mirror, distance of object from plane mirror is equal to distance of image of the object from the plane mirror.
∴Distance of the image of insect from plane mirror = v = u – 1 m
(ii) Distance between insect and its image = u + v = 1 + 1 = 2 m
(iii) Characteristics of the image formed by a plane mirror :

(a) Image formed is virtual.
(b) Image formed is erect.
(c) Image formed is of same size as that of the object.

Question 17.
(i) Draw a diagram to show reflection of a ray of light using plane mirror. In the diagram label the incident ray, the reflected ray, the normal, the angle of incidence and angle of reflection. (ii) State the laws of reflection.
Answer:

1. See Q. No. (2) of Exercise (1).
2. See Q. No. (3) of Exercise (1).

Question 18.
(i) Parallel rays are incident :

1. on regular surface and
2. on irregular surface. In what respect do reflected rays in (1) differ from those of (2)?

(ii) Write down four characteristics of image formed in a plane mirror.
Answer:
(i) Parallel rays of light after reflection from a regular surface . goes in a particular direction while after reflection from irregular surface go in different directions.
(ii) Characteristics of Image formed by a Plane Mirror.

(a) Image is of the same size as that of object.
(b) Image is laterally inverted.
(c) It is upright.
(d) It is virtual.
(e) Image formed is as far behind the mirror as the object in front of the mirror.

Question 19.
How many images will be formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other? Why do more distant images appear fainter?
Answer:
Angle between two mirrors facing each other = θ = 0°

So, infinite number of images are formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other.
More distant images appear fainter when two plane mirror with their reflecting surfaces facing each other because,
After every successive reflection, some amount of light energy is absorbed. Thus luminosity of images goes on decreasing and hence they appear fainter.

Question 20.

(a) Write down the letters of the word ‘POLEX’ as seen in a plane mirror, held parallel to the plane of this paper.
(b) Distinguish between real and virtual image.

Answer:
(a)

(b)
Real image :

1. Can be taken on the screen.
2. It is inverted.
3. It is formed when light rays after reflection (or refraction) actually meet.

Virtual image :

1. Cannot be taken on screen.
2. It is erect.
3. It is formed when rays of light after reflection (or refraction) appear to meet.

Question 21.

(a) Describe the principle of simple periscope through an outline ray diagram. Give one of its uses.
(b) Draw diagrams to show difference between regular and irregular reflection.

Answer:
(a) Simple periscope is based upon the principle of reflection. It consists of a cardboard or wooden tube, bent twice at right angles and is provided with two openings as shown in figure. Two plane mirrors are fixed at the bends of the tube at an angle 45° to the framework, such that the mirrors face each other. The tube is completely blackened from inside to avoid any reflection from its sides.

The parallel rays coming from an object at a higher plane, strike the plane mirror at an angle of 45° and hence are reflected through an angle of 45°.
These reflected rays strike the second mirror at an angle of 45° and hence are further reflected through an angle of 45°. These reflected rays on reaching the eye form the image on retina.
Use of periscope : It is used by soldiers in trench warfare.
(b)

Question 22.
An object is placed 2 cm from a plane mirror. If the object is moved by 1 cm towards the mirror, what will be the distance between the object and its new image?
Answer:
Distance of the object from plane mirror = 2 cm
If object is moved by 1 cm towards the mirror, then distance between the object and plane mirror = u = 1 cm
We know in case of plane mirror,
Distance of object from plane mirror = Distance of the image of the object from plane mirror
⇒ v = u = 1 cm
∴ Distance between its object and new image = u + v = 1 + 1 = 2 cm

Unit II

Practice Problem 1

Question 1.
An object 3 cm high produces a real image 4.5 cm high, when placed at a distance of 20 cm from a concave mirror. Calculate :

1. the position of image
2. focal length of the concave mirror.

Answer:
Size of object = h_o = 3 cm
Size of image = h₁ = -4.5 cm [Image formed is real]
Distance of the object from the concave mirror = u = – 20 cm
Distance of the image from the concave mirror = v = ?
Focal length of the concave mirror = f = ?

Practice Problem 2

Question 1.
An object 1.5 cm high when placed in front of a concave . mirror, produces a virtual image 3 cm high. If the object is placed at a distance of 6 cm from the pole of the mirror, calculate :

1. the position of the image
2. the focal length of the mirror.

Answer:

Question 2.
A converging mirror, forms a three times magnified virtual image when an object is placed at a distance of 8 cm from it. Calculate :

1. the position of the image
2. the focal length of the mirror.

Answer:
Let size of the object = h_o = x
Size of image = h₁= +3 x (Virtual image)
Distance of the object from the mirror = u = -8 cm
Distance of the image from the mirror = v = ?
Focal length of the mirror = f = ?

Practice Problems 3

Question 1.
An object 5 cm high forms a virtual image of 1.25 cm high, when placed in front of a convex mirror at a distance of 24 cm. Calculate :

1. the position of the image
2. the focal length of the convex mirror.

Answer:
Size of the object = h₀ = 5 cm
Size of the image = h₁ = +1.25 cm
Distance of the object from mirror = u = – 24 cm
Distance of the image from the mirror = v = ?
Focal length of the concave mirror = f = ?

Question 2.
An object forms a virtual image which is 1/8th of the size of the object. If the object is placed at a distance of 40 cm from the convex mirror, calculate :

1. the position of the image
2. the focal length of the convex mirror.

Answer:

Exercise 2

(A) Objective Questions

Multiple Choice Questions.
Select the correct option:

1. A concave mirror is made by cutting a portion of a hollow glass sphere of radius 30 cm. The focal length of the concave mirror is :
(a) 24 cm
(b) 12 cm
(c) 15 cm
(d) 60 cm
Ans. (c) 15 cm
Explanation :
Radius of curvature = R = 30 cm

2. A mirror forms a virtual image (diminished) of an object, whatever be the positiion of object :
(a) it must be a concave mirror
(b) it must be a convex mirror
(c) it must be a plane mirror
(d) it may be (b) or (c) or both
Ans. (b) it must be a convex mirror

3. A ray of light is incident on a concave mirror. If it is parallel to principal axis, the reflected ray will :
(a) pass through its principal focus
(b) pass through its centre of curvature
(c) pass through its pole
(d) retraces its path
Ans. (a) pass through its principal focus

4. If an incident ray passes through the centre of curvature of a spherical mirror, the reflected ray will :
(a) pass through its pole
(b) retraces its path
(c) pass through its focus
(d) be parallel to principal axis
Ans. (b) retraces its path

5. In case of concave mirror, the minimum distance between an object and its real image is :
(a) f
(b) 2f
(c) 4f
(d) zero
Ans. (d) zero

6. Looking into a mirror one finds her image diminished, the mirror is :
(a) concave
(b) convex
(c) cylindrical
(d) parabolic
Ans. (b) convex

7. Which mirror is used in periscope?
(a) Convex mirror
(b) Concave mirror
(c) Plane mirror
(d) Parabolic mirror
Ans. (c) Plane mirror

(B) Subjective Questions

Question 1.
Define the following terms :

1. spherical mirror
2. convex mirror
3. concave mirror

Answer:

1. Spherical mirror : “A mirror which is made from a part of a hollow sphere is called Spherical Mirror.
2. Convex mirror : “A mirror made by silvering the inner surface such that reflection takes place from the bulging surface” is called Convex Mirror.
   Centre of curvature is towards the silvered surface.
3. Concave mirror : “A mirror made by silvering the outer or the bulging surface such that the reflection takes place from the concave surface.” Centre of curvature is towards the reflecting surface.

Question 2.
Define the following terms in relation to concave mirror.

1. Pole
2. Centre of curvature
3. Principal axis
4. Principal focus
5. Focal length
6. Radius of curvature
7. Aperture

Answer:

1. Pole : Pole “is the mid-point of the mirror”.
2. Centre of curvature : The centre of hollow sphere of which the mirror forms a part, is called centre of curvature.
3. Principal axis : An imaginary line passing through the pole and the centre of curvature of a spherical mirror is called principal axis
4. Principal focus : It is a point on the principal axis, where a beam of light, parallel to principal axis, after reflection actually meet.
5. Focal length : The linear distance between the pole and the principal focus is called focal length.
6. Radius of curvature : The linear distance between the pole and the centre of curvature is called radius of curvature.
7. Aperture : The diameter of a spherical mirror is called its aperture.

Question 3.

(a) Define the term principal focus in case of convex mirror. Draw a convex mirror and show its principal focus and focal length clearly.
(b) What is the relation between focal length and radius of curvature of a concave mirror?

Answer:
(a) Focal length is the distance between pole (P) and focus (F)

(F) Focus is a point on principal axis where rays of light appear to meet.
(b) Focal length of a spherical mirror is equal to half of the radius of curvature of spherical mirror.

Question 4.

(a) What do you understand by the term real image?
(b) What type of mirror is used to obtain a real image?
(c) Does the mirror named by your form real image for all locations? Give reason for your answer.
(d) Is real image always inverted?

Answer:

(a) Real image : When rays of light after reflection or refraction actually meet at some other point” the image is real.
(b) Concave mirror.
(c) No, this mirror does not give real image of the object that lies between principal focus and pole.
(d) Yes. Real image is always ihverted.

Question 5.
Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.

Answer:

Characteristics of the image :

1. Image formed is real.
2. Image formed is inverted.
3. Image formed is enlarged.
4. Image is formed beyond centre of curvature in front of concave mirror.

Question 6.
Draw a neat two ray diagram to illustrate how a concave mirror is used as a shaving mirror.
Answer:
We know that when an object is placed between P and F of a concave mirror, it forms a virtual and enlarged image. Thus, by using concave mirror we can have a proper shave, as the tiny hairs are clearly visible.

Question 7.
Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.

Answer:

Characteristics of image :

1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus, behind the convex mirror.

Question 8.
Why do automobile drivers prefer convex mirror as a rear vew mirror? Illustrate your answer.
Answer:
A convex mirror always forms a small and upright image between pole and focus. That means in small area of mirror driver can see all the traffic coming from behind.

Question 9.
Give two uses of

1. convex mirror
2. concave mirror.

Answer:
(i) Two uses of convex mirror are :

(a) It is used as a rear view mirror in automobile to see the traffic behind.
(b) It is used as reflector for street light bulbs.

(ii) Two uses of concave mirror are :

(a) Concave mirror is used as A Reflector in head lights of cars and in search light. The source of light (bulb) is placed at the principal focus and the reflector forms parallel beam of light.
(b) For doctors to examine throat, ear, nose and eyes, light is focused with the help of concave mirror.

Question 10.
You are provided a convex mirror, a concave mirror and a plane mirror. How will you distinguish between them, without touching or using any other apparatus?
Answer:
We can identify convex mirror, concave mirror and a plane mirror by looking at them one by one.

1. If the size of the image of an object is of the same size as that of the object, then that mirror is a plane mirror.
2. If the size of image of an object increases as the object is brought closer to the mirror and size of the image of the object decreases when the object is taken away from the mirror, then that mirror is a concave mirror.
3. If the size of the iamge of an object remains diminished, either the object is moved away from the mirror or moved towards the mirror, then that mirror is a convex mirror.

Question 11.
Compare the characteristics of an image formed by a convex mirror and a concave mirror, when object is beyond centre of curvature, but not at infinity.
Answer:
Characteristics of the image formed by concave mirror when object is beyond centre of curvature, but not at infinity are :

1. Image formed is real.
2. Image formed is inverted.
3. Image formed is diminished.
4. Image is formed between centre of curvature (C) and principal focus (F), in front of the concave mirror.

Characteristics of the image formed by convex mirror when object is beyond centre of curvature, but not at infinity are :

1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus and behind the convex mirror.

Question 12.

1. Why does a driver use a convex mirror as a rear view mirror?
2. Illustrate your answer with the help of ray diagram.

Answer:
See Q. No. 8 of Exercise (2).

Question 13.

1. What is a real image?
2. What type of mirror is used to obtain a real image-of an object?
3. Does the mirror named by you above give real images for all locations of object?

Answer:

1. Real image : When the rays of light diverging from a point, after reflection of refraction, actually converge at some point, then that point is the real image of the object.
2. Concave mirror is used to obtain the real image of an object.
3. Concave mirror can not give real image for all the locations of object.

Question 14.
In the figure is shown a concave mirror. A is a point on the principal axis. If an object O is kept at A, image is formed on A itself. Copy the diagram. Draw the image in the diagram. Is the image real or virtual?

Measure the distance PA and write it in the diagram. What is the distance PA called?
Mark a point B on the principal axis, at which, if a point source of light is kept, the rays travel parallel to principal axis after reflection from M. What is point B called?
Answer:
As the object is placed at A and its image is also formed at A, so object must be at centre of curvature (C).

PA is called radius of curvature and on measuring PA = 4.8 cm. Point B marked on principal axis is called principal focus.

Question 15.
An object OA is placed on the principal axis of a concave mirror as shown in the figure. Copy and complete the diagram to show the formation of image.

Answer:

Question 16.
Copy the figure and complete it, by drawing two rays to show the formation of the image of the object AB. State the size, position and nature of image formed.

Answer:

• Nature : Image formed is virtual and erect.
• Position : Image is formed between principal focus (F) and pole.
• Size : Image formed is diminished.

More Resources

• A New Approach to ICSE Physics Part 1 Class 9 Solutions
• ICSE Solutions for Class 9 Hindi
• ICSE Solutions for Class 9 Geography
• ICSE Solutions for Class 9 History and Civics
• ICSE Solutions for Class 9 English Literature and Language
• Selina ICSE Solutions for Class 9 Maths
• Selina ICSE Solutions for Class 9 Physics
• Selina ICSE Solutions for Class 9 Chemistry
• Selina ICSE Solutions for Class 9 Biology
• New Simplified Chemistry Class 9 ICSE Solutions

Hope given A New Approach to ICSE Physics Part 1 Class 9 Solutions Light are helpful to complete your science homework.

If you have any doubts, please comment below. APlusTopper try to provide online science tutoring for you.