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 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.

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