Chapter 1 : Measurements and Experimentation - Selina Solutions for Class 9 Physics ICSE

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Chapter 1 - Measurements and Experimentation Excercise Ex 1(C)

Question 1

A simple pendulum completes 40 oscillations in a minute.

Find its (a) Frequency and (b) Time period.

Solution 1

(a) Frequency = Oscillations per second

 = (40/60) s-1

 = 0.67 s-1

 

(b) Time period = 1/frequency

 = (1/0.67) s

 = 1.5 s

Question 2

If the length of a simple pendulum is made one-fourth, then its time period becomes

1. Four times

2. One-fourth

3. Double

4. Half.

Solution 2

Half

Question 3

What is a simple pendulum? Is the pendulum used in a pendulum clock as a simple pendulum? Give reason to your answer.

Solution 3

A simple pendulum is a heavy point mass (known as bob) suspended from a rigid support by a massless and inextensible string.

No, the pendulum used in pendulum clock is not a simple pendulum because the simple pendulum is an ideal case. We cannot have a heavy mass having the size of a point and string having no mass.

Question 4

The time period of a simple pendulum is 2 s. What is its frequency?

Solution 4

Time period = 2 s

Frequency = 1/time period

 = (½)s-1

 = 0.5 s-1

Such a pendulum is called the seconds' pendulum.

Question 5

The time period of a pendulum clock is

1. 1 s

2. 2 s

3. 1 min

4. 12 h

Solution 5

2 s

Question 6

Define the terms: (i) oscillation, (ii) amplitude, (iii) frequency and (iv) time period as related to a simple pendulum.

Solution 6

(i) Oscillation: One complete to and fro motion of the pendulum is called one oscillation.

(ii) Amplitude: The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation. It is measured in metres (m).

(iii) Frequency: It is the number of oscillations made in one second. Its unit is hertz (Hz).

(iv)Time period: This is the time taken to complete one oscillation. It is measured in second (s).

Question 7

A seconds pendulum is taken to a place where acceleration due to the gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reasons. What will be its new time period?

Solution 7

Time period of 'a' is inversely proportional to the square root of acceleration due to gravity.

i.e.

Now, if the acceleration due to gravity falls to one-fourth, the time period will be doubled.

Let the new time period be T' and let g' be the acceleration due to gravity.

Then,

Question 8

The length of a second's pendulum clock is :

 

a. 0.5 m

b. 9.8 m

c. 1.0m

d. 2.0 m

Solution 8

The time period of a second's pendulum is T = 2 s.

 

The time period is related to the length as

 

  

 

Question 9

Draw a neat diagram of a simple pendulum.

Show the effective length and one oscillation of the pendulum.

Solution 9

Simple Pendulum:

 

  

 

Question 10

Find the length of a second's pendulum at a place where g = 10 m s-2 (Take π = 3.14)

Solution 10

Given, g= 10 m/s2 and time period (T) = 2s

Let 'l' be the length of the seconds' pendulum.

We know that

  

 

 

Question 11

Name two factors on which the time period of a simple pendulum depends. Write the relation for the time period in terms of the above named factors.

Solution 11

Two factors on which the time period of a simple pendulum depends are

 

(i) Length of pendulum (l)

(ii) Acceleration due to gravity (g)

 

Question 12

Compare the time periods of the two pendulums of lengths 1 m and 9 m, respectively.

Solution 12

Let T1 and T2 be the time periods of the two pendulums of lengths 1m and 9m, respectively.

  

 

 

Question 13

Name two factors on which the time period of a simple pendulum does not depend.

Solution 13

Two factors on which the time period of a simple pendulum does not depend are

(i) Material of the bob

(ii) Amplitude

Question 14

A pendulum completes two oscillations in 5 s. What is its time period? If g = 9.8 m s-2, find its length.

Solution 14

Time period = Total time/total no. of oscillations

 = (5/2) s

 = 2.5 s

Let 'l' be the length. Then,

   

 

 

 

Question 15

How is the time period of a simple pendulum affected, if at all, in the following situations:

(a) The length is quadrupled.

(b) The acceleration due to gravity is reduced to one-fourth.

Solution 15

We know that,

(a) If length quadruples then,

 

Therefore, the time period is doubled.

 

(b) If the acceleration due to gravity is reduced to one-fourth,

 

  

 

 

Therefore, the time period is doubled.

Question 16

The time periods of two simple pendulums at a place are in the ratio 2:1. What will be the corresponding ratio of their lengths?

Solution 16

Let T1 and T2 be the time periods of the two pendulums of lengths l1 and l2, respectively.

Then, we know that the time period is directly proportional to the square root of the length of the pendulum.

 

Question 17

How is the time period T and frequency f of a simple pendulum related to each other?

Solution 17

Time period of a simple pendulum is inversely proportional to its frequency.

Question 18

It takes 0.2 s for a pendulum bob to move from the mean position to one end. What is the time period of the pendulum?

Solution 18

Time period = Time taken to complete 1 oscillation

 = (4 × 0.2) s

 = 0.8 s

Question 19

How do you measure the time period of a given pendulum? Why do you note the time for more than one oscillation?

Solution 19

Measurement of time period of a simple pendulum:

(i) To measure the time period of a simple pendulum, the bob is slightly displaced from its mean position and is then released. This gives a to and fro motion about the mean position to the pendulum.

(ii) The time 't' for 20 complete oscillations is measured with the help of a stop watch.

(iii)Time period 'T' can be found by dividing 't' by 20.

To find the time period, the time for the number of oscillations more than 1 is noted because the least count of stop watch is either 1 s or 0.5 s. It cannot record the time period in fractions such as 1.2 or 1.4 and so on.

Question 20

How much time does the bob of a second's pendulum take to move from one extreme to the other extreme of its oscillation?

Solution 20

Time period of a seconds' pendulum = 2 s

Time taken to complete half oscillation, i.e. from one extreme to the other extreme = 1 s.

 

Question 21

How does the time period (T) of a simple pendulum depend on its length (l)? Draw a graph showing the variation of T2 with l. How will you use this graph to determine the value of g (acceleration due to gravity)?

Solution 21

The time period of a simple pendulum is directly proportional to the square root of its effective length.

  

From this graph, the value of acceleration due to gravity (g) can be calculated as follows.

 

The slope of the straight line can be found by taking two points P and Q on the straight line and drawing normals from these points on the X- and Y-axis, respectively. Then, the value of T2 is to be noted at a and b, the value of l at c and d. Then,

  

This slope is found to be constant at a place and is equal to   , where g is the acceleration due to gravity at that place. Thus, g can be determined at a place from these measurements using the following relation:

 

Question 22

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? Give reasons for your answer.

Solution 22

The ratio of their time periods would be 1:1 because the time period does not depend on the weight of the bob.

Question 23

Two simple pendulums A and B have lengths 1.0 m and 4.0 m, respectively, at a certain place. Which pendulum will make more oscillations in 1 min? Explain your answer.

Solution 23

Pendulum A will take more time (twice) in a given time because the time period of oscillation is directly proportional to the square root of the length of the pendulum. Therefore, the pendulum B will have a greater time period and thus making lesser oscillations.

Question 24

State how does the time period of a simple pendulum depend on (a) length of pendulum, (b) mass of bob, (c) amplitude of oscillation and (d) acceleration due to gravity.

Solution 24

(a) The time period of oscillations is directly proportional to the square root of the length of the pendulum.

(b) The time period of oscillations of simple pendulum does not depend on the mass of the bob.

(c) The time period of oscillations of simple pendulum does not depend on the amplitude of oscillations.

(d) The time period of oscillations of simple pendulum is inversely proportional to the square root of acceleration due to gravity.

Question 25

What is a second's pendulum?

Solution 25

A pendulum with the time period of oscillation equal to two seconds is known as a seconds pendulum.

Question 26

State the numerical value of the frequency of oscillation of a second's pendulum. Does it depend on the amplitude of oscillations?

Solution 26

The frequency of oscillation of a seconds' pendulum is 0.5 s-1. It does not depend on the amplitude of oscillation.

Chapter 1 - Measurements and Experimentation Excercise Ex 1(A)

Question 1

What is meant by measurement?

Solution 1

Measurement is the process of comparing a given physical quantity with a known standard quantity of the same nature.

Question 2

The fundamental unit present in the list mentioned below is

1. Newton

2. Pascal

3. Hertz

4. Second

Solution 2

Second

Question 3

The wavelength of light of a particular colour is 5800 Å. (a) Express it in (i) nanometre and (ii) metre. (b) What is its order of magnitude in metre?

Solution 3

Wavelength of light of particular colour = 5800  

(a)

(i) 1 = 10-1 nm

  5800  = 5800 × 10-1 nm

 = 580 nm

 

(ii) 1 = 10-10 m

  5800  = 5800 × 10-10 m

 = 5.8 × 10-7 m

(b) The order of its magnitude in metre is 10-6 m because the numerical value of 5.8 is more than 3.2.

Question 4

What do you understand by the term unit?

Solution 4

Unit is a quantity of constant magnitude which is used to measure the magnitudes of other quantities of the same manner.

Question 5

Which of the following unit is not a fundamental unit:

1. Metre 

2. Litre

3. Second

4. Kilogramme

 

Solution 5

Litre

Question 6

The size of bacteria is 1 µ. Find the number of bacteria present in 1 m length.

Solution 6

Size of a bacteria = 1 µ

Since 1 µ = 10-6 m

  Number of the particle = Total length/size of 

 one bacteria

 = 1 m/10-6 m

 = 106

Question 7

What are the three requirements for selecting a unit of a physical quantity?

Solution 7

The three requirements for selecting a unit of a physical quantity are

(i) It should be possible to define the unit without ambiguity.

(ii) The unit should be reproducible.

(iii) The value of units should not change with space and time.

Question 8

The unit of time is

1. Light year

2. Parsec

3. Leap year

4. Angstrom.

Solution 8

Leap year

Question 9

The distance of a galaxy is 5·6 × 1025 m.

Assuming the speed of light to be 3 × 108 m s ". (i) Find the time taken by light to travel this distance and (ii) express its order of magnitude.

 [Hint : Time taken = distance/speed   ]

Solution 9

Distance of galaxy = 5.6 × 1025 m

Speed of light = 3 × 108 m/s

 

(a) Time taken by light = Distance travelled/speed of light

 = (5.6 × 1025 / 3 × 108) s

 = 1.87 × 1017 s

 

(b) Order of magnitude = 100 × 1017 s = 1017 s

(This is because the numerical value of 1.87 is less than the numerical value 3.2)

Question 10

Define the three fundamental quantities.

Solution 10

Definitions of three fundamental quantities:

 

S.I. unit of length (m): A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy made of 90% platinum and 10% iridium) rod kept at 0°C in the International Bureau of Weights and Measures at serves near Paris.

S.I. unit of mass (kg): In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at International Bureau of Weights and Measures at serves near Paris.

 

S.I. unit of time (s): A second is defined as 1/86400th part of a mean solar day, i.e.

1s =   × one mean solar day.

Question 11

1 Å is equal to

1. 0.1 nm

2. 10-10cm

3. 10-8m

4. 104 µ.

Solution 11

0.1 nm

Question 12

Name the three systems of unit and state their various fundamental units.

Solution 12

Three systems of unit and their fundamental units:

 

(i) C.G.S. system (or French system): In this system, the unit of length is centimeter (cm), unit of mass is gramme (g) and unit of time is second (s).

 

(ii) F.P.S. system (or British system): In this system, the unit of length is foot (ft), unit of mass is pound (lb) and unit of time is second (s).

 

(iii) M.K.S. system (or metric system): In this system, the unit of length is metre (m), unit of mass is kilogramme (kg) and unit of time is second (s).

Question 13

Light year (ly) is the unit of

1. Time

2. Length

3. Mass

4. None of these.

Solution 13

Length

Question 14

Define a fundamental unit.

Solution 14

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.

Question 15

What are the fundamental units in S.I. system? Name them along with their symbols.

Solution 15

Fundamental quantities, units and symbols in S.I. system are

 

Quantity

Unit

Symbol

Length

metre

m

Mass

kilogramme

kg

Time

second

s

Temperature

kelvin

K

Luminous intensity

candela

cd

Electric current

ampere

A

Amount of substance

mole

mol

Angle

radian

rd

Solid angle

steradian

st-rd

 

Question 16

Explain the meaning of a derived unit with the help of one example.

Solution 16

The units of quantities other than those measured in fundamental units can be obtained in terms of the fundamental units, and thus the units so obtained are called derived units.

Example:

Speed = Distance/time

Hence, the unit of speed = fundamental unit of distance/fundamental unit of time

Or, the unit of speed = metre/second or ms-1.

As the unit of speed is derived from the fundamental units of distance and time, it is a derived unit.

Question 17

Define standard metre.

Solution 17

A metre was originally defined in 1889 as the distance between two marks drawn on a platinum-iridium (an alloy with 90% platinum and 10% iridium) rod kept at 0o C in the International Bureau of Weights and Measures at serves near Paris.

Question 18

Name two units of length which are bigger than a metre. How are they related to the unit metre?

Solution 18

Astronomical unit (A.U.) and kilometer (km) are units of length which are bigger than a metre.

1 km = 1000 m

1 A.U. = 1.496 × 1011 m

Question 19

Write the names of two units of length smaller than a metre. Express their relationship with metre.

Solution 19

Centimeter (cm) and millimeter (mm) are units of length smaller than a metre.

1 cm = 10-2 m

1 mm = 10-3 m

Question 20

How is nanometer related to Angstrom?

Solution 20

1 nm = 10

Question 21

Name three convenient units used to measure lengths ranging from very short to very long value. How are they related to each other?

Solution 21

Three convenient units of length and their relation with the S.I. unit of length:

(i) 1 Angstrom (Å) = 10-10 m

(ii) 1 kilometre (km) = 103 m

(iii) 1 light year (ly) = 9.46 × 1015 m

Question 22

Name the S.I. unit of mass and define it.

Solution 22

S.I. unit of mass is 'kilogramme'.

In 1889, one kilogramme was defined as the mass of a cylindrical piece of a platinum-iridium alloy kept at the International Bureau of Weights and Measures at serves near Paris.

Question 23

Complete the following

(a) 1 light year = ...........m

(b) 1 m   = ...........Å

(c) 1 m   = ...........µ (micron)

(d) 1 micron = ...........Å

(e) 1 fermi = ...........m

Solution 23

(a) 1 light year = 9.46 × 1015 m

(b) 1 m = 1010

(c) 1 m = 106 µ (micron)

(d) 1 micron = 104

(e) 1 fermi = 10-15 m

Question 24

State two units of mass smaller than a kilogram. How are they related to the unit kilogramme?

Solution 24

The units 'gramme' (g) and 'milligramme' (mg) are two units of mass smaller than 'kilogramme'.

1 g = 10-3 kg

1 mg = 10-6 kg

Question 25

State two units of mass bigger than a kilogram. Give their relationship with the unit kilogramme.

Solution 25

The units 'quintal' and 'metric tonne' are two units of mass bigger than 'kilogramme'.

1 quintal = 100 kg

1 metric tonne = 1000 kg

Question 26

Complete the following

(a) 1 g    = ..........kg

(b) 1 mg    = ..........kg

(c) 1 quintal  = ..........kg

(d) 1 a.m.u (or u) = ..........kg

Solution 26

(a) 1 g = 10-3 kg

(b) 1 mg = 10-6 kg

(c) 1 quintal = 100 kg

(d) 1 a.m.u (or u) = 1.66 x 10-27 kg

Question 27

Name the S.I. unit of time and define it.

Solution 27

The S.I. unit of time is second (s).

 

A second is defined as 1/86400th part of a mean solar day, i.e.

1 s =   × one mean solar day

Question 28

'The year 2016 will have February of29 days'. Is this statement true?

Solution 28

Yes, this statement is true. The year 2016 was a leap year.

Question 29

What is a leap year?

Solution 29

A leap year is the year in which the month of February has 29 days.

Question 30

'The year 2016 will have February of29 days'.

Is this statement true?

Solution 30

Yes, the given statement is true.

Question 31

What is a lunar month?

Solution 31

One lunar month is the time in which the moon completes one revolution around the earth. A lunar month is made of nearly 4 weeks.

Question 32

Complete the following

(a) 1 nanosecond =.........s.

(b) 1 µs   =.........s.

(c) 1 mean solar day =.........s.

(d) 1 year = .........s.

Solution 32

(a) 1 nanosecond = 10-9 s

(b) 1 µs = 10-6 s

(c) 1 mean solar day = 86400 s

(d) 1 year = 3.15 × 107 s

Question 33

Name the physical quantities which are measured in the following units

(a) u  (b) ly  (c) ns  (d) nm

Solution 33

(a) Mass (b) Distance (or length) (c) Time (d) Length

Question 34

Write the derived units of the following

(a) Speed  (b) Force

(c) Work  (d) Pressure.

Solution 34

(a) ms-1 (b) kg ms-2 (c) kg m2s-2 (d) kg m-1s-2

Question 35

How are the following derived units related to the fundamental units?

(a) Newton   (b) Watt

(c) Joule   (d) Pascal.

Solution 35

(a) kg ms-2 (b) kg m2s-3

(c) kg m2s-2 (d) kg m-1s-2

Question 36

Name the physical quantities related to the following units:

(a) km2 (b) Newton (c) Joule

(d) Pascal (e) Watt

Solution 36

(a) Area (b) Force (c) Energy

(d) Pressure (f) Power

Chapter 1 - Measurements and Experimentation Excercise Ex 1(B)

Question 1

Explain the meaning of the term 'least count of an instrument' by taking a suitable example.

Solution 1

The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.

Question 2

The least count of a vernier calipers is :

a. 1 cm

b. 0.001 cm

c. 0.1 cm

d. 0.01 cm

Solution 2

The least count of a vernier calipers is 0.01 cm

Question 3

A stopwatch has 10 divisions graduated between the 0 and 5 s marks. What is its least count?

Solution 3

Range of the stop watch = 5s

Total number of divisions = 10

L.C. = 5/10 = 0.5 s

Question 4

A boy makes a scale with graduations in cm on it, i.e. 100 divisions in 1 m. To what accuracy can this scale measure? How can this accuracy be increased?

Solution 4

Total length of the scale = 1 m = 100 cm

No. of divisions = 100

Length of each division = Total length/total no. of divisions

 = 100 cm/100

 = 1 cm

Thus, this scale can measure with an accuracy of 1 cm.

 

To increase the accuracy, the total number of divisions on the scale must be increased.

Question 5

A microscope has its main scale with 20 divisions in 1 cm and vernier scale with 25 divisions, the length of which is equal to the length of 24 divisions of main scale. The least count of microscope is

1. 0.002 cm

2. 0.001 cm

3. 0.02 cm

4. 0.01 cm

Solution 5

0.002 cm

Question 6

A vernier has 10 divisions and they are equal to 9 divisions of the main scale in length. If the main scale is calibrated in mm, what is its least count?

Solution 6

Value of 1 m.s.d. = 1 mm

10 vernier divisions = 9 m.s.d.

L.C. = Value of 1 m.s.d./number of divisions on vernier scale

 = 1 mm/10

 = 0.1 mm or 0.01 cm

Question 7

A microscope is provided with a main scale graduated with 20 divisions in 1 cm and a vernier scale with 50 division on it of length same as of 49 divisions of main scale. Find the least count of the microscope.

Solution 7

 

Thus, the least count of the microscope is 0.001 cm.

Question 8

A boy measures the length of a pencil by a metre rule and expresses it to be 2.6 cm. What is the least count of the metre rule? Can he write it as 2.60 cm?

Solution 8

The least count of a metre rule is 1 cm.

The length cannot be expressed as 2.60 cm because a metre scale measures length correctly only up to one decimal place of a centimeter.

 

Question 9

The diameter of a thin wire can be measured by

1. A vernier calipers

2. A metre rule

3. A screw gauge

4. Any of these.

Solution 9

A screw gauge

Question 10

Define the least count of vernier callipers. How do you determine it?

Solution 10

The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.

 

Let n divisions on vernier callipers be of length equal to that of (n - 1) divisions on the main scale and the value of 1 main scale division be x. Then,

Value of n divisions on vernier = (n - 1) x

Alternatively, value of 1 division on vernier =

Hence,

Least count =   

L.C. = (Value of one main scale division)/(Total no. of divisions on vernier callipers)

Value of one main scale division = 1 mm

Total no. of divisions on vernier = 10

Therefore, L.C. =

Question 11

A boy uses a vernier callipers to measure the thickness of his pencil. He measures it to be 1.4 mm. If the zero error of vernier callipers is +0.02 cm, what is the correct thickness of the pencil?

Solution 11

Thickness of the pencil (observed reading) = 1.4 mm

Zero error = + 0.02 cm = + 0.2 mm

Correct reading = observed reading - zero error (with sign)

 = 1.4 mm - 0.2 mm

 = 1.2 mm

Question 12

Define the term 'Vernier constant'.

Solution 12

Vernier constant is equal to the difference between the values of one main scale division and one vernier scale division. It is the least count of vernier callipers.

Question 13

A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of the vernier scale is ahead of the zero of the main scale and the 3rd division of the vernier

scale coincides with a main scale division.

Find : (i) The least count and

(ii) The zero error of the vernier callipers.

Solution 13

(i) Value of 1 m.s.d. =1 mm

10 vernier divisions = 9 m.s.d.

L.C. = Value of 1 m.s.d./number of divisions on the vernier scale

 = 1 mm/10

 = 0.1 mm or 0.01 cm

 

(ii) On bringing the jaws together, the zero of the vernier scale is ahead of zero of the main scale, the error is positive.

3rd vernier division coincides with a main scale division.

Total no. of vernier divisions = 10

Zero error = +3 × L.C.

 = +3 × 0.01 cm

 = +0.03 cm

Question 14

When is a vernier calipers said to be free from zero error?

Solution 14

A vernier calipers is said to be free from zero error, if the zero mark of the vernier scale coincides with the zero mark of the main scale.

 

Question 15

The main scale of a vernier callipers is calibrated in mm and 19 divisions of main scale are equal in length to 20 divisions of the vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of vernier scale coincides with a main scale division. Find

(i) Least count and (ii) Radius of a cylinder.

Solution 15

(i) Value of 1 m.s.d = 1 mm = 0.1 cm

20 vernier divisions = 19 m.s.d.

L.C. = Value of 1 m.s.d./number of divisions on the vernier scale

 = 1mm/20

 = (0.1/20) cm

 = 0.005 cm

 

(ii) Main scale reading = 35 mm = 3.5 cm

Since 4th division of the main scale coincides with the main scale, i.e. p = 4.

Therefore, the vernier scale reading = 4 × 0.005 cm = 0.02 cm

Total reading = Main scale reading + vernier scale reading

 = (3.5 + 0.02) cm

 = 3.52 cm

Radius of the cylinder = Diameter (Total reading) / 2

 = (3.52/2) cm

   = 1.76 cm

Question 16

What is meant by zero error of vernier callipers ? How is it determined? Draw neat diagrams to explain it. How is it considered to get the correct measurement?

Solution 16

Due to mechanical errors, sometimes the zero mark of the vernier scale does not coincide with the zero mark of the main scale, the vernier callipers is said to have zero error.

It is determined by measuring the distance between the zero mark of the main scale and the zero mark of the vernier scale.

The zero error is of two kinds

(i) Positive zero error

(ii) Negative zero error

 

(i) Positive zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the right of the zero mark of the main scale, the error is said to be positive.

  

To find this error, we note the division of the vernier scale, which coincides with any division of the main scale. The number of this vernier division when multiplied by the least count of the vernier callipers, gives the zero error.

For example, for the scales shown, the least count is 0.01 cm and the 6th division of the vernier scale coincides with a main scale division.

 

Zero error = +6 × L.C. = +6 × 0.01 cm

 = +0.06 cm

 

 

(ii) Negative zero error: On bringing the two jaws together, if the zero mark of the vernier scale is on the left of the zero mark of the main scale, then the error is said to be negative.

  

To find this error, we note the division of the vernier scale coinciding with any division of the main scale. The number of this vernier division is subtracted from the total number of divisions on the vernier scale and then the difference is multiplied by the least count.

For example, for the scales shown, the least count is 0.01 cm and the sixth division of the vernier scale coincides with a certain division of the main scale. The total number of divisions on vernier callipers is 10.

 

Zero error = - (10 - 6) × L.C.

 = - 4 × 0.01 cm = - 0.04 cm

 

Correction:

To get correct measurement with vernier callipers having a zero error, the zero error with its proper sign is always subtracted from the observed reading.

 

Correct reading = Observed reading - zero error (with sign)

Question 17

In the vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided into 10 parts. While measuring the length, the zero of the vernier lies just ahead of the 1.8 cm mark and the 4th division of vernier coincides with

a main scale division.

(a) Find the length.

(b) If zero error of vernier callipers is -0.02 cm,

What is the correct length ?

Solution 17

(a) L.C. of vernier callipers = 0.01 cm

Main scale reading = 1.8 cm

Since 4th division of the main scale coincides with the main scale, i.e. p = 4.

Therefore, the vernier scale reading = 4 × 0.01 cm = 0.04 cm

Total reading = Main scale reading + vernier scale reading

 = (1.8 + 0.04) cm

 = 1.84 cm

(b) Observed reading = 1.84 cm

Zero error = -0.02 cm

Correct reading = Observed reading - Zero error (with sign)

 = [1.84 - (-0.02)] cm

 = 1.86 cm

Question 18

A vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.

Solution 18

  

Question 19

While measuring the length of a rod with a vernier callipers, Fig. 2.22 shows the position of its scales. What is the length of the rod? 

  

Solution 19

L.C. of vernier callipers = 0.01 cm

In the shown scale,

Main scale reading = 3.3 mm

6th vernier division coincides with an m.s.d.

Therefore, vernier scale reading = 6 × 0.01 cm = 0.06 cm

Total reading = m.s.r. + v.s.r.

 = 3.3 + 0.06

   = 3.36 cm

Question 20

Draw a neat and labelled diagram of a vernier callipers. Name its main parts and state their functions.

Solution 20

Diagram of vernier callipers:

  

 

Main parts and their functions:

 

Main scale: It is used to measure length correct up to 1 mm.

Vernier scale: It helps to measure length correct up to 0.1 mm.

Outside jaws: It helps to measure length of a rod, diameter of a sphere, external diameter of a hollow cylinder.

Inside jaws: It helps to measure the internal diameter of a hollow cylinder or pipe.

Strip: It helps to measure the depth of a beaker or a bottle.

Question 21

The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of a screw gauge?

Solution 21

Pitch of a screw gauge = 0.5 mm

No. of divisions on the circular scale = 100

L.C. = (0.5/100) mm

  = 0.005 mm or 0.0005 cm

Question 22

State three uses of the vernier callipers.

Solution 22

Three uses of vernier callipers are

(a) Measuring the internal diameter of a tube or a cylinder.

(b) Measuring the length of an object.

(c) Measuring the depth of a beaker or a bottle.

Question 23

The thimble of a screw gauge has 50 divisions. The spindle advances 1 mm when the screw is turned through two revolutions.

(i) What is the pitch of the screw gauge?

(ii) What is the least count of the screw gauge?

Solution 23

No. of divisions on the circular scale = 50

(i) Pitch = Distance moved ahead in one revolution

 = 1 mm/2 = 0.5 mm

 

(ii) L.C. = Pitch/No. of divisions on the circular head

 = (0.5/50) mm

 = 0.01 mm

Question 24

Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.

Solution 24

Two scales of vernier calipers are

(a) Main scale

(b) Vernier scale

The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.

Value of 1 division on the main scale= 1 mm

Total no. of divisions on the vernier scale = 10

Thus, L.C. = 1 mm /10 = 0.1 mm = 0.01 cm.

 

Hence, a vernier callipers can measure length correct up to 0.01 cm.

Question 25

The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find

(i) The least count and

(ii) The diameter of the wire.

Solution 25

Pitch of the screw gauge = 1mm

No. of divisions on the circular scale = 100

 

(i) L.C. = Pitch/No. of divisions on the circular head

 = (1/100) mm

 = 0.01 mm or 0.001 cm

 

(ii) Main scale reading = 2mm = 0.2 cm

No. of division of circular head in line with the base line (p) = 45

Circular scale reading = (p) × L.C.

 = 45 x 0.001 cm

 = 0.045 cm

Total reading = M.s.r. + circular scale reading

  = (0.2 + 0.045) cm

 = 0.245 cm

Question 26

Describe in steps, how would you use a vernier callipers to measure the length of a small rod ?

Solution 26

Measuring the length of a small rod using vernier calipers:

 

The rod whose length is to be measured is placed in between the fixed end and the vernier scale as shown in the figure.

 

  

In this position, the zero mark of the vernier scale is ahead of 1.2 cm mark on main scale. Thus the actual length of the rod is 1.2 cm plus the length ab (i.e., the length between the 1.2 cm mark on the main scale and 0 mark on vernier scale).

To measure the length ab, we note the pth division of the vernier scale, which coincides with any division of main scale.

Now, ab + length of p divisions on vernier scale = length of p divisions on main scale

Alternatively, ab = length of p divisions on the main scale - length of p divisions on the vernier scale.

 = p (length of 1 division on main scale - length of 1 division on vernier scale)

 = p × L.C.

 

Therefore, total reading = main scale reading + vernier scale reading

   = 1.2 cm + (p × L.C.)

Question 27

When a screw gauge with a least count of 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions.

(i) What is the diameter of the wire in cm?

(ii) If the zero error is +0.005 cm, what is the correct diameter?

Solution 27

(i) L.C. of screw gauge = 0.01 mm or 0.001 cm

Main scale reading = 1 mm or 0.1 cm

No. of division of circular head in line with the base line (p) = 27

Circular scale reading = (p) × L.C.

 = 27 × 0.001 cm

 = 0.027 cm

Diameter (Total reading) = M.s.r. + circular scale reading

 = (0.1 + 0.027) cm

 = 0.127 cm

 

(ii) Zero error = 0.005 cm

Correct reading = Observed reading - zero error (with sign)

 = [0.127 - (+0.005)] cm

 = 0.122 cm

Question 28

Name the part of the vernier callipers which is used to measure the following

(a) External diameter of a tube,

(b) Internal diameter of a mug,

(c) Depth of a small bottle,

(d) Thickness of a pencil.

Solution 28

(a)  Outside jaws

(b)  Inside jaws

(c)  Strip

(d)  Outer jaws

Question 29

A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the screw is in contact with the stud, the zero of the circular scale lies below the base line and 4th division of the circular scale is in line with the base line. Find

(i) The pitch,

(ii) The least count and

(iii) The zero error of the screw gauge

Solution 29

No. of divisions on the circular scale = 50

(i) Pitch = Distance moved ahead in one revolution

 = 1 mm/2 = 0.5 mm

 

(ii) L.C. = Pitch/No. of divisions on the circular head

 = (0.5/50) mm

 = 0.01 mm

 

(iii) Because the zero of the circular scale lies below the base line, when the flat end of the screw is in contact with the stud, the error is positive.

No. of circular division coinciding with m.s.d. = 4

Zero error = + (4 × L.C.)

 = + (4 × 0.01) mm

 = + 0.04 mm

Question 30

Explain the terms (i) Pitch and (ii) Least count of a screw gauge. How are they determined?

Solution 30

(i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation.

(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.

Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.

If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.

Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm

Question 31

Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.  

  

Find: (i) Pitch of the screw gauge,

 (ii) Least count of the screw gauge and

 (iii) The diameter of the wire.

Solution 31

No. of divisions on the circular scale = 50

(i) Pitch = Distance moved ahead in one revolution

 = 1 mm/1 = 1 mm.

 

(ii) L.C. = Pitch/No. of divisions on the circular head

 = (1/50) mm

 = 0.02 mm

 

(iii) Main scale reading = 4 mm

No. of circular division coinciding with m.s.d. (p) = 47

Circular scale reading = p × L.C.

 = (47 × 0.02) mm

 = 0.94 mm

Diameter (Total reading) = M.s.r. + circular scale reading

 = (4 + 0.94) mm

   = 4.94 mm

Question 32

A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head in order to read correctly up to 0.001 mm with its help?

Solution 32

Pitch of the screw gauge = 0.5 mm

L.C. of the screw gauge = 0.001 mm

No. of divisions on circular scale = Pitch / L.C.

 = 0.5 / 0.001

 = 500

Question 33

Draw a neat and labelled diagram of a screw gauge.

Name its main parts and state their functions.

Solution 33

Diagram of screw gauge:

  

 

Main parts and their functions:

 

1. Ratchet: It advances the screw by turning it until the object is gently held between the stud and spindle of screw.

2. Sleeve: It marks the main scale and base line.

3. Thimble: It marks the circular scale.

4. Main scale: It helps to read the length correct up to 1 mm.

5. Circular scale: It helps to read length correct up to 0.01 mm.

Question 34

State one use of a screw gauge.

Solution 34

A screw gauge is used for measuring diameter of circular objects mostly wires with an accuracy of 0.001 cm.

Question 35

State the purpose of ratchet in a screw gauge.

Solution 35

Ratchet helps to advance the screw by turning it until the object is gently held between the stud and spindle of the screw.

Question 36

What do you mean by zero error of a screw gauge? How is it accounted for?

Solution 36

Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of main scale. It is either above or below the base line of the main scale, in which case the screw gauge is said to have a zero error. It can be both positive and negative.

It is accounted by subtracting the zero error (with sign) from the observed reading in order to get the correct reading.

Correct reading = Observed reading - zero error (with sign)

Question 37

A screw gauge has a least count 0.001 cm and zero error +0.007 cm. Draw a neat diagram to represent it.

Solution 37

Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.

 

  

 

Question 38

What is backlash error? Why is it caused? How is it avoided?

Solution 38

Backlash error: If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.

Reason: It happens due to wear and tear of the screw threads.

To avoid the backlash error, while taking the measurements the screw should be rotated in one direction only. If the direction of rotation of the screw needs to be changed, then it should be stopped for a while and then rotated in the reverse direction.

Question 39

Describe the procedure to measure the diameter of a wire with the help of a screw gauge.

Solution 39

Measurement of diameter of wire with a screw gauge:

 

The wire whose thickness is to be determined is placed between the anvil and spindle end, the thimble is rotated until the wire is firmly held between the anvil and spindle. The rachet is provided to avoid excessive pressure on the wire. It prevents the spindle from further movement. The thickness of the wire could be determined from the reading as shown in the figure below.

  

The pitch of the screw = 1 mm

L.C. of screw gauge = 0.01 mm

Main scale reading = 2.5 mm

46th division of circular scale coincides with the base line.

Therefore, circular scale reading = 46 × 0.01 = 0.46 mm

Total reading = Main scale reading + circular scale reading

 = (2.5 + 0.46) mm

 = 2.96 mm

Question 40

Name the instrument which can accurately measure the following

(a) The diameter of a needle,

(b) The thickness of a paper,

(c) The internal diameter of the neck of a water bottle,

(d) The diameter of a pencil.

Solution 40

(a) Screw gauge

(b) Screw gauge

(c) Vernier calipers

(d) Screw gauge

Question 41

Which of the following measures a small length with high accuracy: metre scale, vernier callipers or screw gauge?

Solution 41

Screw gauge measures a length to a high accuracy.

Question 42

Name the instrument which has the least count

(a) 0.1 mm (b) 1 mm (c) 0.01 mm.

Solution 42

(a) Vernier callipers (b) Metre scale (c) Screw gauge.