Measurements and Experimentation

Solutions for Physics, Class 9, ICSE

Exercise 1A Multiple Choice Type

18 questions

Question 1

The condition(s) essential for a unit to be accepted internationally :

  1. The unit should be of convenient size and without any ambiguity.
  2. The unit should be reproducible.
  3. The value of the unit should not change with space and time.
  4. All of the above.
Exercise 1A Multiple Choice Type

Answer:

All of the above

Reason — The condition(s) essential for a unit to be accepted internationally are :

  1. The unit should be of convenient size.
  2. It should be possible to define the unit without any ambiguity.
  3. The unit should be reproducible.
  4. The value of unit should not change with space and time.

Question 2

In mechanics, the three fundamental quantities are:

  1. Length, mass, velocity
  2. Mass, time, density
  3. Mass, time, pressure
  4. Length, mass, time
Exercise 1A Multiple Choice Type

Answer:

Length, mass, time

Reason — In mechanics, length, mass and time are the three fundamental units. Unit of length is metre (m), unit of mass is kilogram (kg), unit of time is second (s).

Question 3

How many units are fundamental and supplementary?

  1. 2 fundamental, 7 supplementary
  2. 3 fundamental, 6 supplementary
  3. 7 fundamental, 2 supplementary
  4. 4 fundamental, 5 supplementary
Exercise 1A Multiple Choice Type

Answer:

7 fundamental, 2 supplementary

Reason — Fundamental and supplementary units are as follows :

Fundamental UnitsUnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
TemperaturekelvinK
Luminous intensitycandelacd
Electric currentampereA
Amount of substancemolemol
Supplementary UnitsUnitSymbol
Angleradianrd
Solid anglesteradianst-rd

Question 4

The fundamental unit is —

  1. newton
  2. pascal
  3. hertz
  4. second
Exercise 1A Multiple Choice Type

Answer:

second

Reason — Out of the given units, only 'second' is a fundamental unit.

Question 5

Which of the following unit is not a fundamental unit:

  1. Metre
  2. Litre
  3. Second
  4. Kilogram
Exercise 1A Multiple Choice Type

Answer:

Litre

Reason — Metre, second and kilogram are the fundamental units whereas litre is a derived unit.

Question 6

One astronomical unit is equal to:

  1. 1.496 x 1011 metre
  2. 1.596 x 1012 metre
  3. 1.498 x 1010 metre
  4. 1.398 x 1011 metre
Exercise 1A Multiple Choice Type

Answer:

1.496 x 1011 metre

Reason — One astronomical unit is equal to the mean distance between the earth and the sun. i.e.,

1 A.U. = 1.496 x 1011 metre

Question 7

The distance of stars from earth is generally measured in:

  1. Kilometre
  2. Light year
  3. Angstrom
  4. Fermi
Exercise 1A Multiple Choice Type

Answer:

Light year

Reason — A light year is the distance travelled by light in vacuum, in one year.

Question 8

The unit of time is:

  1. light year
  2. parsec
  3. leap year
  4. angstrom
Exercise 1A Multiple Choice Type

Answer:

leap year

Reason — Leap year is the unit of time. Light year, parsec, angstrom are unit of distance. A leap year is the year in which the month of February is of 29 days.

Question 9

1 Å is equal to:

  1. 0.1 nm
  2. 10-10cm
  3. 10-8m
  4. 104µ
Exercise 1A Multiple Choice Type

Answer:

0.1 nm

Reason — 10 Å = 1nm
∴ 1 Å = 0.1nm

Question 10

One metric tonne is equal to :

  1. 10 quintal
  2. 100 quintal
  3. 1000 quintal
  4. 25 quintal
Exercise 1A Multiple Choice Type

Answer:

10 quintal

Reason — It is one thousand times a kilogram. i.e., 1 metric tonne = 1000 kg = 10 quintal.

Question 11

One solar mass is equal to :

  1. 3 x 1020 kg
  2. 2 x 1027 kg
  3. 2 x 1030 kg
  4. 3 x 1030 kg
Exercise 1A Multiple Choice Type

Answer:

2 x 1030 kg

Reason — The mass of large heavenly bodies is measured in terms of solar mass where 1 solar mass is the mass of the sun i.e., 1 solar mass = 2 x 1030 kg

Question 12

One lunar cycle is nearly equal to :

  1. 28.5 days
  2. 29.5 days
  3. 30 days
  4. 30.5 days
Exercise 1A Multiple Choice Type

Answer:

29.5 days

Reason — One lunar cycle is the amount of time it takes for the Moon to complete one orbit around the Earth and it is nearly equal to 29.5 days.

Question 13

One shake is equal to :

  1. 10-9 s
  2. 10-8 s
  3. 10-10 s
  4. 108 s
Exercise 1A Multiple Choice Type

Answer:

10-8 s

Reason — Shake is a smaller unit of time where, 1 shake = 10-8 s and 1 nano shake = 10-9 s

Question 14

Which of the following unit is equivalent to watt?

  1. Volt x Ampere
  2. Volt/Ampere
  3. Ampere/Volt
  4. Ohm
Exercise 1A Multiple Choice Type

Answer:

Volt x Ampere

Reason — watt is the unit of electrical power and is defined as potential x current = volt x ampere.

Question 15

Which one of the following is a derived unit?

  1. Metre
  2. Kelvin
  3. Metre3
  4. Mole
Exercise 1A Multiple Choice Type

Answer:

Metre3

Reason — Volume is a derived unit. Its formula is :

Volume = length x breadth x height = metre x metre x metre = metre3

Question 16

The S.I. unit of energy is:

  1. Joule
  2. Kg m2
  3. Kg m2s-3
  4. Kg m2s-1
Exercise 1A Multiple Choice Type

Answer:

Joule

Reason — Work done or energy = force x displacement = kgm2s-2 or J

Question 17

The S.I. unit of pressure is:

  1. Kg m-2s-2
  2. Kg m-1s-2
  3. Kg m2
  4. Kg ms-2
Exercise 1A Multiple Choice Type

Answer:

Kg m-1s-2

Reason — Pressure = forcearea\dfrac{\text{force}}{\text{area}} = Kg m-1s-2

Question 18

Which of the following is the smallest unit?

  1. Millimetre
  2. Angstrom
  3. Fermi
  4. Metre
Exercise 1A Multiple Choice Type

Answer:

Fermi

Reason — Fermi is 10-15th part of a metre. i.e., 1 fermi = 10-15 m, hence it is the smallest unit.

Angstrom = 10-10th part of a metre.
Millimetre = 10-3th part of a metre.

Exercise 1A Numericals

7 questions

Question 1

The wavelength of light of a particular colour is 5800 Å.

Express it in —

(a) nanometer and

(b) metre

Exercise 1A Numericals

Answer:

As we know, 1nm = 10 Å

Given,

The wavelength of light = 5800 Å

Substituting the value of wavelength in the relation above, we get,

10A˚=1nm1A˚=110nm5800A˚=110×5800nm5800A˚=580nm10 \text Å = 1\text {nm} \\[0.5em] 1 \text Å = \dfrac{1}{10}\text{nm} \\[0.5em] 5800 \text Å = \dfrac{1}{10} \times 5800 \text{nm}\\[0.5em] \Rightarrow 5800 \text Å = 580 \text{nm}\\[0.5em]

Hence, wavelength of 5800 Å in nm = 580 nm.

(b) As we know,
1m = 1010 Å

Given,

The wavelength of light = 5800 Å

Substituting the value of wavelength in the relation above, we get,

1010A˚=1m1A˚=1010m5800A˚=5800×1010m5800A˚=5.8×107m10^{10} \text Å = 1\text m \\[0.5em] 1 \text Å = 10^{-10} \text m \\[0.5em] 5800 \text Å = 5800 \times 10^{-10} \text m \\[0.5em] \Rightarrow 5800 \text Å = 5.8 \times 10^{-7} \text m \\[0.5em]

Hence, wavelength of 5800 Å in m = 5.8 x 10 -7m.

Question 2

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

Exercise 1A Numericals

Answer:

Given,

Size of a bacteria = 1 µ

Total length = 1m

∴ Number of bacteria in 1m length = 1m1µ\dfrac{1 m}{ 1 µ}

As we know, 1 µ = 10-6m, substituting the value in the relation above we get:

Number of bacteria in 1m length = 1m106m\dfrac{1 \text m}{10^{-6}\text m}

Hence,
number of bacteria in 1m length = 106 bacteria.

Question 3

The distance of a galaxy from the earth is 5.6 x 1025m. Assuming the speed of light to be 3 x 108 ms-1, find the time taken by light to travel this distance.
[Hint : Time taken=distance travelledspeed\text{Time taken} =\dfrac{\text{distance travelled}}{\text{speed}}]

Exercise 1A Numericals

Answer:

Given,

Time taken=distance travelledspeed\text{Time taken} =\dfrac{\text{distance travelled}}{\text{speed}}

distance = 5.6 x 1025m

speed = 3 x 108 ms-1

Substituting the values in the formula above we get,

Time taken=5.6×10253×108Time taken=1.87×1017s\text{Time taken} =\dfrac{5.6 \times 10^{25}}{3 \times 10^8} \\[0.5em] \Rightarrow \text{Time taken} ={1.87 \times 10^{17}} \text s \\[0.5em]

∴, time taken by light = 1.87 x 1017s.

Question 4

The wavelength of light is 589 nm. What is its wavelength in Å?

Exercise 1A Numericals

Answer:

As we know, 1nm = 10 Å

Given,

The wavelength of light = 589 nm

Substituting the value of wavelength in the relation above, we get,

1nm=10A˚589 nm=589×10A˚589 nm=5890A˚1\text nm = 10 \text Å \\[0.5em] 589\ \text nm = 589 \times 10 \text Å \\[0.5em] \Rightarrow 589\ \text nm = 5890 \text Å \\[0.5em]

Hence, the wavelength of light in Å is 5890 Å.

Question 5

The distance of the nearest star, Proxima Centauri, from the Earth is 4.0 x 1013 km. Express it in light year.

Exercise 1A Numericals

Answer:

1 Light year = 9.46 x 1012 km

or 9.46 x 1012 km = 1 Light year

∴ 4.0 x 1013 km = 19.46×1012\dfrac{1}{9.46 \times 10^{12}} x 4.0 x 1013 = 4.2 light years.

Hence, distance of the nearest star from earth = 4.2 light years.

Question 6

It takes time 8 min for light to reach from the sun to the earth surface. If speed of light is taken to be 3 x 108ms-1, find the distance from the sun to the earth in km.

Exercise 1A Numericals

Answer:

As we know, Distance = speed x time

Given,

Speed = 3 x 108ms-1

Time = 8 min = 8 x 60s = 480s

Substituting the values in the formula above we get,

Distance = 3 x 108 x 480 = 1440 x 108 m.

Converting the distance to km :

Distance in km = 1440×1081000\dfrac{1440 \times 10^8}{1000} km
= 1.44 x 108 km

∴,
the distance from the sun to the earth is 1.44 x 108 km.

Question 7

'The distance of a star from the earth is 8.33 light minutes'. What do you mean by this statement? Express the distance in metre.

Exercise 1A Numericals

Answer:

'The distance of a star from the earth is 8.33 light minutes' implies, it takes 8.33 minutes for light to reach the earth from the star.

As we know,
Distance = speed x time

Given,

Speed = 3 x 108ms-1

Time = 8.33 min
= 8.33 x 60s
= 499.8s
≈ 500s

Substituting the values in the formula above we get,

Distance = 3 x 108 x 500
= 1500 x 108
= 1.5 x 1011

∴, the distance from the star to the earth is 1.5 x 1011 m.

Exercise 1A Short Answer Type

10 questions

Question 1

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

Exercise 1A Short Answer Type

Answer:

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

  1. The unit should be of convenient size.
  2. It should be possible to define the unit without any ambiguity.
  3. The unit should be reproducible.
  4. The value of unit should not change with space and time. (i.e. it must always remain same everywhere).

Question 2

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

Exercise 1A Short Answer Type

Answer:

The fundamental units in S.I. system along with their symbols are as follows —

QuantityUnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
TemperaturekelvinK
Luminous intensitycandelacd
Electric currentampereA
Amount of substancemolemol
Angleradianrd
Solid anglesteradianst-rd

Question 3

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

Exercise 1A Short Answer Type

Answer:

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

Thus, derived units are those which depend on the fundamental units or which can be expressed in terms of the fundamental units.

Example – For the measurement of area, we need to measure length and breadth in the unit of length and then express area in a unit which is:

length x length or (length)2.

Question 4

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

Exercise 1A Short Answer Type

Answer:

The units of length which are bigger than a metre are —

  1. Astronomical Unit (A.U.) — One astronomical unit is equal to the mean distance between the earth and the sun. Relation between metre and astronomical unit is expressed as :
        A.U. = 1.496 x 1011m

  2. Light year (ly) — A light year is the distance travelled by light in vacuum, in one year. Relation between metre and light year is expressed as:
        1 light year = 9.46 x 1015m

Question 5

Name the three convenient units used to measure length ranging from very short to very long value. How are they related to the S.I. unit ?

Exercise 1A Short Answer Type

Answer:

The 3 convenient units used to measure length ranging from very short to very long value are —

  1. Centimeter (cm)
  2. Metre (m)
  3. Kilometer (km)

S.I. unit of length is meter (m). Relation between meter (m) and centimeter is —

    1 m = 100cm

Relation between meter (m) and kilometer is —

    1 km = 1000m

Question 6

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

Exercise 1A Short Answer Type

Answer:

The S.I. unit of mass is Kilogram (Kg).

One kilogram is defined as the mass of a cylindrical piece of platinum-iridium alloy kept at International Bureau of Weights and Measures at Serves near Paris.

Question 7

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

Exercise 1A Short Answer Type

Answer:

The two units of mass smaller than a kilogram (kg) are :

  1. gram (g) — Relation between gram and kilogram is :
        1 g = 10-3 kg
  2. milligram (mg) — Relation between miligram and kilogram is :
        1 mg = 10-6 kg

Question 8

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

Exercise 1A Short Answer Type

Answer:

The two units of mass bigger than a kilogram (kg) are :

  1. Quintal — It is one hundred times a kilogram. Relation between quintal and kilogram is :
        1 quintal = 100 kg
  2. Metric tonne — It is one thousand times a kilogram. Relation between metric tonne and kilogram is :
        1 metric tonne = 1000 kg

Question 9

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

Exercise 1A Short Answer Type

Answer:

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

A second can be defined as 1/86400th part of a mean solar day,

i.e.,

1s=186400×one mean solar day1s = \dfrac{1}{86400} \times \text{one mean solar day}

Question 10

Name two units of time bigger than a second. How are they related to second?

Exercise 1A Short Answer Type

Answer:

The two units of time bigger than a second (s) are :

  1. Minute (min) — One minute is the duration of 60 seconds. Relation between minute and second is :
        1 min = 60s
  2. Hour (h) — One hour is the duration of 60m minutes. Relation between hour and second is :
        1 h = 3600s

Exercise 1A Very Short Answer Type

20 questions

Question 1

What is meant by measurement?

Exercise 1A Very Short Answer Type

Answer:

Measurement is the process of comparison of the given quantity with the known standard quantity of the same nature.

Question 2

What do you understand by the term unit?

Exercise 1A Very Short Answer Type

Answer:

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

Question 3

How is a physical quantity expressed?

Exercise 1A Very Short Answer Type

Answer:

A physical quantity is expressed in terms of the following parameters :

  1. The unit in which the quantity is being measured, and
  2. The numerical value which expresses, how many times the above selected unit is contained in the given quantity.

Thus, the magnitude of physical quantity is expressed as : Physical quantity = (numerical value) x (unit)

Question 4

Name the three fundamental quantities.

Exercise 1A Very Short Answer Type

Answer:

The three fundamental quantities are —

  1. Mass
  2. Length
  3. Time

Question 5

What is the S.I. unit of Luminous intensity?

Exercise 1A Very Short Answer Type

Answer:

The S.I. unit of Luminous intensity is candela (cd).

Question 6

Define one parsec.

Exercise 1A Very Short Answer Type

Answer:

One parsec is the distance from where the semi major axis of orbit of earth (1 A.U.) subtends an angle of one second.

Question 7

Define a fundamental unit.

Exercise 1A Very Short Answer Type

Answer:

A fundamental 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.

Example – Units of mass, length, time, temperature, current etc.

Question 8

Define a derived unit.

Exercise 1A Very Short Answer Type

Answer:

Derived units are those which depend on the fundamental units or which can be expressed in terms of fundamental units.

Example: For the measurement of area, we need to measure length and breadth in the unit of length and then express area in a unit which is length x length or (length)2

Question 9

Define standard metre.

Exercise 1A Very Short Answer Type

Answer:

The standard metre is defined in terms of speed of light, according to which, one metre is the distance travelled by light in 1/299,792,458 of a second in air (or vacuum).

Question 10

How is nanometer related to Angstrom ?

Exercise 1A Very Short Answer Type

Answer:

Relation between nanometer (nm) and Angstrom (Å) is expressed as :
    1 nanometer = 10 Å

Question 11

Complete the following —

  1. 1 light year = ________ m
  2. 1 m = ________ Å
  3. 1 m = ________ µ
  4. 1 micron = ________ Å
  5. 1 fermi = ________ m
Exercise 1A Very Short Answer Type

Answer:

  1. 1 light year = 9.46 x 1015 m
  2. 1 m = 1010 Å
  3. 1 m = 106 µ
  4. 1 micron = 104 Å
  5. 1 fermi = 10-15 m

Question 12

Complete the following —

  1. 1g = ______kg
  2. 1mg = ________kg
  3. 1 quintal = ________kg
  4. 1 a.m.u (or u) = ________kg
Exercise 1A Very Short Answer Type

Answer:

  1. 1 g = 10-3 kg
  2. 1 mg = 10-6 kg
  3. 1 quintal = 100 kg
  4. 1 a.m.u (or u) = 1.66 x 10-27 kg

Question 13

What is a leap year?

Exercise 1A Very Short Answer Type

Answer:

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

1 Leap year = 366 days

Every fourth year (i.e., the year divisible by 4) has one day extra in the month of february (i.e., February has 29 days) and so it is the leap year.

Question 14

'The year 2024 will have February of 29 days'. Is this statement true?

Exercise 1A Very Short Answer Type

Answer:

Yes the statement is true.

We know that, if any year is divisible by 4, then it is a leap year and in a leap year, February has 29 days. As, the year 2024 is divisible by 4, so it will have 29 days in February.

Question 15

What is a lunar month?

Exercise 1A Very Short Answer Type

Answer:

A lunar month is the time of one lunar cycle, i.e., it is the amount of time it takes for the Moon to complete one orbit around the Earth and it is nearly equal to 29.5 days.

Question 16

Complete the following —

  1. 1 nano second = ________s
  2. 1 µs = _______s
  3. 1 mean solar day = ________s
  4. 1 year = ________s
Exercise 1A Very Short Answer Type

Answer:

  1. 1 nano second = 10-9 s
  2. 1 µs = 10-6 s
  3. 1 mean solar day = 86400 s
  4. 1 year = 3.15 x 107 s

Question 17

Name the physical quantities which are measured in the following units —

  1. u
  2. ly
  3. ns
  4. nm
Exercise 1A Very Short Answer Type

Answer:

Physical quantity related to the unit are as follows —

UnitPhysical quantity
uMass
lyDistance
nsTime
nmLength

Question 18

Write the derived units of the following —

  1. Speed
  2. Force
  3. Work
  4. Pressure
Exercise 1A Very Short Answer Type

Answer:

The derived units of the following quantities are as follows —

QuanityDerived unit
Speedms-1
Forcekg m s-2
Workkg m2s-2
Pressurekg m-1s-2

Question 19

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

  1. Newton
  2. Watt
  3. Joule
  4. Pascal
Exercise 1A Very Short Answer Type

Answer:

Derived unitFundamental unit
Newtonkg m s-2
Wattkg m2s-3
Joulekg m2s-2
Pascalkg m-1s-2

Question 20

Name the physical quantities related to the following units —

  1. km2
  2. newton
  3. joule
  4. pascal
  5. watt
Exercise 1A Very Short Answer Type

Answer:

The physical quantities related to the following units are —

UnitPhysical quantity
km2area
newtonforce
jouleenergy
pascalpressure
wattpower

Exercise 1B Long Answer Type

5 questions

Question 1

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

Exercise 1B Long Answer Type

Answer:

Sometimes due to mechanical error in vernier callipers, the zero mark of the vernier scale does not coincide with the zero mark on the main scale. Hence, the vernier callipers is said to have a zero error.

In order to find the zero 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 gives the zero error.

What is meant by zero error of a vernier callipers? How is it determined? Draw neat diagrams to explain it. How is it taken in account to get the correct measurement? Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

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

we get,

Zero Error=6×L.C.Zero Error=6×0.01cmZero Error=0.06cm\text{Zero Error} = 6 \times \text{L.C.} \\[0.5em] \text{Zero Error}= 6 \times 0.01 \text{cm} \\[0.5em] \Rightarrow \text{Zero Error}= 0.06 \text {cm} \\[0.5em]

Hence, zero error = 0.06 cm.

In order to correct the measurement of the vernier callipers with zero error, the zero error with proper sign is subtracted from the observed reading.

Hence, Correct reading = observed reading – zero error (with sign)

Question 2

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

Exercise 1B Long Answer Type

Answer:

A neat labelled diagram of a vernier callipers is shown below:

Draw a neat labelled diagram of a vernier callipers. Name its main parts and state their functions. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

The main parts of the vernier callipers are as follows —

PartFunction
Outside jawsTo measure the length of a rod, diameter of a sphere, external diameter of a hollow cylinder.
Inside jawsTo measure the internal diameter of a hollow cylinder or pipe.
StripTo measure the depth of a beaker or a bottle.
Main scaleTo measure length correct up to 1 mm.
Vernier scaleHelps to measure length correct upto 0.1 mm.

Question 3

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

Exercise 1B Long Answer Type

Answer:

In order to measure the length of a small rod, we follow the following steps:

Describe in steps, how would you use a vernier callipers to measure the length of a small rod? Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.
  1. Find the least count and zero error of the vernier callipers.

  2. The rod is placed between the fixed end and the vernier scale as shown in the diagram.

  3. Note the main scale reading.

  4. Note that division p on vernier scale which coincide or is in line with any division of the main scale. Multiply this vernier division p with the least count. This is the vernier scale reading i.e., Vernier scale reading = p x L.C.

  5. Repeat it two times and record the observation.

Observations —

Total number of divisions on vernier scale (n) = ...........

Value of one division on main scale (x) = ........... cm

Least count (L.C.)=xn=..........cm\text {Least count (L.C.)} = \dfrac{\text{x}}{\text{n}} = ..........\text{cm}

Zero error = ....... cm

S No.Main scale reading a (in cm)Vernier division coinciding (p)Vernier scale division b = p x L.C.(in cm)Observed length = a + b (in cm)
(1)             
(2)             
(3)             

Mean observed length = ......... cm

From the mean observed length, subtract zero error, if any, with its proper sign to obtain the true measurement of the length of the given project.

Thus we get,

Observed length = main scale reading + (vernier division p coinciding with any division on the main scale) x least count.

True length = observed length - zero error (with sign).

Question 4

Draw a neat and labelled diagram of a screw gauge. Name its main parts and state their functions.

Exercise 1B Long Answer Type

Answer:

Below is the diagram of a screw gauge with all its parts labelled:

Draw a neat and labelled diagram of a screw gauge. Name its main parts and state their functions. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

The main parts and their functions of a screw gauge are as follows —

PartFunction
RatchetTo advance the screw by turning it till the object is gently held between the stud and the spindle of the screw.
SleeveTo mark main scale and base line.
ThimbleTo mark circular scale.
Main scaleTo read length correct up to 1 mm.
Circular scaleHelps to read length correctly to 0.01 mm.

Question 5

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

Exercise 1B Long Answer Type

Answer:

In order to measure the diameter of a wire with the help of a screw gauge we follow the following steps —

  1. Find the least count and the zero error of the screw gauge.

  2. Turn the ratchet anticlockwise, so as to obtain a gap between the stud A and the flat end B. Place the wire in the gap between the stud A and the flat end B. Then turn the ratchet clockwise so as to hold the given wire gently between the stud A and the flat end B of the screw.

  3. Note the main scale reading.

  4. Note that division p of the circular scale that coincides with the base line of the main scale. This circular scale division p when multiplied by the least count, gives the circular scale reading i.e., Circular scale reading = p x L.C.

  5. Add the circular scale reading to the main scale reading to obtain the total reading (i.e., the observed diameter of the wire).

  6. Repeat it by keeping the wire in perpendicular direction. Take two more observations at different places of the wire and record them in a table.

Observations —

Pitch of the screw = ......... cm

Total number of divisions on the circular scale (n) = ...........

L.C. of screw gauge=PitchTotal no. of circular scale div=..........cm\text {L.C. of screw gauge} = \dfrac{\text{Pitch}}{\text{Total no. of circular scale div}} = ..........\text{cm}

Zero error = ....... cm

S No.Main scale reading a (in cm)Circular scale reading b = number of division of circular scale in line with the base line, p x L.C. (in cm)Observed diameter = a + b (in cm)
(1) (i) in one direction
    (ii) in perpendicular direction
             
(2) (i)
     (ii)
             
(3) (i)
     (ii)
             

Mean observed length = ......... cm

From the mean observed length, subtract zero error, if any, with its proper sign to obtain the true measurement of the length of the given project.

Thus we get,

Thus,

Observed diameter = main scale reading + (circular scale division p coinciding the coinciding the base line of main scale x least count).

True diameter = observed diameter - zero error (with sign)

Exercise 1B Multiple Choice Type

6 questions

Question 1

The strip of a vernier callipers is used to measure :

  1. Length of a wire
  2. Depth of a beaker
  3. Internal diameter of a sphere
  4. Diameter of a hollow sphere
Exercise 1B Multiple Choice Type

Answer:

Depth of a beaker

Reason — The function of Vernier callipers' strip is to measure the depth of a beaker.

Question 2

On bringing the two jaws of a vernier callipers together, the 6th division of vernier scale coincides with the main scale. If the least count of the vernier callipers is 0.01 cm, then its zero error is:

  1. +0.06 cm
  2. +0.6 cm
  3. +0.006 cm
  4. +0.001 cm
Exercise 1B Multiple Choice Type

Answer:

+0.06 cm

Reason — Zero error = +6 x Least count = +6 x 0.01 = 0.06 cm

Question 3

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
Exercise 1B Multiple Choice Type

Answer:

0.002 cm

Reason — The value of one main scale division x = 120\dfrac{1}{20} cm

The number of divisions on vernier scale n = 25

Using the formula given below,

Least count (L.C.)= Value of 1 main scale div (x)total no. of div on vernier(n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text {total no. of div on vernier(n)}}

we get,

L.C.=xnL.C.=12025 cmL.C.=1500 cmL.C.=0.002 cm\text {L.C.} = \dfrac{\text{x}}{\text{n}} \\[0.5em] \text{L.C.} = \dfrac{\dfrac{\text{1}}{\text{20}}}{25}\text{ cm} \\[0.5em] \text{L.C.} = \dfrac{\text{1}}{\text{500}}\text{ cm} \\[0.5em] \Rightarrow \text{L.C.} = 0.002\text{ cm} \\[0.5em]

Hence, least count of microscope is 0.002 cm.

Question 4

Identify the correct use of the thimble of a screw gauge :

  1. To read length correct up to 0.01 mm
  2. To read length correct up to 1 mm
  3. To mark main scale and base line
  4. To mark circular scale
Exercise 1B Multiple Choice Type

Answer:

To mark circular scale

Reason — Use of the thimble of a screw gauge is to mark the circular scale.

Question 5

The circular head of a screw gauge is divided into 50 divisions and the screw moves 1 mm ahead in two revolutions of the circular head. Its pitch would be :

  1. 0.05 cm
  2. 0.001 cm
  3. 0.5 cm
  4. 0.01 cm
Exercise 1B Multiple Choice Type

Answer:

0.05 cm

Reason — Pitch = distance moved ahead in 1 revolution

Given, distance moved in 2 revolutions = 1 mm

∴ Distance moved in 1 revolution = 12\dfrac{1}{2} mm = 0.5 mm

Converting 0.5 mm to cm,

0.5 mm = 0.510\dfrac{0.5}{10} cm = 0.05 cm

∴ Pitch of screw gauge is 0.05 cm

Question 6

Identify the incorrect statement from the following :

(a) A screw gauge is used to measure very small dimensions like the thickness of a wire or sheet.

(b) Vernier callipers is used to measure larger dimensions like the diameter of a sphere, internal or external diameter lengths, etc.

(c) Vernier callipers works on the principle of a sliding scale to measure fractional parts of a main scale.

(d) Vernier callipers has a higher precision with a larger least count whereas screw gauge has a lower precision with a smaller least count.

Exercise 1B Multiple Choice Type

Answer:

Vernier callipers has a higher precision with a larger least count whereas screw gauge has a lower precision with a smaller least count.

Reason — A screw gauge has higher precision (smaller least count) than vernier callipers, not lower and screw gauge can typically measure up to 0.01 mm or 0.001 cm, whereas a vernier caliper usually has a least count of 0.1 mm or 0.01 cm. So, the screw gauge is more precise.

Exercise 1B Numericals

15 questions

Question 1

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

Exercise 1B Numericals

Answer:

Using the formula given below,

Least count (L.C.)= Value of one div (x)Total no. of div on stop watch (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of one div (x)}}{\text {Total no. of div on stop watch (n)}}

Given,

Total number of divisions on stop watch = 10

we get,

L.C.=xnL.C.=510L.C.=12L.C.=0.5s\text{L.C.} = \dfrac{\text{x}}{\text{n}} \\[0.5em] \text{L.C.} = \dfrac{5}{10} \\[0.5em] \text{L.C.} = \dfrac{1}{2} \\[0.5em] \Rightarrow \text{L.C.} = 0.5\text{s}

Hence, least count of stop watch is 0.5s.

Question 2

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

Exercise 1B Numericals

Answer:

(i) As we know,

L.C.=Value of 1 main scale div (x)Total no. of div on vernier (n)\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier (n)}}

Given,

Total number of divisions on vernier = 10

Value of one main scale division(x) = 1 mm

Substituting the values in the formula given above we get,

L.C.=1mm10L.C.=0.1mmL.C.=0.01cm\text{L.C.} = \dfrac{\text{1mm}}{\text{10}} \\[0.5em] \Rightarrow \text{L.C.} = 0.1 \text{mm} \\[0.5em] \Rightarrow \text{L.C.} = 0.01 \text{cm} \\[0.5em]

Hence, least count of the vernier callipers is 0.01cm.

Question 3

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

Exercise 1B Numericals

Answer:

i) As we know,

L.C.=Value of 1 main scale div (x)Total no. of div on vernier(n)\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier(n)}}

Given,

20 divisions = 1 cm

∴,

1 division=1201 division=0.05cm\text{1 division} = \dfrac{1}{20} \\[0.5em] \Rightarrow \text{1 division} = 0.05 \text{cm}

Hence, value of one main scale division = 0.05 cm

Total number of divisions = 50

Substituting the values in the formula above, we get,

L.C.=0.05cm50L.C.=0.001cm\text{L.C.} = \dfrac{\text{0.05} \text{cm}}{\text{50}} \\[0.5em] \text{L.C.} = 0.001 \text{cm} \\[0.5em]

Hence, least count of the microscope is 0.001 cm.

Question 4

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

Exercise 1B Numericals

Answer:

As we know,

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

Given,

Thickness of the pencil = 1.4 mm

Zero error of the vernier callipers = +0.02cm = 0.2 mm

Substituting the values in the formula above we get,

Correct reading = observed reading - zero error (with sign)
Correct reading = 1.4mm - 0.2mm
Correct reading = 1.2mm

Hence, correct thickness of pencil is 1.2 mm

Question 5

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 vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with a main scale division.

Find —

(i) the least count and

(ii) the zero error of the vernier callipers.

Exercise 1B Numericals

Answer:

(i) As we know,

L.C.=Value of 1 main scale div (x)Total no. of div on vernier(n)\text{L.C.} = \dfrac{\text{Value of 1 main scale div (x)}}{\text{Total no. of div on vernier(n)}}

Given,

Total number of divisions on vernier (n) = 10

Value of one main scale division (x) = 1 mm

Substituting the values in the formula above, we get,

L.C.=1mm10L.C.=0.1mmL.C.=0.01cm\text{L.C.} = \dfrac{\text{1mm}}{\text{10}} \\[0.5em] \text{L.C.} = 0.1 \text{mm} \\[0.5em] \text{L.C.} = 0.01 \text{cm} \\[0.5em]

Hence, least count of the vernier callipers is 0.01cm.

(ii) As we know,

Zero error = L.C. x Coinciding division (with sign)

and

Coinciding division = 3

L.C. = 0.01 cm

Substituting the values in the formula above we get,

Zero error=0.01 cm x 3Zero error=0.03 cm\text {Zero error} = \text {0.01 cm x 3} \\[0.5em] \Rightarrow \text {Zero error} = \text {0.03 cm} \\[0.5em]

Hence, the zero error of the vernier callipers = + 0.03 cm.

Question 6

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 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 cylinder.

Exercise 1B Numericals

Answer:

(i) As we know,

Least count (L.C.)= Value of 1 main scale div (x)total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{total no. of div on vernier (n)}}

Given,

Value of one main scale division(x) = 1 mm

Total number of divisions on vernier callipers (n) = 20

Substituting the values in the formula above we get,

L.C.=120L.C.=0.05mmL.C.=0.005cm\text{L.C.} = \dfrac{1}{\text{20}} \\[0.5em] \text{L.C.} = 0.05 \text{mm} \\[0.5em] \text{L.C.} = 0.005 \text{cm} \\[0.5em]

Hence, least count of the vernier callipers is 0.005 cm.

(ii) As we know,

circular scale reading = L.C. x Coinciding division

Given,

Coinciding division = 4

L.C. = 0.005 cm

Substituting the values in the formula above we get,

circular scale reading=0.005 cm x 4circular scale reading=0.02 cm\text {circular scale reading} = \text {0.005 cm x 4} \\[0.5em] \Rightarrow \text {circular scale reading} = \text {0.02 cm} \\[0.5em]

Hence, the circular scale reading = 0.02 cm. ...... (1)

Main scale reading = 3.5 cm (as it reads 35th division on mm scale)

Hence, the main scale reading = 3.5 cm. ........ (2)

Substituting the values 1 and 2 in the formula for total reading we get,

Total reading=main scale reading + vernier scale readingDiameter = Total reading=3.5 cm + 0.02 cmDiameter=3.52 cmRadius=3.522cmRadius=1.76cm\text{Total reading} = \text{main scale reading } + \text{ vernier scale reading} \\[0.5em] \text{Diameter = Total reading} = \text{3.5 cm + 0.02 cm} \\[0.5em] \Rightarrow \text{Diameter} = \text{3.52 cm} \\[0.5em] \Rightarrow \text{Radius} = \dfrac{3.52}{2}\text{cm} \\[0.5em] \Rightarrow \text{Radius} = 1.76\text{cm} \\[0.5em]

Hence, the radius of the cylinder = 1.76 cm.

Question 7

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

  1. Find the length

  2. If zero of vernier callipers is -0.02cm, what is the correct length?

Exercise 1B Numericals

Answer:

(i) As we know,

Least count (L.C.)= Value of 1 main scale div (x)total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{total no. of div on vernier (n)}}

Given,

Total number of divisions on the vernier scale (n) = 10

Value of one main scale division (x) = 1 mm

Coinciding division = 4

Substituting the values in the formula above we get,

L.C.=110L.C.=0.1mmL.C.=0.01cm\text{L.C.} = \dfrac{1}{\text{10}} \\[0.5em] \Rightarrow \text{L.C.} = 0.1\text{mm} \\[0.5em] \Rightarrow \text{L.C.} = 0.01\text{cm} \\[0.5em]

Hence, least count of the vernier callipers is 0.01 cm.

As we know,

Vernier scale reading=Vernier scale division×L.C.Vernier scale reading=4×0.01cmVernier scale reading=0.04cm\text{Vernier scale reading} = \text{Vernier scale division} \times \text{L.C.} \\[0.5em] \text{Vernier scale reading} = 4 \times 0.01 \text{cm} \\[0.5em] \Rightarrow \text{Vernier scale reading} = 0.04 \text{cm} \\[0.5em]

Hence, Vernier scale reading is 0.04 cm.

and

Main scale reading of the vernier callipers = 1.8 cm

∴,

Total reading=main scale reading+vernier scale readingTotal reading=1.8cm+0.04cmTotal reading=1.84cm\text{Total reading} = \text{main scale reading} + \text{vernier scale reading} \\[0.5em] \text{Total reading} = 1.8 \text{cm} + 0.04 \text{cm} \\[0.5em] \Rightarrow \text{Total reading}= 1.84\text{cm} \\[0.5em]

Hence, the length is 1.84 cm.

(b) As we know,

Correct reading=observed reading–zero error\text{Correct reading} = \text {observed reading} – \text{zero error}

Given,

zero error is -0.02cm,

Substituting the values in the formula above we get,

Correct reading=1.84(0.02)Correct reading=1.84+0.02Correct reading=1.86cm\text{Correct reading} = 1.84 – (-0.02) \\[0.5em] \text{Correct reading} = 1.84 + 0.02 \\[0.5em] \text{Correct reading} = 1.86 \text{cm} \\[0.5em]

Hence, the correct length is 1.86 cm.

Question 8

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

While measuring the length of a rod with a vernier callipers, figure below shows the position of its scales. What is the length of the rod? Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.
Exercise 1B Numericals

Answer:

As we know,

Least count (L.C.)= Value of 1 main scale div (x)total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{total no. of div on vernier (n)}}

There are 10 divisions on 1 cm on the main scale

∴,

Value of 1 main scale division(x)=110 cm=0.1cm\text{Value of 1 main scale division(x)} = \dfrac{1}{10}\text{ cm} = 0.1 \text{cm} \\[0.5em]

Total number of divisions on the vernier scale (n) = 10

Least count (L.C.)=0.110L.C.=0.01cm\text{Least count (L.C.)} = \dfrac{0.1}{10} \\[0.5em] \text{L.C.} = 0.01 \text {cm}\\[0.5em]

Hence, the least count = 0.01 cm.

Given,

Main scale reading = 3.3 cm

coinciding division = 6

As we know,

Vernier scale reading=Vernier scale division×L.C.Vernier scale reading=6×0.01cmVernier scale reading=0.06cm\text{Vernier scale reading} = \text{Vernier scale division} \times \text{L.C.} \\[0.5em] \text{Vernier scale reading} = 6 \times 0.01 \text{cm} \\[0.5em] \Rightarrow \text{Vernier scale reading} = 0.06 \text{cm} \\[0.5em]

Hence, Vernier scale reading is 0.06 cm.

Observed reading=main scale reading+vernier scale readingObserved reading=3.3cm+0.06cmObserved reading=3.36cm\text{Observed reading} = \text{main scale reading} + \text{vernier scale reading} \\[0.5em] \text{Observed reading} = 3.3 \text{cm} + 0.06 \text{cm} \\[0.5em] \Rightarrow \text{Observed reading} = 3.36\text{cm} \\[0.5em]

Hence, the length of the rod is 3.36 cm.

Question 9

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 screw gauge ?

Exercise 1B Numericals

Answer:

As we know,

Least Count=PitchTotal no. of div on circular head\text {Least Count} = \dfrac{\text{Pitch}}{\text {Total no. of div on circular head}} \\[0.5em]

Given,

Pitch = 0.5 mm

Number of divisions on circular head = 100

Substituting the values in the formula above we get,

Least count=0.5100Least count=0.005 mm=0.0005 cm\text {Least count} = \dfrac{0.5}{100} \\[0.5em] \text {Least count} = 0.005 \text { mm} = 0.0005 \text { cm} \\[0.5em]

Hence, the least count of the screw gauge = 0.005 mm or 0.0005 cm.

Question 10

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?

Exercise 1B Numericals

Answer:

(i) As we know,

Pitch = distance moved ahead in 1 revolution

Given,

Distance covered in two revolutions = 1 mm

∴, we get,

Pitch=12 mm=0.5 mm\text {Pitch} = \dfrac{1}{2} \text{ mm} = 0.5 \text{ mm} \\[0.5em]

Hence, pitch of the screw gauge = 0.5 mm

(ii) As we know,

Least Count=PitchTotal no. of div on circular head\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]

Given,

Pitch = 0.5 mm

Total number of divisions on circular scale = 50

Substituting the values in the formula above we get,

Least count=0.550Least count=0.01 mm\text {Least count} = \dfrac{0.5}{50} \\[0.5em] \text {Least count} = 0.01 \text { mm} \\[0.5em]

Hence, the least count of the screw gauge = 0.01 mm

Question 11

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

Find —

  1. The least count, and
  2. The diameter of the wire
Exercise 1B Numericals

Answer:

(i) As we know,

Least Count=PitchTotal no. of div on circular head\text{Least Count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]

Given,

Pitch = 1 mm

Number of divisions on circular head = 100

Substituting the values in the formula above we get,

Least count=1100mmLeast count=0.01mm=0.001cm\text {Least count} = \dfrac{1}{100} \text{mm} \\[0.5em] \text {Least count} = 0.01 \text {mm} = 0.001 \text {cm}\\[0.5em]

Hence, the least count of the screw gauge = 0.001 cm

(ii) As we know,

Diameter of the wire = main scale reading + circular scale reading     [Equation 1]

and

Circular scale reading = p x L.C.     [Equation 2]

p = 45

L.C. = 0.001 cm

Substituting the values in the Equation 2 we get,

Circular scale reading = 45 x 0.001 = 0.045

Hence, circular scale reading = 0.045 cm

Given, main scale reads 2 mm = 0.2 cm

Hence, main scale reading = 0.2 cm

Substituting the values in Equation 1 we get,

Diameter of the wire = 0.2 cm + 0.045 cm = 0.245 cm

Hence, the diameter of the wire is 0.245 cm.

Question 12

When a screw gauge of least count 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.

  1. What is the diameter of the wire in cm?

  2. If the zero error is +0.005 cm, what is the correct diameter?

Exercise 1B Numericals

Answer:

As we know,

Diameter of the wire = main scale reading + circular scale reading     [Equation 1]

and

Reading on thimble = p x L.C.     [Equation 2]

Given,

p = 27

L.C. = 0.01 mm = 0.001 cm

Substituting the values in the Equation 2 we get,

reading on thimble = 27 x 0.001 = 0.027 cm

Hence, reading on thimble = 0.027 cm and reading on sleeve = 1 mm = 0.1 cm

Using Equation 1 we get,

Diameter of the wire = 0.1 cm + 0.027 cm = 0.127 cm

Hence, the diameter of the wire is 0.127 cm.

(ii) As we know,

Correct reading = observed reading - zero error

Given,

Zero error = + 0.005 cm

Substituting the value of zero error in the formula above we get,

Correct reading = 0.127 cm - 0.005 cm = 0.122 cm

Hence, the correct diameter of the wire is 0.122 cm

Question 13

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

Exercise 1B Numericals

Answer:

As we know,

Pitch = Distance moved ahead in 1 revolution.

Given,

Total number of divisions on circular scale = 50

Distance covered in two revolutions = 1 mm

∴, we get,

Pitch=12 mm=0.5 mm\text {Pitch} = \dfrac{1}{2} \text{ mm} = 0.5 \text{ mm} \\[0.5em]

Hence, pitch of the screw gauge = 0.5 mm

(ii) As we know,

Least count=PitchTotal no. of div on circular head\text{Least count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]

Here,

Pitch = 0.5 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

Least count=0.550Least count=0.01mm\text {Least count} = \dfrac{0.5}{50} \\[0.5em] \text {Least count} = 0.01 \text {mm} \\[0.5em]

Hence, the least count of the screw gauge = 0.01 mm

(iii) As we know,

Zero error = Coinciding Division x Least Count

Coinciding Division = 4

L.C. = 0.01 mm

Substituting the values in the formula above we get,

Zero error = 4 x 0.01 = + 0.04 mm

Hence, zero error, of the screw gauge = + 0.04 mm

Question 14

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

Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once. Find (i) pitch of the screw gauge (ii) least count of the screw gauge (iii) the diameter of the wire. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

Find —

(i) pitch of the screw gauge,

(ii) least count of the screw gauge, and

(iii) the diameter of the wire.

Exercise 1B Numericals

Answer:

(i) As we know,

Pitch = distance moved ahead in 1 revolution

and given,

Distance covered in one revolutions = 1 mm

Hence, Pitch of the screw gauge = 1 mm

(ii) As we know,

Least count=PitchTotal no. of div on circular head\text{Least count} = \dfrac {\text{Pitch}}{\text{Total no. of div on circular head}}\\[0.5em]

Here,

Pitch = 1 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

Least count=150Least count=0.02 mm\text {Least count} = \dfrac{1}{50} \\[0.5em] \text {Least count} = 0.02 \text { mm} \\[0.5em]

Hence, the least count of the screw gauge = 0.02 mm

(iii) As we know,

Diameter of the wire = main scale reading + circular scale reading     [Equation 1]

and

Circular scale reading = p x L.C.     [Equation 2]

p = 47

L.C. = 0.02 mm

Substituting the values in the Equation 2 we get,

Circular scale reading = 47 x 0.02 = 0.94 mm

Hence, circular scale reading = 0.94 mm and main scale reading = 4 mm.

Using Equation 1 we get,

Diameter of the wire = 4 mm + 0.94 mm = 4.94 mm

Hence, the diameter of the wire is 4.94 mm.

Question 15

A screw has a pitch equal to 0.5 mm. What should be the number of division on its head so as to read correct up to 0.001mm with its help?

Exercise 1B Numericals

Answer:

As we know,

Least count=PitchTotal no. of div on circular head\text {Least count} = \dfrac{\text{Pitch}}{\text{Total no. of div on circular head}} \\[0.5em]

Given,

Pitch = 0.5 mm

L.C. = 0.001 mm

Substituting the values in the formula above we get,

0.001mm=0.5Total no. of div on circular headTotal no. of div on circular head=0.50.001Total no. of div on circular head=5000.001\text{mm} = \dfrac{0.5}{\text{Total no. of div on circular head}} \\[0.5em] \Rightarrow \text{Total no. of div on circular head} = \dfrac{0.5}{0.001} \\[0.5em] \Rightarrow \text{Total no. of div on circular head} = 500

Hence, total number of divisions on circular head = 500.

Exercise 1B Short Answer Type

12 questions

Question 1

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

Exercise 1B Short Answer Type

Answer:

The least count of an instrument is the smallest measurement that can be taken accurately with it.

Example — The least count of a stop watch is 0.5 second, if there are 10 divisions between 0 and 5s marks.

Question 2

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

Exercise 1B Short Answer Type

Answer:

The ruler graduated by the boy, was having its zero mark at one end and 100 cm mark at the other end.

It had 100 subdivisions in one metre length, so the value of its one small division is 1 cm.

Thus, the ruler can be used to measure length accurately up to 1 cm.

In order to increase the accuracy, each cm should be further divided into 10 divisions so that the ruler can have a least count of 1 mm.

Question 3

A boy measures the length of a pencil and expresses it to be 2.6cm. What is the accuracy of his measurement? Can he write it as 2.60cm?

Exercise 1B Short Answer Type

Answer:

Given, the length of the pencil is 2.6 cm.

As the length is measured in cm till one place of decimal, we can assume that it was measured using a metre rule.

So, the measurement is accurate

No, it cannot be written as 2.60, as it would mean that length is measured precisely up to second decimal place using vernier callipers or a screw gauge.

Question 4

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

Exercise 1B Short Answer Type

Answer:

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

It is calculated as below —

Least count (L.C.)= Value of 1 main scale divValue of 1 vernier scale div\text{Least count (L.C.)} = \text{ Value of 1 main scale div} - \text{Value of 1 vernier scale div}

Question 5

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

Exercise 1B Short Answer Type

Answer:

Below diagram shows a vernier callipers with a zero error of +0.06 cm.

A vernier callipers has a zero error +0.06cm. Draw a neat labelled diagram to represent it. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

The least count of the vernier callipers shown in the diagram is 0.01 cm and the 6th division of the vernier scale, coincides with a main scale division.

∴,

Zero Error=6×L.C.Zero Error=6×0.01cmZero Error=0.06cm\text{Zero Error} = 6 \times \text{L.C.} \\[0.5em] \text{Zero Error}= 6 \times 0.01 \text{cm} \\[0.5em] \Rightarrow \text{Zero Error}= 0.06 \text {cm} \\[0.5em]

Hence, we get the zero error = 0.06 cm.

Question 6

State three uses of a vernier callipers.

Exercise 1B Short Answer Type

Answer:

Three uses of vernier callipers are as follows —

  1. It is used to measure the length of a rod.

  2. It is used to measure the diameter of a sphere.

  3. It is used to measure the internal and external diameter of a hollow cylinder.

Question 7

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

Exercise 1B Short Answer Type

Answer:

The two scales of a vernier callipers are —

  1. Main scale - it is fixed
  2. Vernier scale – slides along the main scale.
Name the two scales of a vernier callipers and explain, how it is used to measure a length correct up to 0.01cm. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

In the figure shown above, the main scale is graduated to read up to 1mm and the vernier scale has the length of 10 divisions equal to the length of 9 divisions of main scale.

Value of 1 division of main scale (x) = 1mm.

Total number of divisions on the vernier scale (n) = 10

Using the formula given below,

Least count (L.C.)= Value of 1 main scale div (x)total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text {total no. of div on vernier (n)}}

Substituting the value in the formula above, we get,

Least count (L.C.)=xnLeast count (L.C.)=110mmLeast count (L.C.)=0.1mmLeast count (L.C.)=0.01cm\text{Least count (L.C.)} = \dfrac{\text{x}}{\text{n}} \\[0.5em] \text{Least count (L.C.)} = \dfrac{\text{1}}{\text{10}} \text{mm} \\[0.5em] \text{Least count (L.C.)} = 0.1\text{mm} \\[0.5em] \Rightarrow \text{Least count (L.C.)} = 0.01\text{cm} \\[0.5em]

Hence, a vernier callipers is used to measure a length accurately up to 0.01cm.

Question 8

Explain the terms —

(i) pitch, and

(ii) least count of a screw gauge.

How are they determined?

Exercise 1B Short Answer Type

Answer:

  1. Pitch of a screw gauge — The pitch of a screw gauge is the linear distance moved by its screw on the main scale when the circular scale is given one complete rotation.

  2. Least count of a screw gauge — Least count of a screw gauge is the linear distance moved by its screw along the main scale when the circular scale is rotated by one division on it.

The pitch and least count of the screw gauge are determined by using the formula —

Least Count=Pitch of screw gaugeTotal no. of div on circular scale\text{Least Count} = \dfrac{\text{Pitch of screw gauge}}{\text{Total no. of div on circular scale}}

Question 9

How can the least count of a screw gauge be decreased?

Exercise 1B Short Answer Type

Answer:

The least count can be decreased by —

  1. Decreasing the pitch.

  2. Increasing the total number of divisions on the circular scale.

Question 10

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

Exercise 1B Short Answer Type

Answer:

In an ideal case, when the flat end B of the screw is in contact with the stud A, and if the zero mark of circular scale coincide with the base line of main scale, the screw gauge is said to be free from zero error.

But sometimes, due to the mechanical error, on bringing the stud A in contact with stud B, the zero mark of the circular scale is either below or above the base line of the main scale, then the screw gauge is said to have a zero error.

There are two types of zero error —

  1. Positive zero error and

  2. Negative zero error.

The zero error is accounted by subtracting the zero error with its sign from the observed reading.

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

Question 11

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

Exercise 1B Short Answer Type

Answer:

Below diagram shows a screw gauge with least count 0.001 cm and zero error + 0.007 cm:

A screw gauge has a least count 0.001 cm and zero error + 0.007 cm. Draw a neat diagram to represent it. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

Question 12

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

Exercise 1B Short Answer Type

Answer:

Sometimes, it is observed that on reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction at once, but it remains stationary for a part of rotation. This causes error in the observation which is called backlash error.

The reason for this is the wear and tear of threads of screw.

To avoid the error, while taking the measurements, screw should be rotated in one direction only.

If it is required to change the direction of rotation of screw then do not change the direction of rotation at once. Move the screw still further, stop there for a while and then rotate it in the reverse direction.

Exercise 1B Very Short Answer Type

13 questions

Question 1

Define the least count of an instrument.

Exercise 1B Very Short Answer Type

Answer:

Least count of an instrument is the smallest measurement that can be taken accurately with it.

Question 2

How can you decrease the least count of a vernier callipers?

Exercise 1B Very Short Answer Type

Answer:

The least count of a vernier callipers can be decreased by:

  1. increasing the number of divisions on the vernier scale.
  2. decreasing the value of one division on main scale.

Question 3

Define the term 'Vernier constant'?

Exercise 1B Very Short Answer Type

Answer:

The term 'Vernier constant' is defined as the difference between the values of one main scale division and one vernier scale division. It is also known as least count of vernier.

Question 4

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

Exercise 1B Very Short Answer Type

Answer:

On bringing, the movable jaw in contact with the fixed jaw, the zero mark of the vernier scale should coincide with the zero mark of the main scale. In this position, the tenth division of the vernier callipers coincides with the ninth division of the main scale.

If it is so, the vernier is said to be free from zero error.

Question 5

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

  1. external diameter of a tube,
  2. internal diameter of a mug,
  3. depth of a small bottle,
  4. thickness of a pencil.
Exercise 1B Very Short Answer Type

Answer:

  1. External diameter of a tube — Outside jaws
  2. Internal diameter of a mug — Inside jaws
  3. Depth of a small bottle — Strip
  4. Thickness of a pencil — Outer jaws

Question 6

State the difference between positive and negative zero error of a vernier callipers.

Exercise 1B Very Short Answer Type

Answer:

On bringing the two jaws of a vernier callipers together, if zero mark of the vernier scale is on the right of zero mark of the main scale, the zero error is said to be positive and if zero mark of the vernier scale is to the left of zero mark of the main scale, the zero error is said to be negative.

Question 7

Define the pitch of a screw gauge.

Exercise 1B Very Short Answer Type

Answer:

The pitch of a screw gauge is the linear distance moved by its screw on the main scale when the circular scale is given one complete rotation.

Question 8

State one use of a screw gauge.

Exercise 1B Very Short Answer Type

Answer:

A screw gauge is used to measure the diameter of a wire or thickness of a paper, etc.

Question 9

State the purpose of ratchet in a screw gauge.

Exercise 1B Very Short Answer Type

Answer:

The purpose of a ratchet in a screw gauge is to advance the screw by turning it till the object is gently held between the stud and the spindle of the screw.

Question 10

State the difference between positive and negative zero error of a screw gauge.

Exercise 1B Very Short Answer Type

Answer:

If on bringing the flat end of the screw in contact with the stud, the zero mark on the circular scale is below the base line of the main scale, the zero error is said to be positive and if the zero mark on the circular scale is above the base line of the main scale, the zero error is said to be negative.

Question 11

Name the instrument which can measure accurately the following —

  1. the diameter of a needle,
  2. the thickness of a paper,
  3. the internal diameter of the neck of a water bottle,
  4. the diameter of a pencil.
Exercise 1B Very Short Answer Type

Answer:

The given physical quantities can be measured accurately with the help of the following instruments —

  1. the diameter of a needle — screw gauge.
  2. the thickness of a paper — screw gauge.
  3. the internal diameter of the neck of a water bottle — vernier callipers.
  4. the diameter of a pencil — screw gauge.

Question 12

Which of the following measures a small length to a high accuracy — metre rule, vernier callipers, screw gauge?

Exercise 1B Very Short Answer Type

Answer:

A screw gauge measures a small length to a high accuracy.

Question 13

Name the instrument which has the least count —

  1. 0.1 mm
  2. 1 mm
  3. 0.01 mm
Exercise 1B Very Short Answer Type

Answer:

The following instruments have the given least count —

  1. 0.1 mm — vernier callipers
  2. 1 mm — metre rule
  3. 0.01 mm — screw gauge

Exercise 1C Assertion Reason Type

4 questions

Question 2(i)

Assertion (A) : The unit used to measure speed is an example of a derived unit.

Reason (R) : Derived units can neither be changed nor can be related to any other fundamental unit.

  1. both A and R are true and R is the correct explanation of A
  2. both A and R are true and R is not the correct explanation of A
  3. assertion is false but reason is true
  4. assertion is true but reason is false
Exercise 1C Assertion Reason Type

Answer:

assertion is true but reason is false

Explanation

Assertion (A) is true as Speed=DistanceTime\text {Speed} = \dfrac {\text {Distance}}{\text {Time}} so its unit is metre/second (m/s), which is a derived unit because it is formed from the fundamental units of length (metre) and time (second).

Reason (R) is false because derived units are defined in terms of fundamental units. For example, speed is m/s, force is kg·m/s², etc. so derived units are completely related to fundamental units.

Question 2(ii)

Assertion (A) : Smaller the least count of an instrument, more precise the measurement made by using it.

Reason (R) : Least count of an instrument is the smallest measurement that can be taken accurately with it.

  1. both A and R are true and R is the correct explanation of A
  2. both A and R are true and R is not the correct explanation of A
  3. assertion is false but reason is true
  4. assertion is true but reason is false
Exercise 1C Assertion Reason Type

Answer:

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because an instrument with a smaller least count can measure finer divisions, which means the measurements are more precise.

Reason (R) is true because by definition, least count is the smallest value that an instrument can measure accurately.

The reason correctly explains the assertion because a smaller least count means the instrument can measure smaller changes, resulting in greater precision.

Question 2(iii)

Assertion (A) : The least count of a screw gauge can be decreased by increasing the pitch and decreasing the total number of divisions on circular scale.

Reason (R) : Least count of screw gauge is directly proportional to pitch of the screw gauge.

  1. both A and R are true and R is the correct explanation of A
  2. both A and R are true and R is not the correct explanation of A
  3. assertion is false but reason is true
  4. assertion is true but reason is false
Exercise 1C Assertion Reason Type

Answer:

assertion is false but reason is true

Explanation

Assertion (A) is false because the least count (LC) of a screw gauge is given by :

LC=PitchNo. of divs on circular scale\text {LC}= \dfrac{\text {Pitch}}{\text {No. of divs on circular scale}} ​ So, to decrease the least count for better precision, pitch should be decreased and number of divisions should be increased

Reason (R) is true because from the formula above, least count is directly proportional to pitch, provided the number of divisions remains the same.

Question 2(iv)

Assertion (A) : A pendulum clock goes slow (i.e., the time period of oscillation increases) when it is taken to mines.

Reason (R) : This is due to increase in the value of acceleration due to gravity g.

  1. both A and R are true and R is the correct explanation of A
  2. both A and R are true and R is not the correct explanation of A
  3. assertion is false but reason is true
  4. assertion is true but reason is false
Exercise 1C Assertion Reason Type

Answer:

assertion is true but reason is false

Explanation

Assertion (A) is true because inside mines, the value of acceleration due to gravity (g) decreases because you are closer to the Earth's center.

As,

For a simple pendulum :

T=2πlg\text T=2\text π\sqrt {\dfrac {\text l}{\text g}} ​ If g decreases, T increases — the pendulum swings more slowly, so the clock loses time and goes slow.

Reason (R) is false because actually, in mines, g decreases, not increases.

Exercise 1C Long Answer Type

1 question

Question 1

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)?

Exercise 1C Long Answer Type

Answer:

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

i.e., T ∝ l\sqrt{\text l}

Graph showing the variation of T2 with l is given below:

How does the time period (T) of a simple pendulum depend on its length (l)? Draw a graph showing the variation of T<sup>2</sup> with l. How will you use this graph to determine the value of g (acceleration due to gravity)? Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

In order to find the acceleration due to gravity with the help of the above graph, we follow the following steps —

The slope of the straight line obtained in the T2 vs l graph, as shown in fig, can be obtained by taking two points P and Q on the straight line and drawing normals from these points on the X and Y axes. Then, note the value of T2, say T12 and T22 at a and b respectively, and also the value of l say l1 and l2 respectively at c and d.

Then,

Slope=PRQR=abcd=T12T22l1l2\text{Slope} = \dfrac{\text {PR}}{\text {QR}} = \dfrac{\text {ab}}{\text {cd}} = \dfrac{\text T_1^2 - \text T_2^2}{\text l_1 - \text l_2}

This slope is found to be a constant at a place and,

Slope=4π2g\text{Slope} = \dfrac{4\text π^2}{\text g}

where, g = acceleration due to gravity at that place.

Thus, g can be determined at a place from the graph using the following relation,

g=4π2Slope of T2 vs l graph\text g = \dfrac{4\text π^2}{\text{Slope of } \text T^2 \text{ vs l graph}}

Exercise 1C Multiple Choice Type

10 questions

Question 1(i)

The pendulum used in a clock is:

  1. Simple pendulum
  2. Compound pendulum
  3. Second pendulum
  4. None of these
Exercise 1C Multiple Choice Type

Answer:

Compound pendulum

Reason — Compound pendulums have a more constant period of oscillation over a wider range of amplitudes, making them suitable for use in clocks.

Question 1(ii)

The effective length of a pendulum is related with time period as :

  1. T2 ∝ l2
  2. T3 ∝ l
  3. T2 ∝ l
  4. T ∝ l3
Exercise 1C Multiple Choice Type

Answer:

T2 ∝ l

Reason — The time period of oscillation is directly proportional to the square root of its effective length i.e., T ∝ l\sqrt{\text l} or in other words, the square of time period of oscillation (T2) is directly proportional to its effective length (l) i.e., T2 ∝ l.

Question 1(iii)

A simple pendulum is made by suspending a bob of mass 1 kg by a string of length l. Now if the length of this pendulum is increased to 4l, then its time period T will :

  1. remain the same
  2. become twice
  3. become four times
  4. become half
Exercise 1C Multiple Choice Type

Answer:

become twice

Reason — As T ∝ l\sqrt{\text l}
When l2 = 4l then,
T24l\sqrt{4\text l} = 2 l\sqrt{\text l} = 2T

Hence, time period T will become twice.

Question 1(iv)

The time period of a seconds' pendulum clock is —

  1. 1 s
  2. 2 s
  3. 1 min
  4. 12 h
Exercise 1C Multiple Choice Type

Answer:

2 s

Reason — A seconds' pendulum clock takes 1 second in moving from one extreme to the other extreme, so the time taken for one complete oscillation is 2 seconds. Hence, its time period is 2s.

Question 1(v)

Time period of a simple pendulum is given by :

  1. T=2πgl\text T = 2 \text π \sqrt{\dfrac{\text g}{\text l}}
  2. T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}
  3. T=4πlg\text T = 4 \text π \sqrt{\dfrac{\text l}{\text g}}
  4. T=4πgl\text T = 4 \text π \sqrt{\dfrac{\text g}{\text l}}
Exercise 1C Multiple Choice Type

Answer:

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

Reason — Time period of a simple pendulum is given by T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

where

T = time period
l = effective length of pendulum
g = acceleration due to gravity.

Question 1(vi)

The time period of a simple pendulum depends on :

  1. Length of pendulum
  2. Amplitude of oscillation
  3. Mass of bob
  4. All of the above
Exercise 1C Multiple Choice Type

Answer:

Length of pendulum

Reason — The time period of oscillation is directly proportional to the square root of its effective length i.e., T ∝ l\sqrt{\text l} or in other words, the square of time period of oscillation (T2) is directly proportional to its effective length (l) i.e., T2 ∝ l. Hence, we can say that, time period of a simple pendulum depends on length of pendulum.

Question 1(vii)

The time period of two pendulums of length 1 m and 16 m are in ratio :

  1. 1 : 16
  2. 1 : 4
  3. 16 : 1
  4. 4 : 1
Exercise 1C Multiple Choice Type

Answer:

1 : 4

Reason — As, T ∝ l\sqrt{\text l}

T1T2\dfrac{\text T_1}{\text T_2} = l1l2\dfrac{\sqrt{\text l_1}}{\sqrt{\text l_2}}

T1T2\dfrac{\text T_1}{\text T_2} = 116\dfrac{\sqrt{1}}{\sqrt{16}}

T1T2\dfrac{\text T_1}{\text T_2} = 14\dfrac{1}{4}

Question 1(viii)

The length of a simple pendulum is made one-fourth. Its time period becomes —

  1. four times
  2. one-fourth
  3. double
  4. half
Exercise 1C Multiple Choice Type

Answer:

half

Reason — Time period of a simple pendulum is given by:

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Time period is directly proportional to the square root of the length of the pendulum.

In the case when length is made one-fourth, we see that —

T=2πl4×gT=22×πlgT=πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{4\times \text g}} \\[0.5em] \text T = \dfrac{2}{2} \times \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em] \text T = \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Hence, we can say that when the length is made four times, time period of a simple pendulum is reduced to half.

Question 1(ix)

The length of a seconds' pendulum is nearly —

  1. 0.5 m
  2. 9.8 m
  3. 1.0 m
  4. 2.0 m
Exercise 1C Multiple Choice Type

Answer:

1.0 m

Reason — The effective length of a seconds' pendulum, at a place where g = 9.8 ms-2 is nearly 1 meter.

Question 1(x)

Identify the incorrect statement(s) from the following :

(I) The time period of oscillations depends on the extent of swing on either side.

(II) The time period of oscillations is directly proportional to the square root of acceleration due to gravity.

(III) The time period of oscillations is inversely proportional to the square root of its effective length.

(IV) The time period of oscillations does not depend on the mass or material of the suspended body.

  1. (II) and (III)
  2. Only (II)
  3. Only (IV)
  4. (I), (II) and (III)
Exercise 1C Multiple Choice Type

Answer:

(I), (II) and (III)

Reason — Time period of a simple pendulum is given by : T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

where

T = time period
l = effective length of pendulum
g = acceleration due to gravity.

From above relation it is clear that the time period of oscillations is inversely proportional to the square root of acceleration due to gravity and directly proportional to the square root of its effective length.

Hence, time period of a pendulum only depends upon its effective length and acceleration due to gravity.

Exercise 1C Numericals

9 questions

Question 1

A simple pendulum completes 40 oscillations in one minute.

Find its —

(a) frequency,

(b) time period.

Exercise 1C Numericals

Answer:

(a) Given,

40 oscillations in one minute, so

frequency per second=4060=23=0.67s1\text {frequency per second} = \dfrac{40}{60} \\[0.5em] = \dfrac{2}{3} \\[0.5em] = \text {0.67} \text s^{-1} \\[0.5em]

Hence, frequency of oscillation = 0.67 s-1.

(b) As we know that,

T=1f\text T = \dfrac{1}{\text f}

So, substituting the value of f = 0.67 s-1, in equation above we get,

T=10.67T=1.49253s1.5s\text T = \dfrac{1}{0.67} \\[0.5em] \Rightarrow \text T = 1.49253 \text s \approx 1.5\text s \\[0.5em]

Hence, time period of the simple pendulum is 1.5 s.

Question 2

The time period of a simple pendulum is 2s. What is its frequency? What name is given to such a pendulum?

Exercise 1C Numericals

Answer:

As we know that,

f=1T\text f = \dfrac{1}{\text T}

Given,

T = 2s

So, substituting the value of T in equation above we get,

f=12f=0.5 s1\text f = \dfrac{1}{2} \\[0.5em] \Rightarrow \text f = 0.5\ \text s^{-1}

Hence, the frequency of oscillation of the simple pendulum is 0.5 s-1.

The name given to such a pendulum is seconds' pendulum.

Question 3

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

Exercise 1C Numericals

Answer:

As we know,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

We observe that time period is inversely proportional to the square root of acceleration due to gravity.

Hence, when 'g' falls to one-fourth, time period increases.

When acceleration due to gravity is reduced to one fourth, we see that —

T=2πlg4T=2π4lgT=2×2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\dfrac{\text g}{4}}} \\[0.5em] \text T = 2 π \sqrt{\dfrac{4\text l}{\text g}} \\[0.5em] \text T = 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Hence, we can say that when acceleration due to gravity is reduced to one fourth, time period of a simple pendulum doubles.

As, the given pendulum is a seconds' pendulum so T = 2s

New T = 2 x 2 = 4s

Question 4

Find the length of a seconds' pendulum at a place where g = 10 ms-2 (Take π = 3.14).

Exercise 1C Numericals

Answer:

As we know,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

Given,

g = 10 ms-2

π = 3.14

T = 2 s

Substituting the values in the formula above we get,

2=2×3.14l10(22×3.14)=l10(22×3.14)2=l10(13.14)2=l10(0.3184)2=l100.10142=l10l=1.0142 m2 = 2 \times 3.14 \sqrt{\dfrac{\text l}{10}} \\[0.5em] \Rightarrow (\dfrac{2}{2 \times 3.14}) = \sqrt{\dfrac{\text l}{10}} \\[0.5em] \Rightarrow (\dfrac{2}{2 \times 3.14})^2 = \dfrac{\text l}{10} \\[0.5em] \Rightarrow (\dfrac{1}{3.14})^2 = \dfrac{\text l}{10} \\[0.5em] \Rightarrow (0.3184)^2 = \dfrac{\text l}{10} \\[0.5em] \Rightarrow 0.10142 = \dfrac{\text l}{10} \\[0.5em] \Rightarrow \text l = 1.0142\ \text m\\[0.5em]

Hence, length of a seconds’ pendulum = 1.0142 m

Question 5

Compare the time periods of two pendulums of length 1m and 9m.

Exercise 1C Numericals

Answer:

As we know that,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Time period is directly proportional to the square root of the length of the pendulum.

In the case when length is 1m,

T1=2π1g\text T_1 = 2 \text π \sqrt{\dfrac{1}{\text g}} \\[0.5em]

and

In the case when length is 9m,

T2=2π9g\text T_2 = 2 \text π \sqrt{\dfrac{9}{\text g}} \\[0.5em]

So, comparison of T1 and T2 gives —

T1:T2=2π1g:2π9gT1:T2=19T1:T2=13\text T_1 : \text T_2 = 2 \text π \sqrt{\dfrac{1}{\text g}} : 2 \text π \sqrt{\dfrac{9}{\text g}} \\[0.5em] \text T_1 : \text T_2 = \sqrt{\dfrac{1}{9}} \\[0.5em] \Rightarrow \text T_1 : \text T_2 = \dfrac{1}{3} \\[0.5em]

Hence, T1 : T2 = 1 : 3

Question 6

A pendulum completes 2 oscillations in 5s.

(a) What is its time period?

(b) If g = 9.8 ms-2, find its length.

Exercise 1C Numericals

Answer:

Given,

2 oscillations in 5 seconds, so

frequency per second=25frequency per second=0.4 hertz\text {frequency per second} = \dfrac{2}{5} \\[0.5em] \Rightarrow \text {frequency per second} = \text {0.4 hertz} \\[0.5em]

Hence, frequency of oscillation = 0.4 hertz.

As we know that,

T=1f\text T = \dfrac{1}{\text f}

So, substituting the value of f = 0.4 hertz, in equation above we get,

T=10.4T=2.5 s\text T = \dfrac{1}{0.4} \\[0.5em] \Rightarrow \text T = 2.5\ \text s \\[0.5em]

Hence, time period of pendulum is 2.5 s

(b) As we know,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Given,

g = 9.8 ms-2

and we know,

π = 3.14

T = 2.5 s

Substituting the values in the formula above we get,

2.5=2×3.14l9.8(2.52×3.14)=l9.8(2.52×3.14)2=l9.8(2.56.28)2=l9.8(0.398)2=l9.80.158=l9.8l=9.8×0.158l=1.55 m2.5 = 2 \times 3.14 \sqrt{\dfrac{\text l}{9.8}} \\[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14}) = \sqrt{\dfrac{\text l}{9.8}} \\[0.5em] \Rightarrow (\dfrac{2.5}{2 \times 3.14})^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow (\dfrac{2.5}{6.28})^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow (0.398)^2 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow 0.158 = \dfrac{\text l}{9.8} \\[0.5em] \Rightarrow \text l = 9.8 \times 0.158 \\[0.5em] \Rightarrow \text l = 1.55\ \text m

Hence, length of a seconds’ pendulum = 1.55 m

Question 7

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

Exercise 1C Numericals

Answer:

As we know that,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em]

Time period is directly proportional to the square root of the length of the pendulum.

In the case when,
T1 : T2 = 2 : 1     [Equation 1]

we know that,

T1 : T2 = l1:l2\sqrt{\text l_1} : \sqrt{\text l_2}     [Equation 2]

So we get,

l1:l2=2:1l1l2=21\sqrt{\text l_1} : \sqrt{\text l_2} = 2 : 1 \\[0.5em] \Rightarrow \dfrac{\sqrt{\text l_1}}{\sqrt{\text l_2}} = \dfrac{2}{1}

Squaring both sides we get,

l1l2=2212l1l2=41l1:l2=4:1\dfrac{\text l_1}{\text l_2} = \dfrac{2^2}{1^2} \\[0.5em] \Rightarrow \dfrac{\text l_1}{\text l_2} = \dfrac{4}{1} \\[0.5em] \Rightarrow \text l_1 : \text l_2 = 4 : 1

Hence, ratio of lengths = 4 : 1

Question 8

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

Exercise 1C Numericals

Answer:

As we know that,

Time period = 4 x (time a pendulum bob takes to move from mean position to one end).

Given,

Time a pendulum bob takes to move from mean position to one end = 0.2s

Substituting the value in the equation above we get,

Time taken to complete one oscillation (T) —

=(4×0.2) s=0.8 s= (4 \times 0.2)\ \text s \\[0.5em] = 0.8\ \text s \\[0.5em]

Hence, time period of the pendulum = 0.8s

Question 9

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

Exercise 1C Numericals

Answer:

As we know that, the time period of a seconds' pendulum is 2s.

So, time taken for a seconds’ pendulum to move from one extreme to other is equal to the half of time period.

Thalf=T2Thalf=22=1 s\text T_{\text {half}} = \dfrac{\text T}{2} \\[0.5em] \Rightarrow \text T_{\text {half}} = \dfrac{2}{2} = 1\ \text s \\[0.5em]

Hence, time taken for a seconds' pendulum to move from one extreme to other extreme = 1s

Exercise 1C Short Answer Type

6 questions

Question 1

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

Exercise 1C Short Answer Type

Answer:

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 the pendulum clock is not a simple pendulum, but it is a compound pendulum.(i.e., a body capable of oscillating about a horizontal axis passing through it.)

Question 2

Draw a neat diagram of a simple pendulum. Show on it the effective length of the pendulum and its one oscillation.

Exercise 1C Short Answer Type

Answer:

Below diagram shows the effective length and one oscillation of a simple pendulum :

Draw a neat diagram of a simple pendulum. Show on it the effective length of the pendulum and its one oscillation. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

Question 3

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.

Exercise 1C Short Answer Type

Answer:

Factors affecting the time period of a simple pendulum are —

  1. Effective length of the pendulum — the time period of oscillation is directly proportional to the square root of its effective length.
    i.e., Tl\text T \propto \sqrt{\text l}

  2. Acceleration due to gravity — the time period of oscillations is inversely proportional to the square root of acceleration due to gravity
    i.e., T1g\text T \propto \dfrac{1}{\sqrt {\text g}}

Relation between time period, effective length and acceleration due to gravity is as follows —

T=2πlg\text T = 2\pi \sqrt{\dfrac{\text l}{\text g}}

where,

T = Time period

l = effective length of the pendulum

g = acceleration due to gravity.

Question 4

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

Exercise 1C Short Answer Type

Answer:

To measure the time period of a given pendulum, we record the time for 20 or more oscillations and then divide the recorded time with the number of oscillations. Hence, we get the time period for one oscillation.

The time for more than one oscillation is noted, as the least count of stop watch is either 1s or 0.5s. It cannot record the time period in fraction such as 1.2s or 1.4s and so on.

Hence, it is made possible by noting the time t for 20 oscillations or more and then dividing it by the number of oscillations.

Question 5

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 minute? Explain your answer.

Exercise 1C Short Answer Type

Answer:

Given,

lA = 1.0m
lB = 4.0m

Since,

Tl\text T \propto \sqrt{\text l} \\[0.5em]

∴,

TATB=lAlB\dfrac {\text T_\text A}{\text T_\text B} = \dfrac{\sqrt{\text l_\text A}}{\sqrt{\text l_\text B}} \\[0.5em]

Substituting the values of l in the formula above we get,

So,

TATB=14TATB=12\dfrac {\text T_\text A}{\text T_\text B} = \dfrac{\sqrt{1}}{\sqrt{4}} \\[0.5em] \Rightarrow \dfrac {\text T_\text A}{\text T_\text B} = \dfrac{1}{2} \\[0.5em]

i.e., T1 : T2 = 1 : 2

∴, time period of B is more (twice) than that of A. Hence, A will make more oscillations (twice) in a given time than B.

Question 6

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.

Exercise 1C Short Answer Type

Answer:

The time period of a simple pendulum varies as follows —

(a) Length of pendulum (l) — Time period of a simple pendulum is directly proportional to the square root of the length of the pendulum.

Tl\text T \propto \sqrt{\text l}

(b) Mass of bob — Time period of a simple pendulum is independent of the mass of the bob.

(c) Amplitude of oscillation — Time period of a simple pendulum is independent of the amplitude of oscillation

(d) Acceleration due to gravity (g) — Time period of a simple pendulum is inversely proportional to the the square root of acceleration due to gravity .

T1g\text T \propto \sqrt{\dfrac{1}{\text g}}

Exercise 1C Very Short Answer Type

8 questions

Question 1

Define 'amplitude of oscillation'.

Exercise 1C Very Short Answer Type

Answer:

The maximum displacement of the bob from its mean position on either side is called the amplitude of oscillation.

Question 2

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

Exercise 1C Very Short Answer Type

Answer:

(i) Oscillation – One complete to and fro motion of the bob of 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 denoted by the letter a or A and is measured in metre(m).

(iii) Frequency – The number of oscillations made in one second is called the frequency. It is denoted by f or n. Its unit is per second (s-1) or hertz (Hz).

(iv) Time period – The time taken to complete one oscillation is the time period. It is denoted by the symbol T. Its unit is in second (s).

Question 3

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

Exercise 1C Very Short Answer Type

Answer:

The time period of a simple pendulum does not depend on —

  1. The mass or material of the body suspended (i.e., the bob).

  2. The extent of swing on either side (i.e. on amplitude), provided the swing is not too large.

Question 4

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

(a) The length is made four times,

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

Exercise 1C Very Short Answer Type

Answer:

As we know that,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

where,

T = Time period

l = effective length of the pendulum

g = acceleration due to gravity

(a) In the case when length is made four times, let time period be T1, we see that —

T1=2π4lgT1=2×2πlgT1=2×T\text T_1 = 2 \text π \sqrt{\dfrac{4\text l}{\text g}} \\[0.5em] \text T_1= 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em] \text T_1 = 2 \times \text T \\[0.5em]

Hence, we can say that when the length is made four times, time period of a simple pendulum is doubled.

(b) In the case, when acceleration due to gravity is reduced to one fourth, let time period be T1, we see that —

T1=2πlg4T1=2π4lgT1=2×2πlgT1=2×T\text T_1 = 2 \text π \sqrt{\dfrac{\text l}{\dfrac{\text g}{4}}} \\[0.5em] \text T_1 = 2 \text π \sqrt{\dfrac{4\text l}{\text g}} \\[0.5em] \text T_1 = 2 \times 2 \text π \sqrt{\dfrac{\text l}{\text g}} \\[0.5em] \text T_1 = 2 \times \text T \\[0.5em]

Hence, we can say that when acceleration due to gravity is reduced to one fourth, time period of a simple pendulum is doubled.

Question 5

How are the time period T and frequency f of an oscillation of a simple pendulum related?

Exercise 1C Very Short Answer Type

Answer:

The time period (T) and frequency (f) of oscillation of a simple pendulum are related as stated below —

f=1T\bold{f} = \dfrac{\bold{1}}{\bold{T}}

Question 6

Two simple pendulum A and B have equal lengths, but heir bobs weigh 50 gf and 100 gf respectively. What would be the ratio of their time periods? Give reason for your answer.

Exercise 1C Very Short Answer Type

Answer:

As we know that,

T=2πlg\text T = 2 \text π \sqrt{\dfrac{\text l}{\text g}}

where,

T = Time period

l = effective length of the pendulum

g = acceleration due to gravity.

So we see that, time period does not depend on the weight of the bob.

As, lengths are equal so,

T1 : T2 = 1 : 1

Question 7

What is a seconds' pendulum?

Exercise 1C Very Short Answer Type

Answer:

A pendulum with a time period of oscillation equal to two seconds is known as the seconds pendulum. Its effective length, at a place, where g = 9.8 ms-2 is nearly 1 meter.

Question 8

State the numerical value of the frequency of oscillation of a seconds’ pendulum. Does it depend on the amplitude of oscillation ?

Exercise 1C Very Short Answer Type

Answer:

As we know that,

f=1T\text f = \dfrac{1}{\text T}

and T for a seconds’ pendulum = 2s.

So, substituting the value of T in equation above we get,

f=12f=0.5\text f = \dfrac{1}{2} \\[0.5em] \Rightarrow \text f = 0.5

Hence, the numerical value of the frequency of oscillation of a seconds’ pendulum is 0.5 s-1.

No, it does not depend on the amplitude of oscillation.