Measurements and Experimentation

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.

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).

7 fundamental, 2 supplementary

Reason — Fundamental and supplementary units are as follows :

Second

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

Litre

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

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

Light year

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

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.

0.1 nm

Reason — 10 Å = 1 nm
∴ 1 Å = 0.1 nm

10 quintal

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

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

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.

10^-8 s

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

Volt x Ampere

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

Metre³

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

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

Joule

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

Kg m^-1s^-2

Reason — Pressure = $\frac{\text{Force}}{\text{Area}}$ = Kg m^-1s^-2

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.

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

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

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)

The three fundamental quantities are —

1. Mass
2. Length
3. Time

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

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

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.

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

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).

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

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

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

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.

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.

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.

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

Physical quantity related to the unit are as follows —

The derived units of the following quantities are as follows —

The physical quantities related to the following units are —

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).

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

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.

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 10¹¹ m

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 10¹⁵ m

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 = 100 cm

Relation between meter (m) and kilometer is —

    1 km = 1000 m

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.

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

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

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.,

$1\ \text s = \frac{1}{86400} \times \text{one mean solar day}$

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

As we know, 1 nm = 10 Å

Given,

• The wavelength of light = 5800 Å

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

$$\begin{aligned}10\ \text Å = 1\text { nm} \\[1em] 1\ \text Å = \frac{1}{10}\text{ nm} \\[1em] 5800\ \text Å = \frac{1}{10} \times 5800 \text{ nm}\\[1em] \Rightarrow 5800 \text Å = 580 \text{ nm}\end{aligned}$$

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

(b) As we know,
1 m = 10¹⁰ Å

Given,

• The wavelength of light = 5800 Å

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

$$\begin{aligned}10^{10}\ \text Å = 1\ \text m \\[1em] 1\ \text Å = 10^{-10}\ \text m \\[1em] 5800\ \text Å = 5800 \times 10^{-10}\ \text m \\[1em] \Rightarrow 5800\ \text Å = 5.8 \times 10^{-7}\ \text m\end{aligned}$$

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

Given,

• Size of a bacteria = 1 µ
• Total length = 1m

∴ Number of bacteria in 1m length = $\frac{1\text m}{1\text µ}$

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

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

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

Given,

• Distance = 5.6 x 10²⁵ m
• Speed = 3 x 10⁸ m s^-1

Substituting the values in the formula above we get,

$$\begin{aligned}\text{Time taken} =\frac{5.6 \times 10^{25}}{3 \times 10^8} \\[1em] \Rightarrow \text{Time taken} ={1.87 \times 10^{17}}\ \text s\end{aligned}$$

∴ Time taken by light = 1.87 x 10¹⁷ s

As we know, 1nm = 10 Å

Given,

• The wavelength of light = 589 nm

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

$$\begin{aligned}1\text { nm} = 10\ \text Å \\[1em] 589\ \text { nm} = 589 \times 10\ \text Å \\[1em] \Rightarrow 589\ \text { nm} = 5890\ \text Å\end{aligned}$$

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

1 Light year = 9.46 x 10¹² km

or 9.46 x 10¹² km = 1 Light year

∴ 4.0 x 10¹³ km = $\frac{1}{9.46 \times 10^{12}}$ x 4.0 x 10¹³ = 4.2 light years.

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

As we know, Distance = speed x time

Given,

• Speed = 3 x 10⁸ m s^-1
• Time = 8 min = 8 x 60 s = 480 s

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 = $\frac{1440 \times 10^8}{1000}$ km = 1.44 x 10⁸ km

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

'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 60 s = 499.8 s ≈ 500 s

Substituting the values in the formula above we get,

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

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

Depth of a beaker

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

+0.06 cm

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

0.002 cm

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

The number of divisions on vernier scale n = 25

Using the formula given below,

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

we get,

$$\begin{aligned}\text {L.C.} = \frac{\text{x}}{\text{n}} \\[1em] \text{L.C.} = \frac{\frac{\text{1}}{\text{20}}}{25}\text{ cm} \\[1em] \text{L.C.} = \frac{\text{1}}{\text{500}}\text{ cm} \\[1em] \Rightarrow \text{L.C.} = 0.002\text{ cm}\end{aligned}$$

Hence, least count of microscope is 0.002 cm.

To mark circular scale

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

0.05 cm

Reason — Pitch = distance moved ahead in 1 revolution

Given,

• Distance moved in 2 revolutions = 1 mm

∴ Distance moved in 1 revolution = $\frac{1}{2}$ mm = 0.5 mm

Converting 0.5 mm to cm,

0.5 mm = $\frac{0.5}{10}$ cm = 0.05 cm

∴ Pitch of screw gauge is 0.05 cm.

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.

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

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.

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.

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.

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

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.

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.

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

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.

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.

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.

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

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

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.

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.

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.

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 —

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

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

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

∴,

$$\begin{aligned}\text{Zero Error} = 6 \times \text{L.C.} \\[1em] \text{Zero Error}= 6 \times 0.01 \text{ cm} \\[1em] \Rightarrow \text{Zero Error}= 0.06 \text { cm}\end{aligned}$$

Hence, we get the zero error = 0.06 cm.

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.

The two scales of a vernier callipers are —

1. Main scale - it is fixed
2. Vernier scale – slides along the main scale.

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,

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

Substituting the value in the formula above, we get,

$$\begin{aligned}\text{Least count (L.C.)} = \frac{\text{x}}{\text{n}} \\[1em] \text{Least count (L.C.)} = \frac{\text{1}}{\text{10}} \text{ mm} \\[1em] \text{Least count (L.C.)} = 0.1\text{ mm} \\[1em] \Rightarrow \text{Least count (L.C.)} = 0.01\text{ cm}\end{aligned}$$

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

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 —

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

The least count can be decreased by —

1. Decreasing the pitch.

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

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)

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

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.

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.

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,

$$\begin{aligned}\text{Zero Error} = 6 \times \text{L.C.} \\[1em] \text{Zero Error}= 6 \times 0.01 \text{ cm} \\[1em] \Rightarrow \text{Zero Error}= 0.06 \text { cm}\end{aligned}$$

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)

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

The main parts of the vernier callipers are as follows —

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

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

$\text {Least count (L.C.)} = \frac{\text{x}}{\text{n}} = ............... \text{ cm}$

Zero error = ................ cm

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).

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

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

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) = ...............

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

Zero error = ............... cm

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)

Using the formula given below,

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

Given,

• Total number of divisions on stop watch = 10

we get,

$$\begin{aligned}\text{L.C.} = \frac{\text{x}}{\text{n}} \\[1em] \text{L.C.} = \frac{5}{10} \\[1em] \text{L.C.} = \frac{1}{2} \\[1em] \Rightarrow \text{L.C.} = 0.5\text{ s}\end{aligned}$$

Hence, least count of stop watch is 0.5 s.

(i) As we know,

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

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,

$$\begin{aligned}\text{L.C.} = \frac{\text{1 mm}}{\text{10}} \\[1em] \Rightarrow \text{L.C.} = 0.1 \text{ mm} \\[1em] \Rightarrow \text{L.C.} = 0.01 \text{ cm}\end{aligned}$$

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

As we know,

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

Given,

• 20 divisions = 1 cm

∴,

$$\begin{aligned}\text{1 division} = \frac{1}{20} \\[1em] \Rightarrow \text{1 division} = 0.05 \text{ cm}\end{aligned}$$

Hence, value of one main scale division = 0.05 cm

Total number of divisions = 50

Substituting the values in the formula above, we get,

$$\begin{aligned}\text{L.C.} = \frac{\text{0.05} \text{ cm}}{\text{50}} \\[1em] \text{L.C.} = 0.001 \text{ cm}\end{aligned}$$

Hence, least count of the microscope is 0.001 cm.

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 = +0.02 cm = +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.

(i) As we know,

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

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,

$$\begin{aligned}\text{L.C.} = \frac{\text{1 mm}}{\text{10}} \\[1em] \text{L.C.} = 0.1 \text{ mm} \\[1em] \text{L.C.} = 0.01 \text{ cm}\end{aligned}$$

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,

$$\begin{aligned}\text {Zero error} = \text {0.01 cm x 3} \\[1em] \Rightarrow \text {Zero error} = \text {0.03 cm}\end{aligned}$$

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

(i) As we know,

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

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,

$$\begin{aligned}\text{L.C.} = \frac{1}{\text{20}} \\[1em] \text{L.C.} = 0.05 \text{ mm} \\[1em] \text{L.C.} = 0.005 \text{ cm}\end{aligned}$$

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,

$$\begin{aligned}\text {Circular scale reading} = \text {0.005 cm} \times 4 \\[1em] \Rightarrow \text {Circular scale reading} = \text {0.02 cm}\end{aligned}$$

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,

$$\begin{aligned}\text{Total reading} = \text{main scale reading } + \text{ vernier scale reading} \\[1em] \text{Diameter = Total reading} = \text{3.5 cm + 0.02 cm} \\[1em] \Rightarrow \text{Diameter} = \text{3.52 cm} \\[1em] \Rightarrow \text{Radius} = \frac{3.52}{2}\text{ cm} \\[1em] \Rightarrow \text{Radius} = 1.76\text{ cm}\end{aligned}$$

Hence, the radius of the cylinder is 1.76 cm.

(i) As we know,

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

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,

$$\begin{aligned}\text{L.C.} = \frac{1}{\text{10}} \\[1em] \Rightarrow \text{L.C.} = 0.1\text{mm} \\[1em] \Rightarrow \text{L.C.} = 0.01\text{cm}\end{aligned}$$

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

As we know,

$$\begin{aligned}\text{Vernier scale reading} = \text{Vernier scale division} \times \text{L.C.} \\[1em] \text{Vernier scale reading} = 4 \times 0.01 \text{ cm} \\[1em] \Rightarrow \text{Vernier scale reading} = 0.04 \text{ cm}\end{aligned}$$

Hence, vernier scale reading is 0.04 cm.

and

Main scale reading of the vernier callipers = 1.8 cm

∴,

$$\begin{aligned}\text{Total reading} = \text{main scale reading} + \text{vernier scale reading} \\[1em] \text{Total reading} = 1.8 \text{ cm} + 0.04 \text{ cm} \\[1em] \Rightarrow \text{Total reading}= 1.84\text{ cm}\end{aligned}$$

Hence, the length is 1.84 cm.

(b) As we know,

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

Given,

• zero error = -0.02cm

Substituting the values in the formula above we get,

$$\begin{aligned}\text{Correct reading} = 1.84 – (-0.02) \\[1em] \text{Correct reading} = 1.84 + 0.02 \\[1em] \text{Correct reading} = 1.86 \text{ cm}\end{aligned}$$

Hence, the correct length is 1.86 cm.

(a) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.

So, the main scale reading (MSR) = 1.0 cm.

As we know,

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

There are 10 divisions on 1 cm on the main scale

$\therefore \text{Value of 1 main scale division(x)} = \frac{1}{10}\text{ cm} = 0.1 \text{ cm}$

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

$$\begin{aligned}\text{Least count (L.C.)} = \frac{0.1}{10} \\[1em] \text{L.C.} = 0.01 \text { cm}\end{aligned}$$

Thus, the least count = 0.01 cm.

Since 6^th division of vernier scale coincides with the main scale then,

Vernier scale reading (VSR) = 6 × 0.01 = 0.06 cm

Observed reading = MSR + VSR = 1.0 + 0.06 = 1.06 cm

Given

Zero error = +0.02 cm,

So the zero correction = −0.02 cm.

Then,

Actual reading = 1.06 − 0.02 = 1.04 cm

Hence, the actual reading is 1.04 cm.

(b) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.

So, the main scale reading (MSR) = 1.0 cm.

As we know,

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

There are 10 divisions on 1 cm on the main scale

$\therefore \text{Value of 1 main scale division(x)} = \frac{1}{10}\text{ cm} = 0.1 \text{ cm}$

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

$$\begin{aligned}\text{Least count (L.C.)} = \frac{0.1}{10} \\[1em] \text{L.C.} = 0.01 \text { cm}\end{aligned}$$

Thus, the least count = 0.01 cm.

Since 4^th division of vernier scale coincides with the main scale then,

Vernier scale reading (VSR) = 4 × 0.01 = 0.04 cm

Observed reading = MSR + VSR = 1.0 + 0.04 = 1.04 cm

Given

Zero error = -0.03 cm,

So the zero correction = +0.03 cm.

Then,

Actual reading = 1.04 + 0.03 = 1.07 cm

Hence, the actual reading is 1.07 cm.

As we know,

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

There are 10 divisions on 1 cm on the main scale

∴,

$\text{Value of 1 main scale division(x)} = \frac{1}{10}\text{ cm} = 0.1 \text{ cm}$

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

$$\begin{aligned}\text{Least count (L.C.)} = \frac{0.1}{10} \\[1em] \text{L.C.} = 0.01 \text { cm}\end{aligned}$$

Hence, the least count = 0.01 cm.

Given,

• Main scale reading = 3.3 cm
• coinciding division = 6

As we know,

$$\begin{aligned}\text{Vernier scale reading} = \text{Vernier scale division} \times \text{L.C.} \\[1em] \text{Vernier scale reading} = 6 \times 0.01 \text{ cm} \\[1em] \Rightarrow \text{Vernier scale reading} = 0.06 \text{ cm}\end{aligned}$$

Hence, vernier scale reading is 0.06 cm.

$$\begin{aligned}\text{Observed reading} = \text{main scale reading} + \text{vernier scale reading} \\[1em] \text{Observed reading} = 3.3 \text{ cm} + 0.06 \text{ cm} \\[1em] \Rightarrow \text{Observed reading} = 3.36\text{ cm}\end{aligned}$$

Hence, the length of the rod is 3.36 cm.

As we know,

$\text {Least Count} = \frac{\text{Pitch}}{\text {Total no. of div on circular head}}$

Given,

• Pitch = 0.5 mm
• Number of divisions on circular head = 100

Substituting the values in the formula above we get,

$$\begin{aligned}\text {Least count} = \frac{0.5}{100} \\[1em] \text {Least count} = 0.005 \text { mm} = 0.0005 \text { cm}\end{aligned}$$

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

(i) As we know,

Pitch = distance moved ahead in 1 revolution

Given,

Distance covered in two revolutions = 1 mm

∴, we get,

$\text {Pitch} = \frac{1}{2} \text{ mm} = 0.5 \text{ mm}$

Hence, pitch of the screw gauge = 0.5 mm

(ii) As we know,

$\text{Least Count} = \frac {\text{Pitch}}{\text{Total no. of div on circular head}}$

Given,

• Pitch = 0.5 mm
• Total number of divisions on circular scale = 50

Substituting the values in the formula above we get,

$$\begin{aligned}\text {Least count} = \frac{0.5}{50} \\[1em] \text {Least count} = 0.01 \text { mm}\end{aligned}$$

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

(i) As we know,

$\text{Least Count} = \frac {\text{Pitch}}{\text{Total no. of div on circular head}}$

Given,

• Pitch = 1 mm
• Number of divisions on circular head = 100

Substituting the values in the formula above we get,

$$\begin{aligned}\text {Least count} = \frac{1}{100} \text{ mm} \\[1em] \text {Least count} = 0.01 \text { mm} = 0.001 \text { cm}\end{aligned}$$

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 is 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.

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 is 0.027 cm and reading on sleeve is 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.

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,

$\text {Pitch} = \frac{1}{2} \text{ mm} = 0.5 \text{ mm}$

Hence, pitch of the screw gauge = 0.5 mm

(ii) As we know,

$\text{Least count} = \frac {\text{Pitch}}{\text{Total no. of div on circular head}}$

Here,

Pitch = 0.5 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

$$\begin{aligned}\text {Least count} = \frac{0.5}{50} \\[0.5em] \text {Least count} = 0.01 \text { mm}\end{aligned}$$

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 is +0.04 mm.

(a) Given,

• Least count = 0.001 cm
• Zero error = +0.05 mm = +0.005 cm

From the figure,

Main scale reading (MSR) = 3 mm = 0.3 cm

And

21^st division of the circular scale coincides with the main scale then

Circular scale reading (CSR) = Least count x 21 = 0.001 cm x 21 = 0.021 cm

Now,

Observed reading = MSR + CSR = 0.3 + 0.021 = 0.321 mm

The zero correction = -0.005 cm.

Then,

Actual reading = 0.321 - 0.005 = 0.316 cm

Hence, the actual reading is 0.316 cm.

(b) Given,

• Least count = 0.001 cm
• Zero error = -0.021 cm

From the figure,

Main scale reading (MSR) = 6 mm = 0.6 cm

And

41^st division of the circular scale coincides with the main scale then

Circular scale reading (CSR) = Least count x 41 = 0.001 cm x 41 = 0.041 cm

Now,

Observed reading = MSR + CSR = 0.6 + 0.041 = 0.641 cm

The zero correction = +0.021 cm.

Then,

Actual reading = 0.641 + 0.021 = 0.662 cm

Hence, the actual reading is 0.662 cm.

(i) As we know,

Pitch = distance moved ahead in 1 revolution

and given,

Distance covered in one revolution = 1 mm

Hence, Pitch of the screw gauge = 1 mm

(ii) As we know,

$\text{Least count} = \frac {\text{Pitch}}{\text{Total no. of div on circular head}}$

Here,

Pitch = 1 mm

Number of divisions on circular head = 50

Substituting the values in the formula above we get,

$$\begin{aligned}\text {Least count} = \frac{1}{50} \\[1em] \text {Least count} = 0.02 \text { mm}\end{aligned}$$

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.

As we know,

$\text {Least count} = \frac{\text{Pitch}}{\text{Total no. of div on circular head}}$

Given,

• Pitch = 0.5 mm
• L.C. = 0.001 mm

Substituting the values in the formula above we get,

$$\begin{aligned}0.001\text{ mm} = \frac{0.5}{\text{Total no. of div on circular head}} \\[1em] \Rightarrow \text{Total no. of div on circular head} = \frac{0.5}{0.001} \\[1em] \Rightarrow \text{Total no. of div on circular head} = 500\end{aligned}$$

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

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.

T² ∝ l

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

become twice

Reason — As T ∝ $\sqrt{\text l}$

When l₂ = 4l then,

T₂ ∝ $\sqrt{4\text l}$ = 2 $\sqrt{\text l}$ = 2T

Hence, time period T will become twice.

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.

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

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

where

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

Length of pendulum

Reason — The time period of oscillation is directly proportional to the square root of its effective length i.e., T ∝ $\sqrt{\text l}$ 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.

1 : 4

Reason — As, T ∝ $\sqrt{\text l}$

$$\begin{aligned}\frac{\text T_1}{\text T_2} = \frac{\sqrt{\text l_1}}{\sqrt{\text l_2}} \\[1em] = \frac{\sqrt{1}}{\sqrt{16}} \\[1em] = \frac{1}{4}\end{aligned}$$

half

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

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

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 —

$$\begin{aligned}\text T = 2 \text π \sqrt{\frac{\text l}{4\times \text g}} \\[1em] \text T = \frac{2}{2} \times \text π \sqrt{\frac{\text l}{\text g}} \\[1em] \text T = \text π \sqrt{\frac{\text l}{\text g}}\end{aligned}$$

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

1.0 m

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

(I), (II) and (III)

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

where

• T = time period,
• l = effective length of pendulum and
• 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.

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

(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).

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.

As we know that,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

where,

• T = Time period,
• l = effective length of the pendulum and
• g = acceleration due to gravity.

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

$$\begin{aligned}\text T_1 = 2 \text π \sqrt{\frac{4\text l}{\text g}} \\[1em] \text T_1= 2 \times 2 \text π \sqrt{\frac{\text l}{\text g}} \\[1em] \text T_1 = 2 \times \text T \\[1em]\end{aligned}$$

Hence, 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 —

$$\begin{aligned}\text T_1 = 2 \text π \sqrt{\frac{\text l}{\frac{\text g}{4}}} \\[1em] \text T_1 = 2 \text π \sqrt{\frac{4\text l}{\text g}} \\[1em] \text T_1 = 2 \times 2 \text π \sqrt{\frac{\text l}{\text g}} \\[1em] \text T_1 = 2 \times \text T\end{aligned}$$

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

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

$\bold{f} = \frac{\bold{1}}{\bold{T}}$

As we know that,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

where,

• T = Time period,
• l = effective length of the pendulum and
• 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

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.

As we know that,

$\text f = \frac{1}{\text T}$

And T for a seconds’ pendulum = 2s.

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

$$\begin{aligned}\text f = \frac{1}{2} \\[1em] \Rightarrow \text f = 0.5\ \text s^{-1}\end{aligned}$$

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.

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.)

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

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., $\text T \propto \sqrt{\text l}$

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

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

$\text T = 2\pi \sqrt{\frac{\text l}{\text g}}$

where,

• T = Time period,
• l = effective length of the pendulum and
• g = acceleration due to gravity.

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.

Given,

• l_A = 1.0 m
• l_B = 4.0 m

Since,

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

∴,

$$\begin{aligned}\frac {\text T_\text A}{\text T_\text B} = \frac{\sqrt{\text l_\text A}}{\sqrt{\text l_\text B}} \\[0.5em]\end{aligned}$$

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

So,

$$\begin{aligned}\frac {\text T_\text A}{\text T_\text B} = \frac{\sqrt{1}}{\sqrt{4}} \\[1em] \Rightarrow \frac {\text T_\text A}{\text T_\text B} = \frac{1}{2}\end{aligned}$$

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.

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.

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

(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 .

$\text T \propto \sqrt{\frac{1}{\text g}}$

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

i.e., T ∝ $\sqrt{\text l}$

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

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,

$\text{Slope} = \frac{\text {PR}}{\text {QR}} = \frac{\text {ab}}{\text {cd}} = \frac{\text T_1^2 - \text T_2^2}{\text l_1 - \text l_2}$

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

$\text{Slope} = \frac{4\text π^2}{\text g}$

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

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

$\text g = \frac{4\text π^2}{\text{Slope of } \text T^2 \text{ vs l graph}}$

(a) Given,

• Number of oscillations = 40
• Time taken = 1 minute = 60 seconds

$$\begin{aligned}\text {frequency per second} = \frac{40}{60} \\[1em] = \frac{2}{3} \\[1em] = \text {0.67} \text s^{-1}\end{aligned}$$

Hence, frequency of oscillation = 0.67 s^-1.

(b) As we know that,

$\text T = \frac{1}{\text f}$

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

$$\begin{aligned}\text T = \frac{1}{0.67} \\[1em] = 1.49253\ \text s \\[1em] \approx 1.5\ \text s\end{aligned}$$

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

As we know that,

$\text f = \frac{1}{\text T}$

Given,

• T = 2 s

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

$$\begin{aligned}\text f = \frac{1}{2} \\[1em] \Rightarrow \text f = 0.5\ \text s^{-1}\end{aligned}$$

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

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

As we know,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

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 —

$$\begin{aligned}\text T = 2 \text π \sqrt{\frac{\text l}{\frac{\text g}{4}}} \\[1em] \text T = 2 π \sqrt{\frac{4\text l}{\text g}} \\[1em] \text T = 2 \times 2 \text π \sqrt{\frac{\text l}{\text g}}\end{aligned}$$

Hence, 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 = 2 s

∴ New T = 2 x 2 = 4s

As we know,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

Given,

• g = 10 m s^-2
• π = 3.14
• T = 2 s

Substituting the values in the formula above we get,

$$\begin{aligned}2 = 2 \times 3.14 \sqrt{\frac{\text l}{10}} \\[1em] \Rightarrow \left(\frac{2}{2 \times 3.14}\right) = \sqrt{\frac{\text l}{10}} \\[1em] \Rightarrow \left(\frac{2}{2 \times 3.14}\right)^2 = \frac{\text l}{10} \\[1em] \Rightarrow \left(\frac{1}{3.14}\right)^2 = \frac{\text l}{10} \\[1em] \Rightarrow (0.3184)^2 = \frac{\text l}{10} \\[1em] \Rightarrow 0.10142 = \frac{\text l}{10} \\[1em] \Rightarrow \text l = 1.0142\ \text m\end{aligned}$$

Hence, the length of a seconds’ pendulum is 1.0142 m.

As we know that,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

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

In the case when length is 1m,

$\text T_1 = 2 \text π \sqrt{\frac{1}{\text g}}$

and

In the case when length is 9m,

$\text T_2 = 2 \text π \sqrt{\frac{9}{\text g}}$

So, comparison of T₁ and T₂ gives —

$$\begin{aligned}\text T_1 : \text T_2 = 2 \text π \sqrt{\frac{1}{\text g}} : 2 \text π \sqrt{\frac{9}{\text g}} \\[1em] \Rightarrow \text T_1 : \text T_2 = \sqrt{\frac{1}{9}} \\[1em] \Rightarrow \text T_1 : \text T_2 = \frac{1}{3}\end{aligned}$$

Hence, T₁ : T₂ = 1 : 3

Given,

• Number of oscillations = 2
• Time taken = 5 seconds

$$\begin{aligned}\text {Frequency per second} = \frac{2}{5} \\[1em] \Rightarrow \text {Frequency per second} = \text {0.4 hertz}\end{aligned}$$

Hence, frequency of oscillation is 0.4 hertz.

As we know that,

$\text T = \frac{1}{\text f}$

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

$$\begin{aligned}\text T = \frac{1}{0.4} \\[1em] \Rightarrow \text T = 2.5\ \text s\end{aligned}$$

Hence, time period of pendulum is 2.5 s.

(b) As we know,

$$\begin{aligned}\text T = 2 \text π \sqrt{\frac{\text l}{\text g}} \\[0.5em]\end{aligned}$$

Given,

• g = 9.8 m s^-2

And we know,

• π = 3.14
• T = 2.5 s

Substituting the values in the formula above we get,

$$\begin{aligned}2.5 = 2 \times 3.14 \sqrt{\frac{\text l}{9.8}} \\[1em] \Rightarrow \left(\frac{2.5}{2 \times 3.14}\right) = \sqrt{\frac{\text l}{9.8}} \\[1em] \Rightarrow \left(\frac{2.5}{2 \times 3.14}\right)^2 = \frac{\text l}{9.8} \\[1em] \Rightarrow \left(\frac{2.5}{6.28}\right)^2 = \frac{\text l}{9.8} \\[1em] \Rightarrow (0.398)^2 = \frac{\text l}{9.8} \\[1em] \Rightarrow 0.158 = \frac{\text l}{9.8} \\[1em] \Rightarrow \text l = 9.8 \times 0.158 \\[1em] \Rightarrow \text l = 1.55\ \text m\end{aligned}$$

Hence, length of a seconds’ pendulum = 1.55 m

As we know that,

$\text T = 2 \text π \sqrt{\frac{\text l}{\text g}}$

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 = $\sqrt{\text l_1} : \sqrt{\text l_2}$     [Equation 2]

So we get,

$$\begin{aligned}\sqrt{\text l_1} : \sqrt{\text l_2} = 2 : 1 \\[1em] \Rightarrow \frac{\sqrt{\text l_1}}{\sqrt{\text l_2}} = \frac{2}{1}\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\frac{\text l_1}{\text l_2} = \frac{2^2}{1^2} \\[1em] \Rightarrow \frac{\text l_1}{\text l_2} = \frac{4}{1} \\[1em] \Rightarrow \text l_1 : \text l_2 = 4 : 1\end{aligned}$$

Hence, the ratio of lengths is 4 : 1

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,

$$\begin{aligned}\text {Time taken to complete one oscillation (T)} = (4 \times 0.2)\ \text s \\[1em] = 0.8\ \text s\end{aligned}$$

Hence, time period of the pendulum = 0.8 s

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.

$$\begin{aligned}\text T_{\text {half}} = \frac{\text T}{2} \\[1em] \Rightarrow \text T_{\text {half}} = \frac{2}{2} = 1\ \text s\end{aligned}$$

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

Assertion is true but reason is false

Explanation

Assertion (A) is true as $\text {Speed} = \frac {\text {Distance}}{\text {Time}}$ 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 s^-2, etc. so derived units are completely related to fundamental units.

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.

Assertion is false but reason is true

Explanation

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

$\text {LC} = \frac{\text {Pitch}}{\text {Number of divisions on circular scale}}$ ​ 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.

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 :

$\text T = 2\text π\sqrt {\frac {\text l}{\text g}}$ ​ 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.

(a) Ayush and Reema obtained different values because the two instruments have different least counts (precision) and an instrument with a larger least count measures only up to fewer decimal places, while a more precise instrument can measure smaller fractions of a unit. Therefore, the readings differ even though the length is the same.

(b) Instrument B is more accurate because its reading (25.36 cm) is given up to two decimal places, indicating a smaller least count and higher precision than instrument A.

(c) Instrument A is likely a metre scale (or ruler), which usually measures up to 0.1 cm and instrument B is likely a vernier caliper, which can measure more precisely, typically up to 0.01 cm.

(d) It is important to use S.I. units while recording measurements because they provide a standard and internationally accepted system of units which ensures that measurements are uniform, consistent and easily comparable anywhere in the world.

(a) The time period T is the time taken for one oscillation and since the time given is for 20 oscillations, then time T is given by,

$\text T = \frac{\text {Time for 20 oscillations}}{20}$

$$\begin{matrix} \text{Length (m)} & \text{Time taken (s)} & \text{Time Period} \\ 0.40 & 25.4 & \text T = \frac{\text {25.4}}{20} \\ & & \phantom{T \space} = 1.27 \text{ s} \\ 0.60 & 31.0 & \text T = \frac{\text {31.0}}{20} \\ & & \phantom{T \space} = 1.55 \text{ s} \\ 0.90 & 38.0 & \text T = \frac{\text {38.0}}{20} \\ & & \phantom{T \space} = 1.90 \text{ s} \\ 1.00 & 40.2 & \text T = \frac{\text {40.2}}{20} \\ & & \phantom{T \space} = 2.01 \text{ s} \\ \end{matrix}$$

(b) The table below shows length (l) and T²:

The graph between T² and the length (l) of the pendulum is shown below :

(c) As the graph is a straight line, we can conclude that T² is directly proportional to the length (l) of the pendulum.

(d) From the relation

$\text T^2 = 4\text π^2\frac{\text l}{\text g}$

The slope of the T² vs l graph can be used to calculate the acceleration due to gravity (g).

(e) No, the graph would not remain the same on the Moon because the value of acceleration due to gravity (g) on the Moon is smaller than on Earth. Since $\text T^2 \propto \frac{1}{\text g}$, the time period would increase, changing the slope of the graph.