Practical Unit

If you were to drive from one city to another, you would have to calculate the fuel consumption in litres or gallons, check the tyre pressure in pounds per square inch, or pascals and of course, measure the distance in kilometres or miles. All these are some examples of practical units. These are basically the units that you experience in day-to-day life and you just cannot do without them if you want to get things done. Such practical units facilitate survival in this world and provide a yardstick to standardize and quantify many aspects of our lives, thereby simplifying communication, planning, and execution of activities.

In the present article, we will have a glance at the concept of practical units, their meaning and usage, and how they help standardize and simplify measurement in real life. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept. But no direct question has been asked in NEET from this concept.

What is Practical Unit?

Practical units are units of measurement that are commonly used in everyday applications and are typically easier to relate to in practical scenarios compared to more abstract units. Examples include meters for length, kilograms for mass, and seconds for time in the SI system.

Practical units of length

$\begin{array}{|ll|}\hline \text{Name}& \text{Symbol Conversion in}\mathrm{m}\\ 1\text{fermi}& 1\mathrm{fm}& {10}^{-15}\mathrm{m}\\ 1\text{X-ray unit}& 1\mathrm{XU}& {10}^{-13}\mathrm{m}\\ 1\text{Angstrom}& 1A^o& {10}^{-10}\mathrm{m}\\ 1\text{micron}& 1\mu \mathrm{m}& {10}^{-6}\mathrm{m}\\ 1\text{Astronomical unit}& 1\mathrm{AU}& 1.5\ast {10}^{11}\mathrm{m}\approx {10}^8\mathrm{km}\\ 1\text{Light year}& 1\mathrm{ly}& 9.46\ast {10}^{15}\mathrm{m}\\ 1\text{Parsec}& 1\mathrm{Pc}& 3.26\text{light year}\\ \hline\end{array}$

Practical units of Mass

$\begin{array}{|ccc|}\hline \text{Name}& \text{Symbol}& \text{Conversion in kilogram}(\mathrm{Kg})\\ 1\text{Chandra Shekhar Unit}& 1\mathrm{CSU}& \begin{array}{l}\begin{array}{c}2.8\ast {10}^{30}\mathrm{kg}\\ =1.4\text{times the mass of the sun}\end{array}\end{array}\\ 1\text{Metric tonne}& 1\text{Metric tonne}& 1000\mathrm{kg}\\ 1\text{Quintal}& 1\text{Quintal}& 100\mathrm{kg}\\ 1\text{Atomic mass unit}& 1\mathrm{amu}& 1.67\ast {10}^{-27}\mathrm{kg}\\ \hline\end{array}$

Practical Units of Time

$1\text{year}=365.25\text{days}=3.156\ast {10}^7\mathrm{Sec}$

Lunar Month- 29.53 days (29 days 12 hours and 44 minutes) (It is the time taken by the moon to complete 1 revolution around the earth in its orbit or A lunar month is a duration between successive new moons)

Solar day- It is the time taken by the earth to complete one rotation about its axis with respect to the Sun.

Sidereal day- It is the time taken by the earth to complete one rotation about its axis with respect to a distant star.

1 Solar year- 366.25 sidereal days = 365.25 average solar day

1 Average Solar Day $=\left({\frac{1}{365.25}}^{\text{th}}\right)$ part of the solar year
1 solar second $={\left({\frac{1}{86400}}^{\text{th}}\right)}_{\text{part of the mean solar day}}$

For More Information On Practical Units, Watch The Below Video:

Solved Example Based on Practical Units

Example 1: Chandra Shekhar's unit is related to

1) time

2) length

3) mass

4) amount of substance

Solution:

Chandra Shekhar's unit (CSU) is a unit of mass. According to the information provided earlier:
- $1\mathrm{CSU}=2.8\times {10}^{\wedge }30\mathrm{kg}$

This definition explicitly states that CSU is a unit used to measure mass. Therefore, the correct answer to the question "Chandra Shekhar's unit is related to" is 3 ) mass. This unit is named after the renowned Indian astrophysicist Subrahmanyan Chandrasekhar, who made significant contributions to our understanding of stellar structure and evolution.

Hence, the answer is the option (3).

Example 2: The distance of the Sun from Earth and its angular diameter is (2000)s when observed from the Earth. The diameter of the Sun will be :

1) $2.45\times {10}^{10}\mathrm{m}$
2) $1.45\times {10}^{10}\mathrm{m}$
3) $1.45\times {10}^9\mathrm{m}$
4) $0.14\times {10}^9\mathrm{m}$

Solution:

1 degree = 60 minute
1 minute = 60 second
1 degree = 3600 second$$\begin{array}{rl}& 1\text{second}=\frac{1}{3600}\text{degree}\\ & =\frac{1}{3600}\times \frac{\pi }{180}\mathrm{rad}.\\ & 2000\text{second}=\frac{2000}{3600}\times \frac{\pi }{180}=\frac{2\pi }{36\times 18}\\ & \end{array}$$

$\begin{array}{rl}& \theta =\frac{\mathrm{D}}{\mathrm{d}}\\ & \begin{array}{rl}\text{diameter}& =\frac{2\pi }{36\times 18}\times 1.5\times {10}^{11}\\ & =1.45\times {10}^9\mathrm{m}\end{array}\end{array}$

Hence, the answer is option (3).

Example 3:Mean Solar Second is equal to

$1)\text{1)}{\frac{1}{365.25}}^{\text{th}}$
part of the solar year
2) $\frac{1}{86400}$
part of the solar day
3) ${\frac{1}{43200}}^{\text{th}}$
part of the mean solar day
4) Both 1 and 2

Solution:

As we have learnt,

Units of Time -

1 year = 365.25 days = 3.156 x 10⁷ Sec

Lunar Month- 29.53 days (29 days 12 hours and 44 minutes) (It is the time taken by the moon to complete 1 revolution around the earth in its orbit or A lunar month is a duration between successive new moons)

1 Average Solar Day 1/365.25 the part of a solar year

$1\text{solar second}=\frac{1}{86400}\text{part of the mean solar day}$

Hence, the answer is the option 2.

Example 4: Sidereal Day

1)is the time for the Earth to rotate about its axis so that the sun appears in the same position in the sky

2)is the time taken by the Earth to complete one rotation about its axis with respect to the distant star

3)is 4 minutes shorter than the solar day

4) Both B and C

Solution:

As we have learnt,

Sidereal day -

It is the time taken by Earth to complete one rotation about its axis with respect to a distant star.

Hence, the answer is the option 4.

Example 5: Given below are two statements :
Statements I: Astronomical unit (Au), Parsec (Pc), and Light year (ly) are units for measuring astronomical distances.
Statements II : Au < Parsec (Pc) < ly
In the light of the above statements, choose the most appropriate answer from the options given below :

1) Both Statement I and Statement II are incorrect.

2) Both Statement I and Statement II are correct.

3) Statement I is incorrect but Statement II is correct.

4) Statement I is correct but Statement II is incorrect.

Solution:

Astronomical unit (Au), Parsec (Pc), and Light year (ly) are the units of distance
Light year $\to$ distance travelled by light in one year
$11\mathrm{y}=9.5\times {10}^{15}\mathrm{m}$
parsec $=3.262$ lightyear
$\mathrm{Au}=1.58\times {10}^{-5}$ light year
$\mathrm{Au}<1\mathrm{y}<$ Parsec.
Statement I is correct and statement II is incorrect.

Hence, the answer is the option (4).

Summary

The practical units are those which are used in everyday activities for the measurement of distance, volume, weight, and pressure. As a matter of fact, some practical units are meters, litres, kilograms, and newtons among others. These units of measure are the fundamental procedures that are used to appraise all aspects of life. These units of measure necessarily have to be standardized; otherwise, they would never be able to effectively execute the function for which they have been designed. Consequently, the units of measurement necessarily have to be standardized in such a manner that their manipulation and adjustment is not possible. Complex concepts become easy with practical units, thereby making it applicable in real life. When the units are known and applied correctly, a person can efficiently plan, decide, and execute day-to-day tasks in daily life besides at the professional level.