EASYVIT - VITEEE 2025
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Practical Units - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Practical units is considered one of the most asked concept.
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18 Questions around this concept.
Solve by difficulty
One solar year is equal to :
The time taken by the earth to complete one rotation about its axis with respect to a distant star is called:
Which of the following has units of time ?
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Which of the following is not a unit for length-
Which of the following is the smallest unit of distance-
Which of the following is a unit of length in the CGS system?
Which of the following is not a unit of time?
Parsec is the unit of
Concepts Covered - 1
Practical units
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Practical units of length
|
Name |
Symbol |
Conversion in m |
|
1 fermi |
1 fm |
$10^{-15} \mathrm{~m}$ m |
|
1 X-ray unit |
1 XU |
$10^{-13} \mathrm{~m}$m |
|
1 Angstrom |
|
$10^{-10} \mathrm{~m}$m |
|
1 micron |
|
$10^{-6} \mathrm{~m}$m |
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1 Astronomical unit |
1 AU |
$10^{8} \mathrm{~Km}$ |
|
1 Light year |
1 ly |
$9.46 * 10^{15} \mathrm{~m}$ |
|
1 Parsec |
1 Pc |
3.26 light year |
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Practical units of mass
|
Name |
Symbol |
Conversion in kilogram (Kg) |
|
1 Chandra Shekhar Unit |
1 CSU |
1.4 times the mass of the sun |
|
1 Metric tonne |
1 Metric tonne |
1000 kg |
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1 Quintal |
1 Quintal |
100 kg |
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1 Atomic mass unit |
1 amu |
$1.67 * 10^{-27} \mathrm{~kg}$ |
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Practical units of Time
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- 1 year $=365.25$ days $=3.156 * 10^7$ Sed
- 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-lt is the time taken by the earth to complete one rotation about its axis with respect to 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= $\left(\frac{1}{365.25}^{t h}\right)$ part of the solar year
- 1 solar second $=\left(\frac{1}{86400}^{t h}\right)_{\text {part of the mean solar day }}$
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