The Smart Batch
Question 1:
In SI units, the dimensions of √ε0/μ0 is
A)
A 2 T 3 M − 1 L − 2 A^{2}{\text{\ }T}^{3}M^{- 1}{\text{\ }L}^{- 2}
B)
A T 2 M − 1 L − 1 {AT}^{2}M^{- 1}{\text{\ }L}^{- 1}
C)
A T − 3 M L 3 / 2 {AT}^{- 3}{ML}^{3/2}
D)
A − 1 T M L 3 A^{- 1}{TML}^{3}
Answer:
Solution:
To find: Dimension of ε 0 μ 0 in SI units \text{To\ find:\ Dimension\ of~}\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}\text{~in\ SI\ units}
Unit of permittivity of free space: (by Coulomb's law),
ε 0 = c 2 m 2 F \varepsilon_{0} = \frac{c^{2}}{m^{2}\text{F}}
dimensional formula of
[ ε 0 ] = [ M − 1 L − 3 T 4 A 2 ] , \left\lbrack \varepsilon_{0} \right\rbrack = \left\lbrack M^{- 1}L^{- 3}T^{4}A^{2} \right\rbrack,\
Also,
ε 0 μ 0 = ε 0 2 ε 0 μ 0 = c ε 0 ( as, c = 1 ϵ 0 μ 0 ) \sqrt{\frac{\varepsilon_{0}}{\mu_{0}}} = \sqrt{\frac{\varepsilon_{0}^{2}}{\varepsilon_{0}\mu_{0}}} = c\varepsilon_{0}\ \left( \text{as,~}c = \frac{1}{\sqrt{\epsilon_{0}\mu_{0}}} \right)
From (i) and (ii) the dimensions of ε 0 μ 0 \sqrt{\frac{\varepsilon_{0}}{\mu_{0}}} will be dimensions of c c times dimensions of ε 0 \varepsilon_{0} :
ε 0 μ 0 = [ c ] [ ε 0 ] \sqrt{\frac{\varepsilon_{0}}{\mu_{0}}} = \lbrack c\rbrack\left\lbrack \varepsilon_{0} \right\rbrack
= [ L 1 T − 1 ] [ M − 1 L − 3 T 4 A 2 ] = \left\lbrack L^{1}{\text{\ }T}^{- 1} \right\rbrack\left\lbrack M^{- 1}{\text{\ }L}^{- 3}{\text{\ }T}^{4}{\text{\ }A}^{2} \right\rbrack
= [ M − 1 L − 2 T 3 A 2 ] = \left\lbrack M^{- 1}{\text{\ }L}^{- 2}{\text{\ }T}^{3}{\text{\ }A}^{2} \right\rbrack
Prev Exam(s):
JEE MAIN
- 2019
, 8 Apr Shift-01
Chapter:
ERRORS IN MEASUREMENT
Question Level:
L3
Tags:
Related Topic DIMENSIONS OF PHYSICAL QUANTITIES
Related Concept DIMENSIONS OF PHYSICAL QUANTITIES
