BITSAT 2019 Slot 2 Solved Paper

We know that, velocity of light c is terms of ∈0 and µ0 is given by
c=

1

√µ0ε0

⇒c2=

1

µ0ε0

c2=

1

µ02

.

µ0

ε0

⇒c=

1

µ0

√

µ0

ε0

⇒√

µ0

ε0

=cµ0 .......(i)
Dimension of magnetic field, B=

τ

NIA

=

[ML2T−2]

[A][L2]

=[MA−1T−2]
Again, B=µ0nI
∴µ0=

B

nl

=

[MA−1T−2]

[L−1][A]

=[MLA−2T−2]
µ0=[MLA−2T−2] .......(ii)
and C=[LT−1] .......(iii)
From Eqs. (i), (ii) and (iii), we get
√

µ0

ε0

=[LT−1][MLA−2T−2] =[ML2T−3A−2]