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Let [in(0)] denote the dimensional formu...

Let `[in_(0)]` denote the dimensional formula of the permittivity of the vacuum and `[mu_(0)]` that of the permability of the vacuum.If `M="mass",L="length",T-"time"` and `I="electric current"`

A

`[epsi_(0)]=[M^(-1)L^(-3)T^(2)]`

B

`[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`

C

`[mu_(0)]=[MLT^(-2)I^(-2)]`

D

`[mu_(0)]=[ML^(2)T^(-1)I]`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`epsi_(0)`= Electrical permittivity of vacuum
`:.F=(1)/(4piepsi_(0)).(q_(1)q_(2))/(r^(2)):.epsi_(0)=(1)/(4piF)(q_(1)q_(2))/(r^(2))`
`:.[epsi_(0)]=([q_(1)q_(2)])/([F][r^(2)])=([IT]^(2))/([MLT^(-2)][L^(2)])`
`:.[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`
Again, `mu_(0)` = Magnetic permeability of vacuum.
`:.(1)/(sqrt(mu_(0)epsi_(0)))=c` (velocity of light)
`:.[mu_(0)]=(1)/([epsi_(0)][c^(2)])=(1)/([M^(-1)L^(-3)T^(4)I^(2)][LT^(-1)]^(2))`
or `[mu_(0)]=[MLT^(-2)I^(-2)]`
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