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If mu(0) is permeability of free space a...

If `mu_(0)` is permeability of free space and `epsilon_(0)` is permittivity of free space , the speed of light in vacuum is given by

A

`sqrt((1)/(mu_(0)epsilon_(0)))`

B

`sqrt((epsilon_(0))/(mu_(0)))`

C

`sqrt(mu_(0)epsilon_(0))`

D

`sqrt((mu_(0))/(epsilon_(0)))`

Text Solution

Verified by Experts

The correct Answer is:
A
The velocity of light `c=(1)/(sqrt(mu_(0)epsilon_(0)))`
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Knowledge Check

  • A quantity X is given by (me^(4))/(8epsilon_(0)^(2)ch^(3)) where m is mass of electron, e is the charge of electron, epsilon_(0) is the permittivity of free space, c is the velocity of light and h is the Planck's constant. The dimensional formula for X is the same as that of :

    A
    length
    B
    frequency
    C
    velocity
    D
    wave number

  • If L has the dimensions of length, V that of potential and epsilon_(0) is the permittivity of free space then quantity epsilon_(0) LV have the dimensions of :

    A
    Current
    B
    Charge
    C
    Resistance
    D
    Voltage

  • Which of the following does not have the dimensions of velocity ? (Given, epsilon_(0) = permittivity of free space, mu_(0) = permeability of free space, v = frequency , is the wavelength, P is the pressure and = density, k = wave number, omega is the angular frequency) :

    A
    `omega k`
    B
    `vlambda`
    C
    `sqrt(epsi_(0)mu_(0))`
    D
    `sqrt(P/rho.)`

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    If x=(epsilon_(0)lV)/(t) where epsilon_(0) is permittivity of free space, l is the length, V is the potential difference and t is the time, then dimensions of x are the same as that of :

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