The dimensional formula of sqrt{mu_r K} is eq | vedclass

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Question

(A) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$.In vacuum,the speed of light is $c = \frac{1}{\sq

Vedclass · Accepted Answer

$M^0 L^0 T^0$

Vedclass · Answer

(A) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \epsilon}}$.In vacuum,the speed of light is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.The refractive index $n$ of a medium is defined as $n = \frac{c}{v} = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}} = \sqrt{\mu_r \epsilon_r}$.Here,$\mu_r$ is the relative permeability and $\epsilon_r$ (often denoted as $K$,the dielectric constant) is the relative permittivity.Thus,$n = \sqrt{\mu_r K}$.Since the refractive index $n$ is the ratio of two speeds,it is a dimensionless quantity.Therefore,the dimensional formula of $\sqrt{\mu_r K}$ is $M^0 L^0 T^0$.

$LIST$-$I$	$LIST$-$II$
$(A)$ Surface tension	$(I)$ $kg m^{-1} s^{-1}$
$(B)$ Pressure	$(II)$ $kg m s^{-1}$
$(C)$ Viscosity	$(III)$ $kg m^{-1} s^{-2}$
$(D)$ Impulse	$(IV)$ $kg s^{-2}$

List-$I$	List-$II$
$(A)$ Young's Modulus	$(I)$ $[ML^{-1}T^{-1}]$
$(B)$ Torque	$(II)$ $[ML^{-1}T^{-2}]$
$(C)$ Coefficient of Viscosity	$(III)$ $[M^{-1}L^3T^{-2}]$
$(D)$ Gravitational Constant	$(IV)$ $[ML^2T^{-2}]$

The dimensional formula of $\sqrt{\mu_r K}$ is equal to:

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The unit of "impulse per unit area" is the same as that of

Match List-$I$ with List-$II$:

$LIST$-$I$ $LIST$-$II$

$(A)$ Surface tension $(I)$ $kg m^{-1} s^{-1}$

$(B)$ Pressure $(II)$ $kg m s^{-1}$

$(C)$ Viscosity $(III)$ $kg m^{-1} s^{-2}$

$(D)$ Impulse $(IV)$ $kg s^{-2}$

Choose the correct answer from the options given below:

Match List-$I$ with List-$II$:
List-$I$ List-$II$
$(A)$ Young's Modulus $(I)$ $[ML^{-1}T^{-1}]$
$(B)$ Torque $(II)$ $[ML^{-1}T^{-2}]$
$(C)$ Coefficient of Viscosity $(III)$ $[M^{-1}L^3T^{-2}]$
$(D)$ Gravitational Constant $(IV)$ $[ML^2T^{-2}]$

Choose the correct answer from the options given below:

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