Question

if g is the acceleration due to gravity and Lambda is wavelength then which physical quantity does Lambda root G represents

Answer 1

λ = m
g =m/s²

(λ g)² = [(m) (m/s²)]
(λ g)² = m^4 x s^4

Answer 2

Answer:

The physical quantity of $\lambda\sqrt{g}\ is [LT^{-1}\sqrt{L}]$

Explanation:

Given that,

$\lambda\sqrt{g}$.....(I)

If g is the acceleration due to gravity and Lambda is wavelength.

We need to calculate the physical quantity of $\lambda\sqrt{g}$,

We know the dimension of acceleration and wavelength

$g = [LT^{-2}]$

$\lambda=[L]$

Put the value of g and lambda in equation (I)

$\lambda\sqrt{g}=[L]\times\sqrt{[LT^{-2}]}$

$\lambda\sqrt{g}=[LT^{-1}\sqrt{L}]$

Hence, The physical quantity of $\lambda\sqrt{g}\ is [LT^{-1}\sqrt{L}]$