JEE Main 2022 (27 Jul, Shift 2) Physics Q31: An expression…

Question

An expression of energy density is given by

u=\frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right)

, where

\alpha

,

\beta

are constants,

x

is displacement,

k

is Boltzmann constant and t is the temperature. The dimensions of

\beta

will be :

Official answer

Option 4

NTA final key.

Accepted Answer

Option 4: $[M^0L^2T^0]$

Answer

$[ML^2T^{-2}	heta^{-1}]$

Answer

$[M^0L^2T^{-2}]$

Answer

$[M^0L^0T^0]$

Answer

$[M^0L^2T^0]$

JEE Main 27 July 2022, Shift 2, Physics Q31