An expression of energy density is given by u | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The argument of the sine function must be dimensionless,so the dimensions of $\frac{\alpha x}{kt}$ must be $[M^0L^0T^0]$.Since $[kt] = [	ext{Ener

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The argument of the sine function must be dimensionless,so the dimensions of $\frac{\alpha x}{kt}$ must be $[M^0L^0T^0]$.Since $[kt] = [	ext{Energy}] = [ML^2T^{-2}]$,we have $[\alpha] = \frac{[kt]}{[x]} = \frac{[ML^2T^{-2}]}{[L]} = [MLT^{-2}]$.The energy density $u$ has dimensions of energy per unit volume,so $[u] = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.From the given equation $u = \frac{\alpha}{\beta} \sin(\dots)$,the dimensions of $\beta$ are given by $[\beta] = \frac{[\alpha]}{[u]}$.Substituting the dimensions: $[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [M^0L^2T^0]$.Therefore,the correct option is $D$.

An expression of energy density is given by $u = \frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{kt}\right)$,where $\alpha$ and $\beta$ are constants,$x$ is displacement,$k$ is the Boltzmann constant,and $t$ is the temperature. The dimensions of $\beta$ will be $-$

Similar Questions

In the half-wave rectifier circuit shown,which one of the following waveforms is true for $V_{CD}$,the output across $C$ and $D$?

Heat energy is incident on the surface at the rate of $1000 ~J/min$. If the coefficient of absorption is $0.8$ and the coefficient of reflection is $0.1$,then the heat energy transmitted by the surface in $5 ~min$ is: (in $~J$)

Unisexuality of flowers prevents

Identify the final product $Y$ in the following reaction sequence:
$CH_2 = CH_2$ $\xrightarrow{O_2/Ag} X$ $\xrightarrow{473 \ K} Y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module