The dimensions of sigma b^4, where sigma is S | vedclass

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

Question

(B) According to Stefan's law, the energy $(E)$ emitted per unit area per unit time is given by $E = \sigma T^4$, where $\sigma$ is Stefan's constant

Vedclass · Accepted Answer

$[M L^4 T^{-3}]$

Vedclass · Answer

(B) According to Stefan's law, the energy $(E)$ emitted per unit area per unit time is given by $E = \sigma T^4$, where $\sigma$ is Stefan's constant and $T$ is temperature.Dimensional analysis for $\sigma$:$\frac{[M L^2 T^{-2}]}{[L^2 T]} = [\sigma] [K^4]$$[M T^{-3}] = [\sigma] [K^4]$$[\sigma] = [M T^{-3} K^{-4}]$Wien's displacement law is given by $b = \lambda T$, where $\lambda$ is wavelength and $T$ is temperature.Dimension of $b = [L K]$.Now, calculating the dimensions of $\sigma b^4$:$[\sigma b^4] = [M T^{-3} K^{-4}] 	imes [L K]^4$$[\sigma b^4] = [M T^{-3} K^{-4}] 	imes [L^4 K^4]$$[\sigma b^4] = [M L^4 T^{-3}]$.

The dimensions of $\sigma b^4$, where $\sigma$ is Stefan's constant and $b$ is Wien's constant, are:

Similar Questions

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now, due to a change in the temperature of that body, it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature, the emissive power is:

Three identical rods $AB$,$CD$ and $PQ$ are joined as shown. $P$ and $Q$ are midpoints of $AB$ and $CD$ respectively. Ends $A, B, C$ and $D$ are maintained at $0^{\circ} C, 100^{\circ} C, 30^{\circ} C$ and $60^{\circ} C$ respectively. The direction of heat flow in $PQ$ is

Three rods of the same dimension have thermal conductivities $3K$,$2K$,and $K$. They are arranged as shown in the figure,with their ends at $100^oC$,$50^oC$,and $20^oC$. The temperature of their junction is ......... $^oC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module