Two spherical black bodies of radius r_{1} an | vedclass

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

Question

(A) According to the Stefan-Boltzmann Law,the power $P$ radiated by a spherical black body of radius $r$ and temperature $T$ is given by $P = \sigma A

Vedclass · Accepted Answer

$\left(\frac{T_{2}}{T_{1}}\right)^{2}$

Vedclass · Answer

(A) According to the Stefan-Boltzmann Law,the power $P$ radiated by a spherical black body of radius $r$ and temperature $T$ is given by $P = \sigma A T^{4}$,where $A = 4 \pi r^{2}$ is the surface area.Since both bodies radiate the same power,we have $P_{1} = P_{2}$.Substituting the formula,we get $4 \pi r_{1}^{2} \sigma T_{1}^{4} = 4 \pi r_{2}^{2} \sigma T_{2}^{4}$.Canceling the common terms $4 \pi \sigma$,we get $r_{1}^{2} T_{1}^{4} = r_{2}^{2} T_{2}^{4}$.Rearranging the terms to find the ratio $\frac{r_{1}^{2}}{r_{2}^{2}}$,we get $\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{T_{2}^{4}}{T_{1}^{4}}$.Taking the square root on both sides,we get $\frac{r_{1}}{r_{2}} = \frac{T_{2}^{2}}{T_{1}^{2}} = \left(\frac{T_{2}}{T_{1}}\right)^{2}$.

Two spherical black bodies of radius $r_{1}$ and $r_{2}$ with surface temperatures $T_{1}$ and $T_{2}$ respectively,radiate the same power. Then the ratio $r_{1}: r_{2}$ is:

Similar Questions

Two spheres of the same material have radii $1 \; m$ and $4 \; m$ and temperatures $4000 \; K$ and $2000 \; K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

$A$ black body at $227^{\circ}C$ radiates heat at the rate of $20 \, cal \, m^{-2} \, s^{-1}$. When its temperature is raised to $727^{\circ}C$,the rate of heat radiation will be $x \, cal \, m^{-2} \, s^{-1}$. Find $x$.

Two spheres $S_{1}$ and $S_{2}$ have radii $R$ and $3R$ and temperatures $T$ and $T/3$ respectively. If they are coated with a material of the same emissivity, and the rate of radiation of $S_{1}$ is $E$, then the rate of radiation of $S_{2}$ is:

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ} C$. If length and breadth are reduced to $1/3$ of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module