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Standard XII

Physics

Emissive and Absorptive Power

Two spherical...

Question

Two spherical black bodies of radii $r_{1}$ and $r_{2}$ are with surface temperatures $T_{1}$ and $T_{2}$ respectively radiate the same power. $r_{1} / r_{2}$ must be equal to

(T_{1} / T_{2})^{2}

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(T_{2} / T_{1})^{2}

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(T_{1} / T_{2})^{4}

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(T_{2} / T_{1})^{4}

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Solution

The correct option is B $(T_{2} / T_{1})^{2}$
$E = ε σ A T^{4}$
Given that both spherical body radiate with same power.
So equating $E_{1} = E_{2}$
$A = 4 π r^{2}$
$r_{1}^{2} T_{1}^{2} = r_{2}^{2} T_{2}^{4}$

$\frac{r_{1}}{r_{2}} = {(\frac{T_{2}}{T_{1}})}^{2}$

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Two spherical black bodies of radii r1 and r2 are with surface temperatures T1 and T2 respectively radiate the same power. r1/r2 must be equal to

The correct option is B (T2/T1)2E=εσAT4Given that both spherical body radiate with same power.So equating E1=E2A=4πr2r21T21=r22T42r1r2=(T2T1)2

Two spherical black bodies of radii $r_{1}$ and $r_{2}$ are with surface temperatures $T_{1}$ and $T_{2}$ respectively radiate the same power. $r_{1} / r_{2}$ must be equal to

The correct option is B $(T_{2} / T_{1})^{2}$
$E = ε σ A T^{4}$
Given that both spherical body radiate with same power.
So equating $E_{1} = E_{2}$
$A = 4 π r^{2}$
$r_{1}^{2} T_{1}^{2} = r_{2}^{2} T_{2}^{4}$

$\frac{r_{1}}{r_{2}} = {(\frac{T_{2}}{T_{1}})}^{2}$