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Standard XII

Physics

Emissive and Absorptive Power

Two copper sp...

Question

Two copper spheres of radii $6 c m$ and $12 c m$ respectively are suspended in an evacuated enclosure. Each of them are at a temperature $15^{0} C$ above the surroundings. The ratio of their rate of loss of heat is ___________.

2 : 1

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1 : 4

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1 : 8

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8 : 1

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Solution

The correct option is C $1 : 4$
According to the Stefan-Boltzmann law, the rate of loss of heat from a hot body is given as
$P = σ A (T^{4} - T_{c}^{4})$
$= σ 4 π r^{2} (T^{4} - T_{c}^{4})$
Hence $\frac{P_{1}}{P_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}}$
$= \frac{6^{2}}{12^{2}}$
$= \frac{1}{4}$

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Two copper spheres of radii 6cm and 12cm respectively are suspended in an evacuated enclosure. Each of them are at a temperature 150C above the surroundings. The ratio of their rate of loss of heat is ___________.

The correct option is C 1:4According to the Stefan-Boltzmann law, the rate of loss of heat from a hot body is given as P=σA(T4−T4c)=σ4πr2(T4−T4c)Hence P1P2=r21r22=62122=14

Two copper spheres of radii $6 c m$ and $12 c m$ respectively are suspended in an evacuated enclosure. Each of them are at a temperature $15^{0} C$ above the surroundings. The ratio of their rate of loss of heat is ___________.

The correct option is C $1 : 4$
According to the Stefan-Boltzmann law, the rate of loss of heat from a hot body is given as
$P = σ A (T^{4} - T_{c}^{4})$
$= σ 4 π r^{2} (T^{4} - T_{c}^{4})$
Hence $\frac{P_{1}}{P_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}}$
$= \frac{6^{2}}{12^{2}}$
$= \frac{1}{4}$