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Standard XII

Physics

Wave Equation of EM Waves

The temperatu...

Question

The temperatures of two bodies A and B are $727^{0}$ C and $127^{0}$ C. The ratio of rate of emission of radiations will be

727/127

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625/16

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1000/400

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100/16

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Solution

The correct option is B
625/16

$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{727 + 273}{127 + 273})}^{4} = (\frac{(1000)^{4}}{(400)^{4}}) = \frac{10^{4}}{4^{4}} = \frac{625}{16}$

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The temperatures of two bodies A and B are 7270C and 1270 C. The ratio of rate of emission of radiations will be

The correct option is B 625/16 E1E2=(T1T2)4=(727+273127+273)4=((1000)4(400)4)=10444=62516

The temperatures of two bodies A and B are $727^{0}$ C and $127^{0}$ C. The ratio of rate of emission of radiations will be

The correct option is B
625/16

$\frac{E_{1}}{E_{2}} = {(\frac{T_{1}}{T_{2}})}^{4} = {(\frac{727 + 273}{127 + 273})}^{4} = (\frac{(1000)^{4}}{(400)^{4}}) = \frac{10^{4}}{4^{4}} = \frac{625}{16}$