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प्रश्न
Two bodies A and B having equal surface areas are maintained at temperature 10°C and 20°C. The thermal radiation emitted in a given time by A and B are in the ratio
पर्याय
1 : 1.15
1 : 2
1 : 4
1 : 16
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उत्तर
1 : 1.15
From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by `u = σAT^4 `
Here, `σ` is Stefan-Boltzmann constant.
`(uA)/(uB)` = `"T"_"A"^4/(T_B^4)`
`(u_A)/ (u_B) = ( 273 +10 )^4/ (273 + 20 )^4`
`uA/uB = 1/1.15`
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