Question

A spherical black body with a radius 12cm radiated 450W power at 500K. If radius were halved and temperature doubled, the power radiated in watt would be...

Answer 1

Answer:

Radiant Power (P) -

P=frac{Q}{t}=Aepsilon sigma heta^{4}

-

Power of radiation is

P=\sigma A T^{4} (For blackbody)

\frac{P_{2}}{P_{1}}=\left(\frac{A_{2}}{A_{1}} \right ).\left(\frac{T_{2}}{T_{1}} \right )^{4}=\left(\frac{r_{2}}{r_{1}} \right )^{2}.\left(\frac{T_{2}}{T_{1}} \right )^{4}

According to question

\frac{r_{2}}{r_{1}}=\frac{1}{2}\ \: \: \: and\ \: \: \: \frac{T_{2}}{T_{1}}=2

Then \frac{P_{2}}{P_{1}}=\left(\frac{1}{2} \right )^{2}.2^{4}=2^{2}

or P2 = 4P1 = 4 \times 450 W = 1800 W

Correct option is 4.

Option 1)

225

Incorrect

Option 2)

450

Incorrect

Option 3)

1000