Question

A sphere is at temperature \(600\) K. In an external environment of \(200\) K, its cooling rate is R. When the temperature of the sphere falls to \(400\) K then cooling rate R' will become

• A. \(\frac{16}{3}R\)
• B. \(\frac{16}{9}R\)
• C. \(\frac{9}{16}R\)
• D. \(\frac{3}{16}R\)

Hint:

Rate of cooling by radiation is proportional to T^4 minus T0^4.

Answer 1

The Correct Option is D

Step 1: Use real numbers:
$600^4 = 1.296\times10^{11}$, $400^4 = 2.56\times10^{10}$, $200^4 = 1.6\times10^{9}$.

Step 2: Net radiative terms:
Initial: $1.296\times10^{11} - 1.6\times10^9 = 1.280\times10^{11}$. Final: $2.56\times10^{10} - 1.6\times10^9 = 2.40\times10^{10}$.

Step 3: Ratio:
$\frac{2.40\times10^{10}}{1.280\times10^{11}} = 0.1875 = \frac{3}{16}$. So $R' = \frac{3}{16}R$.

Final Answer:
R-prime is 3/16 of R, option (D). \[ \boxed{\frac{3}{16}R} \]