Question

In an external environment of temperature (\(T\)) kelvin, a sphere at temperature (\(3T\)) kelvin has cooling rate \(R_1\). When the temperature of that sphere falls to (\(2T\)) kelvin, the cooling rate \(R_2\) of the sphere will become

• A. \(\frac{15}{16}R_1\)
• B. \(\frac{11}{16}R_1\)
• C. \(\frac{7}{16}R_1\)
• D. \(\frac{3}{16}R_1\)

Hint:

Radiation rate \(\propto T^4-T_0^4\) for a body in surroundings at \(T_0\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
A hot sphere in a surrounding at temperature \(T\) loses heat by radiation. By Stefan's law the net rate is proportional to \(T_s^4-T_0^4\).

Step 2: Key Formula or Approach
\(R\propto T_s^4-T^4\) with the surroundings at \(T\).

Step 3: Detailed Explanation
At \(3T\): \(R_1\propto(3T)^4-T^4=80T^4\).
At \(2T\): \(R_2\propto(2T)^4-T^4=15T^4\).
\[ R_2=\frac{15}{80}R_1=\frac{3}{16}R_1 \]

Final Answer:
The cooling rate becomes \(\frac{3}{16}R_1\), option (D). \[ \boxed{\dfrac{3}{16}R_1\ \text{(D)}} \]