Question

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of \(0.1\) gram of ice per second. If the rod is replaced by another with half the length and double the radius of the first, the thermal conductivity of the material of second rod is \(\left(\frac{1}{4}\right)^{th}\) that of first, the rate at which ice melts in gram per second will be (Latent heat of ice \(=80\) cal/g)

• A. \(3.2\)
• B. \(1.6\)
• C. \(0.2\)
• D. \(0.1\)

Hint:

The melting rate is proportional to kA divided by L, so compare the factors for the new rod.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Heat conducted per second is \(\dfrac{kA\Delta T}{L}\), and the ice melts at the rate \(\dfrac{dm}{dt}=\dfrac{kA\Delta T}{LL_f}\). So the rate is proportional to \(\dfrac{kA}{L}\), where \(A\propto r^{2}\).

Step 2: Factors for the new rod
\(k'=\dfrac{k}{4}\), \(r'=2r\Rightarrow A'=4A\), \(L'=\dfrac{L}{2}\).

Step 3: Calculation
\(\dfrac{k'A'/L'}{kA/L}=\dfrac{1}{4}\times4\times2=2\)

Step 4: New rate
\(0.1\times2=0.2\) g/s

Final Answer:
The ice melts at 0.2 gram per second. \[ \boxed{\text{(C) }0.2} \]