A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?

Options

A. `(Q)/(2)`
B. `(Q)/(4)`
C. `(Q)/(16)`
D. `2Q`

Answer

To solve the problem step-by-step, we will analyze the heat conduction through the original rod and the new rod formed after melting.

### Step 1: Understand the heat conduction formula
The heat conducted through a rod can be expressed using the formula:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \cdot t \]
where:
- \( Q \) is the amount of heat conducted,
- \( k \) is the thermal conductivity of the material,
- \( A \) is the cross-sectional area of the rod,
- \( \Delta T \) is the temperature difference between the two ends,
- \( L \) is the length of the rod,
- \( t \) is the time.

### Step 2: Calculate the heat conducted by the original rod
For the original rod with radius \( r \) and length \( L \):
- The cross-sectional area \( A_1 \) is given by:
\[ A_1 = \pi r^2 \]
- The heat conducted in time \( t \) is:
\[ Q_1 = \frac{k \cdot \pi r^2 \cdot \Delta T}{L} \cdot t \]

### Step 3: Analyze the new rod
The new rod has half the radius of the original rod, so:
- New radius \( r' = \frac{r}{2} \)
- The cross-sectional area \( A_2 \) of the new rod is:
\[ A_2 = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

### Step 4: Determine the length of the new rod
Since the volume of the material remains constant when the rod is melted and reshaped, we can equate the volumes:
\[ \text{Volume of original rod} = \text{Volume of new rod} \]
\[ \pi r^2 L = \pi \left(\frac{r}{2}\right)^2 L' \]
This simplifies to:
\[ r^2 L = \frac{r^2}{4} L' \]
Cancelling \( r^2 \) from both sides gives:
\[ L = \frac{L'}{4} \quad \Rightarrow \quad L' = 4L \]

### Step 5: Calculate the heat conducted by the new rod
Using the new dimensions:
- The heat conducted by the new rod \( Q_2 \) is:
\[ Q_2 = \frac{k \cdot A_2 \cdot \Delta T}{L'} \cdot t \]
Substituting \( A_2 \) and \( L' \):
\[ Q_2 = \frac{k \cdot \left(\pi \frac{r^2}{4}\right) \cdot \Delta T}{4L} \cdot t \]
This simplifies to:
\[ Q_2 = \frac{k \cdot \pi r^2 \cdot \Delta T}{16L} \cdot t \]

### Step 6: Relate \( Q_1 \) and \( Q_2 \)
From the original heat conduction equation:
\[ Q_1 = \frac{k \cdot \pi r^2 \cdot \Delta T}{L} \cdot t \]
Now, we can relate \( Q_1 \) and \( Q_2 \):
\[ \frac{Q_1}{Q_2} = \frac{\frac{k \cdot \pi r^2 \cdot \Delta T}{L} \cdot t}{\frac{k \cdot \pi r^2 \cdot \Delta T}{16L} \cdot t} = 16 \]
Thus:
\[ Q_1 = 16 Q_2 \quad \Rightarrow \quad Q_2 = \frac{Q_1}{16} \]

### Step 7: Substitute \( Q_1 \) with \( Q \)
Given that \( Q_1 = Q \):
\[ Q_2 = \frac{Q}{16} \]

### Final Answer
The amount of heat conducted by the new rod in time \( t \) is:
\[ Q_2 = \frac{Q}{16} \]