Question

Two rods of same length, radius and material transfer a given amount of heat in 't' second when they are joined as shown in fig (1). But when they are joined as shown in fig (2) then they will transfer same heat in same condition in time

• A. \(2t\)
• B. \(3t\)
• C. \(4t\)
• D. \(6t\)

Hint:

Parallel doubles the conducting area; series doubles the length. Heat rate ratio is 4.

Answer 1

The Correct Option is C

Step 1: Understanding the Figures:
Figure 1 shows two identical rods side by side, so heat flows through both at once (parallel). Figure 2 shows the same rods joined end to end (series), so heat must flow through twice the length.

Step 2: Heat flow formula:
Heat conducted in time \(t\) is \(Q = \frac{KA(\Delta T)}{L}t\). Let one rod have area \(A\) and length \(L\).

Step 3: Compare:
Parallel: effective area \(2A\), length \(L\), so \(Q = \frac{K(2A)\Delta T}{L}t\).
Series: area \(A\), length \(2L\), so \(Q = \frac{KA\Delta T}{2L}t'\).

Step 4: Equate the same heat Q:
\[ \frac{2KA\Delta T}{L}t = \frac{KA\Delta T}{2L}t' \Rightarrow t' = 4t \]

Step 5: Why the other options are wrong.
\(2t\) results from changing only the length or only the area. \(3t\) and \(6t\) do not follow from the factor of 2 in both area and length.

Final Answer:
The time is \(4t\), option (C). \[ \boxed{4t} \]