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Two identical slabs are welded end to end and 20 cal of heat flows through it for 4 min. If the two slabs are now welded by placing them one above the other, and the same heat is flowing through two ends under the same difference of temperatures, the time taken is

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Detailed Solution

In series (end-to-end)

$K_{eff} = \frac{2 K \times K}{K + K} = K$
In parallel (place one above the other)

$K_{eff} = \frac{K + K}{2} = K$
$Q = \frac{KA (Δθ) t}{l}$ here Q, K, $Δθ$ are same in both cases.
$∴ \frac{t_{2}}{t_{1}} = \frac{l_{2}}{l_{1}} \times \frac{A_{1}}{A_{2}}$

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