Two identical slabs are welded end to end and 20cal of heat flows through it for 4min. If the 2 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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we know that:
[tex]{ \red{ \boxed{ \tt \frac{dQ}{dT} = \frac{kAΔT}{L} }}}[/tex]
Given: Arrangement A is series, so
Q = 20 cal, t = 4 min,
Arrangement B is parallel, so
Q = 20 cal,
let time of flow be t
Since in both arrangements rate of conduction is same
For series
[tex]\tt \dfrac{dQ}{dT} = \dfrac{kA}{L} ( ΔT) = \dfrac{20}{4} ...(i)[/tex]
for parallel
[tex] \tt \dfrac{20}{t} = \dfrac{k×2A}{L} ΔT...(ii)[/tex]
dividing equation (i) and (ii)
[tex] \tt \dfrac{t}{4} \times \dfrac{1}{4} [/tex]
[tex]{ \boxed{ \bold{t = 1 \: min}}}[/tex]
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[tex]{ \orange{ \sf \: hope \: it \: helps \: you}}[/tex]