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NEET · Physics · STD 11 - 10.2 transmission of heat
A deep rectangular pond of surface area \(A,\) containing water (denstity \(=\rho,\) specific heat capactly \(=s\) ), is located In a region where the outside air temperature is at a steady value of \(-26^{\circ} {C}\). The thickness of the frozen ice layer In this pond, at a certaln Instant Is \(x\). Taking the thermal conductivity of Ice as \({K}\), and its specific latent heat of fusion as \(L\), the rate of Increase of the thickness of ice layer, at this instant would be given by
- A \(26 \mathrm{K} / \mathrm{\rho r}(\mathrm{L}-4 \mathrm{s})\)
- B \(26 \mathrm{K} /\left(\rho \mathrm{x}^{2}-\mathrm{L}\right)\)
- C \(26 K /(\rho x L)\)
- D \(26 \mathrm{K} / \mathrm{\rho r}(\mathrm{L}+4 \mathrm{s})\)
Answer & Solution
Correct Answer
(C) \(26 K /(\rho x L)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{KA} \frac{[0-(-26)]}{\mathrm{x}} \mathrm{dt}=\mathrm{A}(\mathrm{dx}) \rho \mathrm{L} \) \(\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{26 \mathrm{K}}{\rho \mathrm{Lx}}\)
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