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JEE Advanced · Physics · 12. Thermal Properties

A metal rod \(A B\) of length \(10 x\) has its one end \(A\) in ice at \(0^{\circ} \mathrm{C}\) and the other end \(B\) in water at \(100^{\circ} \mathrm{C}\). If a point \(P\) on the rod is maintained at \(400^{\circ} \mathrm{C}\), then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is \(540 \mathrm{calg}^{-1}\) and latent heat of melting of ice is \(80 \mathrm{calg}^{-1}\). If the point \(P\) is at a distance of \(\lambda x\) from the ice end \(A\), find the value of \(\lambda\). [Neglect any heat loss to the surrounding.]

  1. A 3
  2. B 6
  3. C 9
  4. D 12
Verified Solution

Answer & Solution

Correct Answer

(C) 9

Step-by-step Solution

Detailed explanation


Heat will flow both sides from point \(P\).
\(
\begin{aligned}
L_1 \frac{d m_1}{d t} & =\left(\frac{\text { Temperature difference }}{\text { Thermal resistance }}\right)_1 \\
& =\frac{400}{(\lambda x) / k A} \\
L_1 \frac{d m_2}{d t} & =\frac{400-100}{(100-\lambda) x / k A}
\end{aligned}
\)
In about two equations,
\(
\begin{aligned}
& \frac{d m_1}{d t}=\frac{d m_2}{d t} \quad \text { (given) } \\
& L_1=80 \mathrm{calg}^{-1} \text { and } L_2=540 \mathrm{calg}^{-1}
\end{aligned}
\)
Solving these two equations we get \(\lambda=9\).
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