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A metal rod $\mathrm{AB}$ of length $10 x$ has its one end $\mathrm{A}$ in ice at $0^{\circ} \mathrm{C}$ and the other end $\mathrm{B}$ in water at $100^{\circ} \mathrm{C}$. If a point $\mathrm{P}$ 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 $540 \ \mathrm{cal} / \mathrm{g}$ and latent heat of melting of ice is $80 \ \mathrm{cal} / \mathrm{g}$. If the point $\mathrm{P}$ is at a distance of $\lambda x$ from the ice end $\mathrm{A}$, find the value of $\lambda$.
[Neglect any heat loss to the surrounding.|
$4$
$5$
$9$
$2$
Solution
The correct option is $B 9$
Let $m$ gram of ice melt and water evaporate per unit time
$( m \times 80)=\left(\frac{\Delta \theta}{\Delta t }\right)_{\text {ice }}$
$( m \times 540)=\left(\frac{\Delta \theta}{\Delta t }\right)_{\text {water }}$
Now
$\frac{\Delta \theta}{\Delta t }=\frac{ KA (\Delta T )}{ x }$
$\therefore\left(\frac{\Delta \theta}{\Delta t }\right)_{\text {ice }}=\frac{ KA \left(400^{\circ}- o \right)}{\lambda x }= m \times 80 \ldots \ldots \text { (i) }$
$\left(\frac{\Delta \theta}{\Delta t }\right)_{\text {water }}=\frac{ KA \left(400^{\circ}-100^{\circ}\right)}{(10-\lambda) x }= m \times 540 \ldots . .$
divide $(i)$ from $(ii)$
$\frac{540}{80}=\frac{300}{400} \times \frac{(\lambda)}{(10-\lambda)}$
$\Rightarrow 90.9 x =10 \times \lambda$
$\Rightarrow \lambda=9 \ m$
