Question

A metal rod AB of length $10x$ has its one end A in ice at $0^o$C and the other end B in water at ${100}^o$C. If a point P on the rod is maintained at ${400}^o$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 $540$ cal/g and latent heat of melting of ice is $80$ cal/g. 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.]

• A. 7
• B. 9
• C. 6
• D. 5

Answer 1

The correct option is B 9

Let $m$ gram of ice melt and water evaporate per unit time

$(m\times 80)={\left(\frac{\Delta \theta }{\Delta t}\right)}_{ice}$

$(m\times 540)={\left(\frac{\Delta \theta }{\Delta t}\right)}_{water}$

Now

$\frac{\Delta \theta }{\Delta t}=\frac{KA\left(\Delta T\right)}{x}$

$\therefore {\left(\frac{\Delta \theta }{\Delta t}\right)}_{ice}=\frac{KA({400}^o-0^{})}{\lambda x}=m\times 80......\left(i\right)$

${\left(\frac{\Delta \theta }{\Delta t}\right)}_{water}=\frac{KA({400}^o-{100}^o)}{(10-\lambda )x}=m\times 540.....\left(ii\right)$

divide $\left(i\right)$ from $\left(ii\right)$

$\frac{540}{80}=\frac{300}{400}\times \frac{\left(\lambda \right)}{(10-\lambda )}$

$\Rightarrow 90.9x=10\times \lambda$

$\Rightarrow \lambda =9m$

Hence Correct answer is $B$