Ice starts forming in a lake with water at 0^ | vedclass

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(D) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K 	heta}{ho L x}$,where $K$ is thermal conductivity,$ho$ is

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More than $7$ hours

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(D) The rate of formation of ice of thickness $x$ is given by $\frac{dx}{dt} = \frac{K 	heta}{ho L x}$,where $K$ is thermal conductivity,$ho$ is density,$L$ is latent heat,and $	heta$ is the temperature difference.Integrating this,the time $t$ taken to grow from thickness $x_1$ to $x_2$ is $t = \frac{ho L}{2K	heta} (x_2^2 - x_1^2)$.This implies $t \propto (x_2^2 - x_1^2)$.For the first $1 \; cm$ ($0$ to $1 \; cm$): $7 \propto (1^2 - 0^2) = 1$.For the next $1 \; cm$ ($1 \; cm$ to $2 \; cm$): $t' \propto (2^2 - 1^2) = 4 - 1 = 3$.Comparing the two,$\frac{t'}{7} = \frac{3}{1}$,which gives $t' = 21$ hours.Since $21 > 7$,the time taken is more than $7$ hours.

Ice starts forming in a lake with water at $0^{\circ}C$ when the atmospheric temperature is $-10^{\circ}C$. If the time taken for $1 \; cm$ of ice to form is $7$ hours,then the time taken for the thickness of ice to change from $1 \; cm$ to $2 \; cm$ is:

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