A candle of diameter d is floating on a liqui | vedclass

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(B) Let the density of the candle be $\rho_C$ and the density of the liquid be $\rho_L$. Let the total length of the candle be $2L$. Initially,the can

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Fall at the rate of $1 \ cm/hour$

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(B) Let the density of the candle be $\rho_C$ and the density of the liquid be $\rho_L$. Let the total length of the candle be $2L$. Initially,the candle floats with length $L$ submerged.By the principle of flotation,the weight of the candle equals the weight of the displaced liquid:$\rho_C \cdot A \cdot (2L) = \rho_L \cdot A \cdot L$$\Rightarrow \frac{\rho_C}{\rho_L} = \frac{1}{2} \dots (i)$When the candle burns by a length $\Delta h = 2 \ cm$,let its new total length be $2L - 2$ and the new submerged length be $L'$.By the principle of flotation: $\rho_C \cdot A \cdot (2L - 2) = \rho_L \cdot A \cdot L'$Using $(i)$,we get: $\frac{1}{2} \rho_L \cdot A \cdot (2L - 2) = \rho_L \cdot A \cdot L'$$L - 1 = L'$The change in submerged length is $\Delta L' = L - L' = 1 \ cm$.The top of the candle is at a height $h_{top} = L' + (2L - 2 - L') = 2L - 2$ above the bottom of the candle. However,relative to the liquid surface,the top is at height $H = 2L - 2 - L' = L - 1$.Since $L$ is constant,the change in height of the top relative to the liquid surface is $\Delta H = 1 \ cm$.Thus,the top of the candle falls at a rate of $1 \ cm/hour$.

$A$ candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D$ $(D >> d)$ as shown in the figure. If it is burning at the rate of $2 \ cm/hour$,then the top of the candle will

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