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 $L_0$ and the submerged

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Fall at the rate of $1 	ext{ cm/hour}$

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(B) Let the density of the candle be $ho_c$ and the density of the liquid be $ho_l$. Let the total length of the candle be $L_0$ and the submerged length be $L_s$. According to the principle of floatation,the weight of the candle equals the weight of the displaced liquid:$ho_c \cdot A \cdot L_0 = ho_l \cdot A \cdot L_s$where $A$ is the cross-sectional area of the candle.This implies $L_s = (ho_c / ho_l) L_0$. Let $k = ho_c / ho_l$. Then $L_s = k L_0$.The height of the top of the candle above the liquid surface is $h = L_0 - L_s = L_0(1 - k)$.As the candle burns,its total length $L_0$ decreases at a rate $v = dL_0/dt = 2 	ext{ cm/hour}$.The rate of change of the height of the top of the candle is $dh/dt = (dL_0/dt)(1 - k)$.Assuming the candle is half-submerged (as suggested by the figure),$k = 0.5$.Therefore,$dh/dt = 2 	ext{ cm/hour} 	imes (1 - 0.5) = 1 	ext{ cm/hour}$.Since $dh/dt$ is positive,the top of the candle falls at a rate of $1 	ext{ cm/hour}$.

$A$ 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 $2 \text{ cm/hour}$,then the top of the candle will:

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