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(B) Let $h$ be the total length of the candle and $h_2$ be the length of the candle submerged in the liquid. Let $\rho_w$ be the density of wax and $\

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fall at the rate of $2\ cm/hr$

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(B) Let $h$ be the total length of the candle and $h_2$ be the length of the candle submerged in the liquid. Let $ho_w$ be the density of wax and $ho_{\ell}$ be the density of the liquid.Given: $ho_{\ell} = 2ho_w$ and $\frac{dh}{dt} = 4\ cm/hr$.For a floating candle, the weight of the candle equals the buoyant force:$Weight = 	ext{Buoyant Force}$$A \cdot h \cdot ho_w \cdot g = A \cdot h_2 \cdot ho_{\ell} \cdot g$Since $ho_{\ell} = 2ho_w$, we have:$h \cdot ho_w = h_2 \cdot (2ho_w)$$h = 2h_2 \Rightarrow h_2 = \frac{h}{2}$Differentiating with respect to time $t$:$\frac{dh_2}{dt} = \frac{1}{2} \frac{dh}{dt} = \frac{1}{2} (4\ cm/hr) = 2\ cm/hr$.This $\frac{dh_2}{dt}$ represents the rate at which the bottom of the candle rises relative to the liquid surface.The rate of change of the position of the upper end relative to the liquid surface is given by the rate at which the total length decreases minus the rate at which the submerged part rises:$v_{upper} = \frac{dh}{dt} - \frac{dh_2}{dt} = 4\ cm/hr - 2\ cm/hr = 2\ cm/hr$.Thus, the upper end falls at the rate of $2\ cm/hr$.

$A$ wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of $4\ cm/hr$. Then, with respect to the surface of the liquid, the upper end of the candle will:

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