6. At the bottom of a 15 cm long, uniformly thick cylindrical candle, a thin metal sheet (negligible volume) is fixed, allowing the candle to float vertically in water with 1 cm of its top end above the water surface. Given that the density of the candle is 0.85 g/cm³. Now, if the candle is lit, the total length burned when the candle is submerged is ( ) cm.
(A) $\frac{5}{3}$
(B) $\frac{10}{3}$
(C) $\frac{20}{3}$
(D) $\frac{40}{3}$
6.C.

Let the cross-sectional area of the candle be $\mathrm{S}$ square centimeters, and the weight of the metal plate be $a$ grams. According to the problem, we have $15 \times S \times 0.85 + a = 14 \times S \times 1$.

Suppose that after the 1 centimeter of the candle above the water is burned, the candle rises $x$ centimeters, then we have
$$
14 \times S \times 0.85 + a = (14 - x) S \times 1.
$$

Eliminating $a$ from the two equations, we get $x = 0.85$ centimeters.
Similarly, when the 0.85 centimeter above the water is burned, the candle rises another $0.85^2$ centimeters.
Continuing this process, the sum of the lengths of the burned segments of the candle is
$$
1 + 0.85 + 0.85^2 + \cdots = \frac{1}{1 - 0.85} = \frac{20}{3} \text{ (centimeters). }
$$