A candle is floating in a liquid placed in a container. The container is a cylinder of diameter $D$, and the candle is of width $d$. ($D>d$) The height of the liquid from the bottom of the container is $p$, and the height of the candle flame from the bottom is $h$. The density of the candle material is $p_c$ and that of the liquid is $p_l$.
If the length of the candle changes by $\Delta L$, find the change in the level of the liquid $\Delta p$, and the change in height of the flame $\Delta h$.
My attempt:
I'm trying to use Archimedes principle. Suppose $x$ is the height of the submerged candle when its length is $L$. Then, balancing gravitational and buoyant forces, $$p_c \pi \left(\frac{d}{2}\right)^2 L g = p_l \pi \left(\frac{d}{2}\right)^2 x g $$
So $$x = \frac{p_c}{p_l}L$$
Now suppose the length of the candle changes by $\Delta L$, causing the liquid height to change by $\Delta h$. I wrote: $$p_c \pi \left(\frac{d}{2}\right)^2 (L+\Delta L) g = p_l \pi \left(\frac{d}{2}\right)^2 (x+\Delta p) g$$
Which gives $$\Delta p = \frac{p_c}{p_l}\Delta L$$
Here I assumed that the change in liquid level would contribute to additional buoyant force. However, I'm not getting the right answer, which involves $D$ as well. So I'm not sure how to use Archimedes Principle for the required case. Any suggestions?
Best Answer
Well below is given my attempt assuming what i understood from your question. You are correct till $$x=p_c/p_l L$$ that means $$\Delta x=p_c/p_l\Delta L$$ After that we have to find a relation between $\Delta p$ and $\Delta x$
before the change and after the volume of the liquid remains constant.
$$\pi D^2p-\pi d^2x=\pi D^2(p-\Delta p)-\pi d^2(x-\Delta x)$$
solve this and you shall get $\Delta x=D^2/d^2 \Delta p$
That is what i think the answer should be based on what i understood in your question.