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?

  • $\begingroup$ Where have you calculated $\Delta h$ ? Assuming it to be $\Delta p$ by mistake then that is the equation where you got wrong. ### And if $\Delta p$ is change in corresponding $x$.Then you are not yet finished. You should proceed further in order to calculate $\Delta h$ $\endgroup$ – ABC Apr 11 '13 at 11:20
  • $\begingroup$ Well I thought I had gone wrong somewhere, so I thought I'd ask for help :) I've tried to calculate $\Delta h$ below now. $\endgroup$ – Comp_Warrior Apr 11 '13 at 17:53

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.

  • $\begingroup$ Thanks, I understand now. I'm trying to calculate $\Delta h$ now. I'm using this relationship for the length of candle which is unsubmerged: $$h-p=L-x$$ So $$\Delta h = \Delta p + \Delta x - \Delta x$$ This gives $$\Delta h = \Delta L \left( 1 - \frac{p_c}{p_l} \left(1 - \frac{d^2 p_c}{D^2 p_l}\right) \right)$$ However the answer does not have $\frac{p_c}{p_l}$ in the second bracket. Do you see where I have gone wrong? $\endgroup$ – Comp_Warrior Apr 11 '13 at 17:48
  • $\begingroup$ You were again right till $$\Delta h=\Delta p +\Delta L-\Delta x$$ Afeter that if you substitute the values of $\Delta p$ and $\Delta x$ then you will get $$d^2/D^2 p_c/p_l\Delta L+\Delta L-p_c/p_l\Delta L$$ which you rearrange to get the answer you gave except the density ratio in the final form.$$\Delta h=\Delta L(1-p_c/p_l(1+d^2/D^2))$$ $\endgroup$ – Satwik Pasani Apr 12 '13 at 2:07
  • $\begingroup$ Ah! Looks like a made a silly mistake in my haste. Thanks! $\endgroup$ – Comp_Warrior Apr 12 '13 at 11:17

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