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Q.
A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D (D > > d)$ as shown in figure. If it is burning at the rate of $2$ cm/hour then the top of the candle will
Weight of candle is equal to weight of liquid displaced.
From Archemedes' principle when a body is immersed in a liquid completely or partly then there is an apparent loss in its weight. This apparent loss in weight is equal to the weight of the liquid displaced by the body.
Also, volume of candle $=$ Area $\times$ length
$=\pi\left(\frac{d}{2}\right)^{2} \times 2 L$
weight of candle $=$ weight of liquid displaced
$V \rho g =V' \rho' g$
$\Rightarrow \left(\pi \frac{d^{2}}{4} \times 2 L\right) \rho$
$=\left(\pi \frac{d^{2}}{4} \times L\right) \rho'$
$\Rightarrow \frac{\rho}{\rho'}=\frac{1}{2}$
Since, candle is burning at the rate of $2\, cm / h$, then after an hour, candle length is $2 L-2$
$\therefore (2 L-2) \rho=(L-x) \rho'$
$\therefore \frac{\rho}{\rho'}=\frac{L-x}{2(L-1)}$
$\Rightarrow \frac{1}{2}=\frac{L-x}{2(L-1)}$
$\Rightarrow x=1\, cm$
Hence, in one hour it melts $1\, cm$ and so it falls at the rate of $1\, cm / h$.
