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A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D\left( {D > > d} \right)$ as shown in figure. If it is burning at the rate of $2\ cm/hour$ then the top of the candle will
Remain at the same height
Fall at the rate of $1$ $cm/hour$
Fall at the rate of $2$ $cm/hour$
Go up the rate of $1$ $cm/hour$
Solution
Initial weight of the candle$-$weight of liquid displaced
$\rho_{C} V_{C} g=\rho_{L}$ (volume displaced) $g$
$\Rightarrow \rho_{C} \pi\left(\frac{d}{2}\right)^{2} 2 L=\rho_{L} \pi \frac{d^{2}}{2} L g \Rightarrow \frac{\rho c}{\rho_{L}}=\frac{1}{2} \ldots(i)$
When $2 \mathrm{cm}$ has been burnt, total length $=2 L-2$
But $\rho_{C}(2 L-2)=\rho_{L}(L-x)$
$\rho_{C} 2(L-1)=2 \rho_{C}(L-x)[$Using eqn.$(i)$]
$\therefore x=1 \mathrm{cm}$
Outside also it lias decreased $1 \mathrm{cm}$ as the total decrease is $2 \mathrm{cm} .$ The level of the candle comes down at half the rate of burning.
