Physics - Fluid Question with Solution | TestHub

Initial weight of the candle = weight of liquid displaced $${\rho }_C{V}_{C}g={\rho }_L\text{ (volume displaced) }g$$ $$\Rightarrow {\rho }_C\pi {\left(\frac{d}{2}\right)}^{2}2Lg={\rho }_L\pi \frac{d^2}{2}Lg\Rightarrow \frac{{\rho }_C}{{\rho }_L}=\frac{1}{2}$$ When $2\mathrm{cm}$ has been burnt, total length $=2L-2$ But ${\rho }_C(2L-2)={\rho }_L(L-x)$ $$\begin{array}{rl}& {\rho }_{C}2(L-1)=2{\rho }_C(L-x)\\ \therefore & x=1\mathrm{cm}.\end{array}$$ [Using eqn. (i)] Outside also it has 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.