- Home
- Standard 11
- Physics
A wax candle floats vertically in a liquid of density twice that of wax. The candle burns at the rate of $4\ cm/hr$ . Then, with respect to the surface of the liquid the upper end of the candle will
fall at the rate of $4\ cm/hr$
fall at the rate of $2\ cm/hr$
rise at the rate of $2\ cm/hr$
remain at the same height
Solution
$\frac{\mathrm{dh}}{\mathrm{dt}}=4 \mathrm{cm} / \mathrm{hr}$ given $\rho_{\ell}=2 \rho_{\mathrm{w}}$
$\mathrm{F}_{\mathrm{boyant}}=\rho_{\omega}(\mathrm{Ah}) \mathrm{g}=\left(\mathrm{Ah}_{2}\right) \rho_{\ell} \cdot \mathrm{g}$
$\rho_{\omega} \cdot \mathrm{Ahg}=\rho_{\ell} \cdot \mathrm{Ah}_{2} \mathrm{g}$
$\rho_{\omega} \cdot \mathrm{h}=\rho_{\ell} \cdot \mathrm{h}_{2}=2 \rho_{\omega} \cdot \mathrm{h}_{2}$
$\mathrm{h}=2 \mathrm{h}_{2}$
$\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{2 \mathrm{dh}_{2}}{\mathrm{dt}} \Rightarrow 4=\frac{2 \mathrm{dh}_{2}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dh}_{2}}{\mathrm{dt}}=2 \mathrm{cm} / \mathrm{hr}$
Therefore net fall of upper end $=\frac{\mathrm{dh}}{\mathrm{dt}}-\frac{\mathrm{dh}_{2}}{\mathrm{dt}}$
$=4 \mathrm{cm} / \mathrm{hr}-2 \mathrm{cm} / \mathrm{hr}$
$=2 \mathrm{cm} / \mathrm{hr}[\text { rate of fall of upper end of candle }]$
