Search any question & find its solution
Question:
Answered & Verified by Expert
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 $2 \mathrm{~cm} /$ hour then the top of the candle will
Options:
Solution:
2742 Upvotes
Verified Answer
The correct answer is:
fall at the rate of $1 \mathrm{~cm} /$ hour
Initial weight of the candle = weight of liquid displaced
$$
\rho_C V_C g=\rho_L \text { (volume displaced) } g
$$
$$
\Rightarrow \rho_C \pi\left(\frac{d}{2}\right)^2 2 L g=\rho_L \pi \frac{d^2}{2} L g \Rightarrow \frac{\rho_C}{\rho_L}=\frac{1}{2}
$$
When $2 \mathrm{~cm}$ has been burnt, total length $=2 L-2$ But $\rho_C(2 L-2)=\rho_L(L-x)$
$$
\begin{aligned}
& \rho_C 2(L-1)=2 \rho_C(L-x) \\
\therefore \quad & x=1 \mathrm{~cm} .
\end{aligned}
$$
[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.
$$
\rho_C V_C g=\rho_L \text { (volume displaced) } g
$$
$$
\Rightarrow \rho_C \pi\left(\frac{d}{2}\right)^2 2 L g=\rho_L \pi \frac{d^2}{2} L g \Rightarrow \frac{\rho_C}{\rho_L}=\frac{1}{2}
$$
When $2 \mathrm{~cm}$ has been burnt, total length $=2 L-2$ But $\rho_C(2 L-2)=\rho_L(L-x)$
$$
\begin{aligned}
& \rho_C 2(L-1)=2 \rho_C(L-x) \\
\therefore \quad & x=1 \mathrm{~cm} .
\end{aligned}
$$
[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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.