Search any question & find its solution
Question:
Answered & Verified by Expert
Let ' $W_1$ ' be the work done in blowing a soap bubble of radius ' $r$ ' from soap solution at room temperature. The soap solution is now heated and second soap bubble of radius ' $2 \mathrm{r}$ ' is blown from the heated soap solution. If ' $\mathrm{W}_2$ ' is the work done in forming this bubble then
Options:
Solution:
2212 Upvotes
Verified Answer
The correct answer is:
$\mathrm{W}_2 < 4 \mathrm{~W}_1$
$$
\mathrm{W}_1=8 \pi \mathrm{r}^2 \mathrm{~T}
$$
Where, $\mathrm{T}=$ surface tension
$$
\mathrm{W}_2=8 \pi(2 \mathrm{r})^2 \mathrm{~T}^{\prime}=32 \pi \mathrm{r}^2 \mathrm{~T}^{\prime}
$$
Since soap solution is heated, its surface tension decrease.
$$
\begin{aligned}
& \therefore \mathrm{T}^{\prime} < \mathrm{T} \\
& \therefore \mathrm{W}_2 < 4 \mathrm{~W}_1
\end{aligned}
$$
\mathrm{W}_1=8 \pi \mathrm{r}^2 \mathrm{~T}
$$
Where, $\mathrm{T}=$ surface tension
$$
\mathrm{W}_2=8 \pi(2 \mathrm{r})^2 \mathrm{~T}^{\prime}=32 \pi \mathrm{r}^2 \mathrm{~T}^{\prime}
$$
Since soap solution is heated, its surface tension decrease.
$$
\begin{aligned}
& \therefore \mathrm{T}^{\prime} < \mathrm{T} \\
& \therefore \mathrm{W}_2 < 4 \mathrm{~W}_1
\end{aligned}
$$
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.