JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 27)
A spherical drop of liquid splits into 1000 identical spherical drops. If u$$_\mathrm{i}$$ is the surface energy of the original drop and u$$_\mathrm{f}$$ is the total surface energy of the resulting drops, the (ignoring evaporation), $${{{u_f}} \over {{u_i}}} = \left( {{{10} \over x}} \right)$$. Then value of x is ____________ :
Answer
1
Explanation
Surface Tension $=\mathrm{T}$
$\mathrm{R}$ : Radius of bigger drop
$\mathrm{r}$ : Radius of smaller drop
Volume will remain same
$\frac{4}{3} \pi R^3=1000 \times \frac{4}{3} \pi r^3$
$\mathrm{R}=10 \mathrm{r}$
$\mathrm{u}_{\mathrm{i}}=\mathrm{T} \cdot 4 \pi \mathrm{R}^2$
$\mathrm{u}_{\mathrm{f}}=\mathrm{T} .4 \pi \mathrm{r}^2 \times 1000$
$\frac{\mathrm{u}_{\mathrm{f}}}{\mathrm{u}_{\mathrm{i}}}=\frac{1000 \mathrm{r}^2}{\mathrm{R}^2}$
$\frac{\mathrm{u}_{\mathrm{f}}}{\mathrm{u}_{\mathrm{i}}}=\frac{10}{1}$
So, $x=1$
$\mathrm{R}$ : Radius of bigger drop
$\mathrm{r}$ : Radius of smaller drop
Volume will remain same
$\frac{4}{3} \pi R^3=1000 \times \frac{4}{3} \pi r^3$
$\mathrm{R}=10 \mathrm{r}$
$\mathrm{u}_{\mathrm{i}}=\mathrm{T} \cdot 4 \pi \mathrm{R}^2$
$\mathrm{u}_{\mathrm{f}}=\mathrm{T} .4 \pi \mathrm{r}^2 \times 1000$
$\frac{\mathrm{u}_{\mathrm{f}}}{\mathrm{u}_{\mathrm{i}}}=\frac{1000 \mathrm{r}^2}{\mathrm{R}^2}$
$\frac{\mathrm{u}_{\mathrm{f}}}{\mathrm{u}_{\mathrm{i}}}=\frac{10}{1}$
So, $x=1$
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