QJEEM24Phc076: JEE Main 2024 Physics - Class 11 JEEM24Phc076 Solution: A small liquid drop of radius is divided into 27 identical…

Volume constant

$\begin{aligned}& \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \times \pi r^3 \\& R^3=27 r^3 \\& R=3 r \\& r=\frac{R}{3} \\& r^2=\frac{R^2}{9}\end{aligned}$

$\begin{aligned}& \text { Work done }=T . \Delta A \\& =27 T\left(4 \pi r^2\right)-T 4 \pi R^2 \\& =27 T 4 \pi \frac{R^2}{9}-4 \pi R^2 T \\& =8 \pi R^2 T\end{aligned}$