Two water drops each of radius ‘\( r \)’ coalesce to form a bigger drop. If ‘\( T \)’ is the surface tension, the surface energy released in this process is:
\( 4 \pi r^{2} T \left[ 2 - 2^{1/3} \right] \)
\( 4 \pi r^{2} T \left[ \sqrt{2} - 1 \right] \)
\( 4 \pi r^{2} T \left[ 1 + \sqrt{2} \right] \)
\( 4 \pi r^{2} T \left[ 2 - 2^{2/3} \right] \)