Question

A disc of paper of radius $R$ has a hole of radius $r$. It is floating on the surface of a liquid of surface tension '$T$'. The energy needed to pull it out carefully is:

• A. $T.2\pi (R+r)$
• B. $T.\pi (R^2-r^2)$
• C. $T.\pi (R^2+r^2)$
• D. $T.\pi r^2$

Answer 1

The correct option is B $T.\pi (R^2-r^2)$

Surface area change for liquid is $=\pi (R^2-r^2)$
$\therefore$ Surface energy $=U=\left(\Delta A\right)T$
$=T\pi (R^2-r^2)$

Thus Option B is correct.