Question

The surface tension of a liquid is $5\text{N/m}$ . If a film of this liquid is held on a ring of area $0.02{\text{m}}^2$, it's surface energy is about

• A. $5\times {10}^{-2}\text{J}$
• B. $2.5\times {10}^{-2}\text{J}$
• C. $2\times {10}^{-1}\text{J}$
• D. $3\times {10}^{-1}\text{J}$

Answer 1

The correct option is C $2\times {10}^{-1}\text{J}$

We know that surface energy is given as,
$U_s=T\times$(Area of the film)
Area of ring $\left(A\right)=\pi r^2$
As the film has two free surfaces, we get,

$U_s=T\times 2\times \left(A\right)=5\times 2\times 0.02=0.2\text{J}$

Hence, $\left(C\right)$ is the correct answer.