By force balancing in vertical direction

\(S_F=2 T \sin \theta\)

\(S_F=2 T \theta\)   \(\left\{\begin{array}{l}\because \theta \text { is small } \\ \sin \theta=\theta\end{array}\right.\)

\(S \times 2 r \times 2 \theta=2 \times T \times \theta\)

\(S \times 2 r=\) Tension   

\(S \times d=\) Tension    \(\left\{\begin{array}{l}r \text {-radius } \\ d \text {-diameter }\end{array}\right.\)

\(\because  S=T\)

So, Tension \(=T d\)   \(\left\{\begin{array}{l}\text { Where, } \\ S_F=\text { Force due to surface tension } \\ T=\text { Tension in string } \\ \theta=\text { Small angle } \\ S=\text { Surface tension }\end{array}\right.\)