A steel coin of thickness $d$ and density $\rho$ is floating on water of surface tension $T$. The radius of…
A steel coin of thickness $d$ and density $\rho$ is floating on water of surface tension $T$. The radius of the coin $R$ is [ $g$ = acceleration due to gravity]
$\frac{4 T}{3 \rho g d}$
$\frac{T}{\rho g d}$
$\frac{2 T}{\rho g d}$
$\frac{3 T}{4 \rho g d}$
Solution
Considering, contact angle is zero, and liquid loves the surface. The weight of the coin is balanced by surface tension.
$\begin{aligned} & F=T(2 \pi R)=\rho\left(d \pi R^2\right) g \\ & \Rightarrow R=\frac{2 T}{\rho g d}\end{aligned}$