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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]
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The correct answer is:
$\frac{2 T}{\rho g d}$
Considering, contact angle is zero, and liquid loves the surface. The weight of the coin is balanced by surface tension.
$\begin{aligned} & F=T(2 \pi R)=\rho\left(d \pi R^2\right) g \\ & \Rightarrow R=\frac{2 T}{\rho g d}\end{aligned}$
$\begin{aligned} & F=T(2 \pi R)=\rho\left(d \pi R^2\right) g \\ & \Rightarrow R=\frac{2 T}{\rho g d}\end{aligned}$
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