MHT CET · Physics · Mechanical Properties of Fluids
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]
- A \(\frac{4 T}{3 \rho g d}\)
- B \(\frac{T}{\rho g d}\)
- C \(\frac{2 T}{\rho g d}\)
- D \(\frac{3 T}{4 \rho g d}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 T}{\rho g d}\)
Step-by-step Solution
Detailed explanation
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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