A steel coin of thickness d and density rho i | vedclass

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(C) For the coin to float,the downward force due to gravity (weight) must be balanced by the upward force due to surface tension acting along the circ

Vedclass · Accepted Answer

$\frac{2 T}{\rho g d}$

Vedclass · Answer

(C) For the coin to float,the downward force due to gravity (weight) must be balanced by the upward force due to surface tension acting along the circumference of the coin.Assuming the contact angle is $0^{\circ}$ and the coin is thin,the upward force is $F = T 	imes (2 \pi R)$.The weight of the coin is $W = 	ext{mass} 	imes g = (	ext{density} 	imes 	ext{volume}) 	imes g = ho 	imes (\pi R^2 d) 	imes g$.Equating the two forces:$T(2 \pi R) = ho (\pi R^2 d) g$Dividing both sides by $\pi R$:$2 T = ho R d g$Solving for $R$:$R = \frac{2 T}{ho g d}$

$A$ 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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