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(D) The pressure at the bottom of the water disc due to its weight is given by $P = \rho g h$.This pressure is balanced by the excess pressure due to

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

(D) The pressure at the bottom of the water disc due to its weight is given by $P = ho g h$.This pressure is balanced by the excess pressure due to surface tension at the curved surface,given by the Young-Laplace equation: $P = T \left(\frac{1}{R_1} + \frac{1}{R_2}ight)$.Here,$R_1$ is the radius of the glass (which is very large compared to the thickness $h$) and $R_2$ is the radius of the semicircular edge,which is $h/2$.Since $R_1 \gg R_2$,we have $\frac{1}{R_1} \approx 0$.Thus,the pressure balance equation becomes $ho g h = T \left(0 + \frac{1}{h/2}ight) = \frac{2T}{h}$.Rearranging for $h$,we get $h^2 = \frac{2T}{ho g}$.Substituting the given values: $h = \sqrt{\frac{2 	imes 0.07}{10^3 	imes 10}} = \sqrt{\frac{0.14}{10^4}} = \sqrt{14 	imes 10^{-6}} \ m$.$h = \sqrt{14} 	imes 10^{-3} \ m \approx 3.741 	imes 10^{-3} \ m$.Converting to $mm$,$h \approx 3.741 \ mm$,which is closest to $3.75 \ mm$.

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