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(D) For the water not to leak through the hole, the pressure due to the height of the water column must be balanced by the capillary pressure (excess

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$0.2$

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(D) For the water not to leak through the hole, the pressure due to the height of the water column must be balanced by the capillary pressure (excess pressure) at the hole.The pressure due to the water column is $P = h ho g$.The excess pressure due to surface tension at the hole is $P_s = \frac{2T \cos 	heta}{r}$.Equating these, $h ho g = \frac{2T \cos 	heta}{r}$.Given: $h = 7 	ext{ cm} = 0.07 	ext{ m}$, $T = 0.07 	ext{ N/m}$, $	heta = 0^{\circ}$ (so $\cos 0^{\circ} = 1$), $ho = 1000 	ext{ kg/m}^3$, and $g = 10 	ext{ m/s}^2$.Rearranging for the radius $r$:$r = \frac{2T \cos 	heta}{h ho g}$$r = \frac{2 	imes 0.07 	imes 1}{0.07 	imes 1000 	imes 10}$$r = \frac{0.14}{700} = 0.0002 	ext{ m} = 0.2 	ext{ mm}$.

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

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