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(D) The range of water issuing from a hole at depth $h$ is given by $R = 2\sqrt{h(H-h)}$,where $H$ is the total height of the water column. Assuming t

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$3$

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(D) The range of water issuing from a hole at depth $h$ is given by $R = 2\sqrt{h(H-h)}$,where $H$ is the total height of the water column. Assuming the hole is at the bottom $(h=H)$,the range is $R = 2\sqrt{h^2} = 2h$. For a hole at depth $h=10 \ m$,the velocity of efflux is $v = \sqrt{2gh}$. The range is $R = v \cdot t$,where $t = \sqrt{2(H-h)/g}$. If we apply an extra pressure $P_{ext}$ on the surface,the effective pressure at the hole becomes $P_{eff} = P_{atm} + P_{ext} + ho gh$. The velocity of efflux becomes $v' = \sqrt{2(P_{ext} + ho gh)/ho} = \sqrt{2gh + 2P_{ext}/ho}$. To make the range $2R$,the velocity of efflux must become $2v$ (since time of flight remains constant if the hole position is fixed). Thus,$v' = 2v \implies v'^2 = 4v^2$. $2gh + 2P_{ext}/ho = 4(2gh) = 8gh$. $2P_{ext}/ho = 6gh \implies P_{ext} = 3ho gh$. Given $ho = 10^3 \ kg/m^3$,$g = 10 \ m/s^2$,and $h = 10 \ m$: $P_{ext} = 3 	imes 10^3 	imes 10 	imes 10 = 3 	imes 10^5 \ Pa$. Since $1 \ atm = 10^5 \ Pa$,the extra pressure required is $3 \ atm$.

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