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(D) The pressure at the base of the cylinder is the sum of the atmospheric pressure $(P_0)$ and the pressure exerted by the weight of the cylinder $(P

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$30$

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(D) The pressure at the base of the cylinder is the sum of the atmospheric pressure $(P_0)$ and the pressure exerted by the weight of the cylinder $(P_{solid} = h ho g)$.Given that the solid breaks when the total pressure exceeds $P_{max} = 13 \ atm$,we have:$P_{total} = P_0 + h ho g \leq P_{max}$$h ho g \leq P_{max} - P_0$$h \leq \frac{P_{max} - P_0}{ho g}$Here,$P_{max} = 13 \ atm = 13 	imes 10^5 \ Pa$,$P_0 = 1 \ atm = 1 	imes 10^5 \ Pa$,and the density of the solid $ho = 4 	imes 10^3 \ kg/m^3$ (since specific gravity is $4$).$h = \frac{(13 - 1) 	imes 10^5}{4 	imes 10^3 	imes 10} = \frac{12 	imes 10^5}{4 	imes 10^4} = 30 \ m$.

If a solid breaks under a pressure greater than $13 \ atm$ and that solid has a specific gravity of $4$,what is the maximum height of a cylinder made from the solid that can be built at the Earth's surface (in $m$)? (Note: $1 \ atm = 10^5 \ Pa$ and take $g = 10 \ m/s^2$.)

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