There is a hole in the bottom of tank having water. If total pressure at bottom is $ 3 atm$ ($1 atm = 10^5N/m^2$) then the velocity of water flowing from hole is
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(a) Pressure at the bottom of tank $P = h\rho g = 3 \times {10^5}\frac{N}{{{m^2}}}$ Pressure due to liquid column
$P_l = 3 \times {10^5} - 1 \times {10^5} = 2 \times {10^5}$
and velocity of water $v = \sqrt {2gh} $
$\therefore \;v = \sqrt {\frac{{2{P_l}}}{\rho }} = \sqrt {\frac{{2 \times 2 \times {{10}^5}}}{{{{10}^3}}}} = \sqrt {400} \,m/s$
$P_l = 3 \times {10^5} - 1 \times {10^5} = 2 \times {10^5}$
and velocity of water $v = \sqrt {2gh} $
$\therefore \;v = \sqrt {\frac{{2{P_l}}}{\rho }} = \sqrt {\frac{{2 \times 2 \times {{10}^5}}}{{{{10}^3}}}} = \sqrt {400} \,m/s$
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