A tank of height 15 m and cross-section area | vedclass

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(A) Let the pressure applied by the piston be $P$ and the atmospheric pressure be $P_a$. The excess pressure is $\Delta P = P - P_a$. The height of th

Vedclass · Accepted Answer

$233$

Vedclass · Answer

(A) Let the pressure applied by the piston be $P$ and the atmospheric pressure be $P_a$. The excess pressure is $\Delta P = P - P_a$. The height of the water column above the hole is $h = 15 \ m - 12 \ m = 3 \ m$. The height of the hole from the ground is $H = 12 \ m$.Using Bernoulli's equation at the top surface and the hole, the velocity of efflux $v$ is given by $v = \sqrt{2gh + \frac{2\Delta P}{ho}}$.The time taken for the water to reach the ground is $t = \sqrt{\frac{2H}{g}}$.The horizontal range $R$ is given by $R = v 	imes t = \sqrt{2gh + \frac{2\Delta P}{ho}} 	imes \sqrt{\frac{2H}{g}}$.Given $R = 16 \ m$, $H = 12 \ m$, $h = 3 \ m$, $ho = 1000 \ kg/m^3$, and $g = 10 \ m/s^2$:$16 = \sqrt{2(10)(3) + \frac{2\Delta P}{1000}} 	imes \sqrt{\frac{2(12)}{10}}$$16 = \sqrt{60 + \frac{\Delta P}{500}} 	imes \sqrt{2.4}$$256 = (60 + \frac{\Delta P}{500}) 	imes 2.4$$106.67 = 60 + \frac{\Delta P}{500}$$46.67 = \frac{\Delta P}{500} \Rightarrow \Delta P = 23335 \ Pa \approx 23.3 \ kPa$.Wait, re-calculating: $256 / 2.4 = 106.666...$, so $106.666 - 60 = 46.666$. $\Delta P = 46.666 	imes 500 = 23333 \ Pa$.Force $F = \Delta P 	imes A = 23333 	imes 10 = 233330 \ N = 233.3 \ kN \approx 233 \ kN$.

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