question_answer 21)

A tank with a square base of area \[F=mg+mr{{\omega }^{2}}=mg+mr4{{\pi }^{2}}{{v}^{2}}=\] is divided by a vertical partition in the middle. The bottom of the partition has a small hinged door of area \[=14\cdot 5\times 9\cdot 8+14\cdot 5\times 1\times 4\times {{\left( 22/7 \right)}^{2}}\times {{2}^{2}}\]The tank is filled with water and an acid (of relative density \[=142\cdot 1+2291\cdot 6=2433\cdot 7N\]) in the other, both to a height of 4.0 m. Compute the force necessary to keep the door closed.

Answer:

For compartment containing water, \[\vartriangle l=\frac{Fl}{AY}\] \[=\frac{2433\cdot 7\times 1}{\left( 0\cdot 065\times {{10}^{-4}} \right)\times \left( 2\times {{10}^{11}} \right)}\] The pressure exerted by water at the door provided at bottom, \[1\cdot 87\times {{10}^{-3}}m=1\cdot 87mm\] For compartment containing acid, \[=100\cdot 0\] \[=100\cdot 5\] The pressure exerted by acid at the door provided at bottom, \[=1\cdot 013\times {{10}^{5}}pa\].'. Difference of pressure \[V=100\] \[=100\times {{10}^{-3}}{{m}^{3}};p=100atm.=100\] Given, area of door, \[=100\times 1\cdot 013\times {{10}^{5}}pa.\] Force on the door = difference in pressure x area =\[V+\vartriangle V=100\cdot 5\]\[\vartriangle V=\left( V+\vartriangle V \right)-\]\[V=100\cdot 5\] \[-100=0\cdot 5\] To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.