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.
You need to login to perform this action.
You will be redirected in
3 sec