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A tall tank filled with water has an irregular shape as shown. The wall $C D$ makes an angle of $45^{\circ}$ with the horizontal, the wall $A B$ is normal to the base $B C$. The lengths $A B$ and $C D$ are much smaller than the height $h$ of water (figure not to scale). Let $p_1, p_2$ and $p_3$ be the pressures exerted by the water on the wall $A B$, base $B C$ and the wall $C D$ respectively. Density of water is $\rho$ and $g$ is acceleration due to gravity. Then, approximately
$p_1=p_2=p_3$
$p_1=0, p_3=\frac{1}{\sqrt{2}} p_2$
$p_1=p_3=\frac{1}{\sqrt{2}} p_2$
$p_1=p_3=0, p_2=h \rho g$
Solution
(a)
Pressure of a fluid column depends only on height of fluid column and as pressure is scalar, its magnitude does not depend on orientation of surface over which pressure acts.
