Answer:
Consider two points
A and B in liquid separated by a distance dx as shown in figure.
Let
P = presure at A
and
P + Pd = pressure at B
\ Force at A ? force at B = ma
or \[PA-(P+dP)A=ma,\]where
A is the area of cross-section of a cylinder AB or \[(-dP)A=ma\]
But \[m=(Adx)\rho
\]
\[\therefore
\] \[(-dP)A=Ad\times \rho a\]
or \[dP=-\rho
a\,dx\] ?. (i)
But \[dP=\rho
g({{h}_{2}}-{{h}_{1}})=-\rho g({{h}_{1}}-{{h}_{2}})\]
\[\therefore
\] eqn. (i),
\[g({{h}_{1}}-{{h}_{2}})=a\,dx\]
or
\[\frac{({{h}_{1}}-\,{{h}_{2}})}{dx}\,=\,\frac{a}{g}\]
Since
\[\frac{{{h}_{ & 1}}-{{h}_{2}}}{dx}\,=\,\tan \,\theta \] \[\therefore
\]\[\tan \,\theta =\,\frac{a}{g}\]
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