Answer:
Let die mercury is
depressed by x in left arm of V-tube so it will rise by \[x\] along the length of the tube in die right
arm of V-tube. The restoring force to the mercury is provided by the pressure
difference in the two arms.
\[\therefore
\] \[F=-(\Delta P)\,A\] ... (i)
Here\[{{P}_{1}}={{h}_{1}}\,\rho
\,g\,\sin {{\theta }_{1}}=(l-x)\sin \,{{\theta }_{1}},\]
\[\rho \,g\,\sin
{{\theta }_{1}}=(l-x)\,\rho \,g\,{{\sin }^{2}}\,{{\theta }_{1}}\]
\[=\frac{(l-\,x)\rho
g}{2}\] \[(\because
\,{{\theta }_{1}}-\,{{45}^{o}})\]
and
\[{{P}_{ & 2}}\,\,(l+x)\,\rho \,g\,{{\sin }^{2}}\,\theta _{2}^{2}\]
\[=\frac{(l+\,x)\,\rho
g}{2}\]
\[\therefore
\] \[\Delta
P=\,({{P}_{2}}-{{P}_{2}})\,=\,\frac{\rho g}{2}\,(l+\,x\,-l+x)\]
\[=\rho
\,gx\]
Hence,
restoring force, \[F=\,-\rho g\,A\,x\]
\[\therefore
\] Acceleration of
mercury column.
\[a=\frac{F}{m}\,=-\,\frac{-\rho
gAx}{m}\] ? (ii)
But,
\[m\,=\,(l\,A\,\rho )\]
\[\therefore
\] \[a=\,\frac{-\rho
g\,Ax}{l\,A\,\rho }=-\left( \frac{g}{l} \right)x\]
Hence,
motion of mercury column is S.H.M.
\[\therefore
\] \[T=2\pi
\,\sqrt{\frac{x}{a}}\,=2\pi \,\sqrt{\frac{l}{g}}\]
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