question_answer 63)

                  One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will die column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces.

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.                 \[\therefore \] \[T=2\pi \,\sqrt{\frac{x}{a}}\,=2\pi \,\sqrt{\frac{l}{g}}\]