A tube of uniform bore of cross-sectional area ' $A$ ' has been set up vertically with open end facing up.…

A tube of uniform bore of cross-sectional area ' $A$ ' has been set up vertically with open end facing up. Now 'M' gram of a liquid of density ' $d$ ' is poured into it. The column of liquid in this tube will oscillate with a period ' T ', which is equal to $[\mathrm{g}=$ acceleration due to gravity]
  1. $2 \pi \sqrt{\frac{\mathrm{MA}}{\mathrm{gd}}}$
  2. $\quad 2 \pi \sqrt{\frac{\mathrm{M}}{2 \mathrm{Adg}}}$
  3. $\quad 2 \pi \sqrt{\frac{M}{g}}$
  4. $2 \pi \sqrt{\frac{M}{g d A}}$

Solution


For a depression of y cm on one side, the level of liquid will be 2 y cm higher on the other side. $\therefore \quad$ Weight of extra liquid on the P -side $=2 \mathrm{Aydg}$ The above weight acts as the restoring force acting on mass M . $\therefore \quad$ Restoring acceleration $=-\frac{2 \mathrm{Aydg}}{\mathrm{M}} \ldots(-\mathrm{ve} \cdot$ sign indicates the acceleration and force are opposite to each other) ... (i) We know for simple harmonic motion, $a=-\left(\frac{k}{m}\right) x$
Comparing (i) and (ii), we see the motion is simple harmonic. $\begin{aligned} \therefore \quad \mathrm{T} & =2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}=2 \pi \sqrt{\frac{\mathrm{y}}{\frac{2 \mathrm{Aydg}}{\mathrm{M}}}} \\ & =2 \pi \sqrt{\frac{\mathrm{M}}{2 \mathrm{Adg}}} \end{aligned}$ .

Asked in: MHT CET 2024 (03 May Shift 1)

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