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Question: The \(9kg\) solution is poured into a glass U-tube as shown in the figure below. The tube’s inner di...

The 9kg9kg solution is poured into a glass U-tube as shown in the figure below. The tube’s inner diameter is 2π5m2\sqrt {\dfrac{\pi }{5}} m and the solution oscillates freely up and down about its position of equilibrium(x=0)\left( {x = 0} \right). The period of oscillation in seconds is: (1m31{m^3}of solution has a mass μ=900kg\mu = 900kg, g=10m/s2g = 10m/{s^2}. Ignore frictional and surface tension effects)

A. 0.10.1
B. 1010
C. π\sqrt \pi
D. 11

Explanation

Solution

Here, we are given a mass of a solution in the volume of 1m31{m^3}which can be taken as the density of the material. We can equate the mass poured in the solution with the density multiplied by the volume of the liquid which is oscillating. By this we can find the oscillation length which is further used to find the period of oscillation.

Formulas used:
V=πr2lV = \pi {r^2}l,
where, VV is the volume, rr is the inner radius of tube and ll is the length of oscillation
ρ=mV\rho = \dfrac{m}{V} ,
where, ρ\rho is the density of the liquid, m is the mass of liquid and VV is the volume of the liquid
T=2πl2gT = 2\pi \sqrt {\dfrac{l}{{2g}}} ,
Where, TT is the period of oscillation, ll is the length of oscillation and gg is the gravitational acceleration

Complete step by step answer:
We know that hh
\rho = \dfrac{m}{V} \\\ \Rightarrow m = \rho V \\\
We will put m=9kgm = 9kg, ρ=900kg/m3\rho = 900kg/{m^3}and V=πr2lV = \pi {r^2}l
9=900πr2l\Rightarrow 9 = 900\pi {r^2}l
We are given that r=2π5mr = 2\sqrt {\dfrac{\pi }{5}} m
9=900π2l5 l=5100π2 9 = \dfrac{{900{\pi ^2}l}}{5} \\\ \Rightarrow l = \dfrac{5}{{100{\pi ^2}}}
Now, to find the period of oscillation we will use the formula:
T=2πl2g T=2π5200π2×10 T=0.1s T = 2\pi \sqrt {\dfrac{l}{{2g}}} \\\ \Rightarrow T = 2\pi \sqrt {\dfrac{5}{{200{\pi ^2} \times 10}}} \\\ \therefore T = 0.1s

Thus, option A is the right answer.

Note: For solving this question, we have used the time period of oscillating fluid. Here we can clearly see that the time period of oscillating fluid is different from the time period of simple oscillation. To be precise, it is 12\dfrac{1}{{\sqrt 2 }}times the normal oscillation such as a simple pendulum.