Question

The external diameter of a test tube weighing 5g is 2cm. 15g of mercury is poured into the tube so that it floats vertically in water. If the tube is depressed vertically through a small distance and then released, determine the period of its oscillations.

Answer 1

Hint:Here when the mercury is poured in the test tube moves vertically inwards and because of that it sets into oscillations. Initially before the mercury was poured in, the test tube was in equilibrium. We need to find out all the forces acting on it. After it gets displaced it sets into simple harmonic motion. We can make use of the formulas to find out the same.

Complete step by step answer:
Total mass= $5+15=20g$.
Given diameter is $2cm$, so radius will be 1 cm= 0.01m.
So, area of the test tube will be $\pi {{r}^{2}}=3.14\times {{0.01}^{2}}=3.14\times {{10}^{-4}}{{m}^{2}}$.
We know the density of water is ${{10}^{3}}kg/{{m}^{3}}$.When the tube is floating, let the tube be depressed in water by a little distance y, and released, it will execute linear SHM with initial floating position as equilibrium position or mean position. Then the spring factor will be given as:
$k=\dfrac{F}{y}=\dfrac{Ay\rho g}{y}=A\rho g$
where $\rho $ is the density of the water
 Putting the value, we get,
$k=3.14\times {{10}^{-4}}\times 1000\times 10$
$\Rightarrow k =3.14N/m$
Also, the mass is $20g=\dfrac{20}{1000}kg=0.02kg$
So, time period can be given using the formula $T=2\pi \sqrt{\dfrac{m}{k}}$
$T =2\pi \sqrt{\dfrac{0.02}{3.14}} \\
\therefore T =0.49s $

So, the time period of oscillation is 0.49 s.

Note:We have made use of an analogy to arrive at the solution of the problem by comparing it to a spring problem. This problem could also have been solved by using Archimedes principle and finding the upthrust acting on the body and weight of the liquid displaced when the mercury was poured into the test tube but that would have been very lengthy.