Question

A test tube weighing 10g and external dismeter 2.5cm is floated vertically in water by placing 20g of mercury at its bottom. The tube is depressed in water a little and then released. Find the time of oscillation. Take `g=10ms^(-2)`.

Answer 1

Total mass of the test tube and mercury,
`m=10+20=30g=.030kg`
Area of cross section of the test tube is
`A=pir^(2)=(22)/(7)xx((2.5)/(200))^(2)=4.91xx10^(-4)m^(2)`
Density of water, `rho=10^(3) kg//sm^(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 spring factor,
`k=(F)/(y)=(Ayrhog)/(y)=Arhog`
`=(4.91xx10^(-4))xx10^(3)xx10=4.91N//m`.
Here inertia factor, `m=.030kg`
Time period of oscialltion,
`T=2pisqrt((m)/(k))=2pisqrt((.030)/(4.91))=0.49s`