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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)`. |
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Answer» 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` |
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