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A test tube weighing 10 g and external diameter 2 cm is floated vertically in water by placing 10 g of mercury at its bottom. The tube is depressed in water a little and then released. Find the time of oscillation. Take g=10 ms^(-1)

Answer» <html><body><p></p>Solution :Total mass of test <a href="https://interviewquestions.tuteehub.com/tag/tube-1428592" style="font-weight:bold;" target="_blank" title="Click to know more about TUBE">TUBE</a> and mercutry, <br/> `m+10+10=20 g=0.02 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a>` <br/> Area of cross- section of the test tube tube <br/> `A= pi <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=(22)/(7)xx((1)/(100))^(2)` <br/> Density of water `rho=10kg m^(-3)` <br/> Let the tube be depresed in water by a little distance y and then released. <br/> Spring factor, <br/> `k=(F)/(y)=(A_(y. rho. g))/(y)=A rho g`<br/> `=(22)/(7)<a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a>^(-1)` <br/> Interia factor, `m=0.02 kg` <br/> `:.T = 2pi sqrt((m)/(k))` <br/> `=2xx(22)/(7)xx sqrt(0.02xx7)/(22)` <br/> `=0.5 `</body></html>


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