A test tube of mass 6 g and uniform area of cross section $10 \mathrm{~cm}^2$ is floating in water vertically when 10 g of mercury is in the bottom. The tube is depressed by a small amount and then released. The time period of oscillation is $\left(\right.$ Acceleration due to gravity $=10…

Question

A test tube of mass 6 g and uniform area of cross section $10 \mathrm{~cm}^2$ is floating in water vertically when 10 g of mercury is in the bottom. The tube is depressed by a small amount and then released. The time period of oscillation is $\left(\right.$ Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{m}=6+10=16 \mathrm{~g}=0.016 \mathrm{~kg} \ & \mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}}=\frac{\delta \mathrm{Axg}}{\mathrm{x}}=\delta^{\mathrm{A}} \mathrm{~g}=10^3 	imes 10 	imes 10^{-4} 	imes 10 \ & =10 \mathrm{Nm}^{-1} \end{aligned}$ $	herefore \quad$ Time period, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{k}}}=2 \pi \sqrt{\frac{0.016}{10}}=\frac{2 \pi 	imes 4}{100}$

A test tube of mass 6 g and uniform area of cross section $10 \mathrm{~cm}^2$ is floating in water…

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