AP EAMCET · PHYSICS · Oscillations
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 \(=10 \mathrm{~ms}^{-2}\) )
- A 0.75 s
- B 0.5 s
- C 0.25 s
- D 0.85 s
Answer & Solution
Correct Answer
(C) 0.25 s
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{m}=6+10=16 \mathrm{~g}=0.016 \mathrm{~kg} \\ & \mathrm{k}=\frac{\mathrm{F}}{\mathrm{x}}=\frac{\delta \mathrm{Axg}}{\mathrm{x}}=\delta^{\mathrm{A}} \mathrm{~g}=10^3 \times 10 \times 10^{-4} \times 10 \\ & =10 \mathrm{Nm}^{-1} \end{aligned}\)…
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