Question

A test tube of mass \(6 \, {g}\) and uniform area of cross section \(10 \, {cm}^2\) 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: (Acceleration due to gravity = \(10 \, {m/s}^2\))

• A. \(0.75 \, {s}\)
• B. \(0.5 \, {s}\)
• C. \(0.25 \, {s}\)
• D. \(0.85 \, {s}\)

Hint:

The time period of oscillation in simple harmonic motion depends inversely on the square root of the spring constant and directly on the square root of the mass. This relationship helps in understanding how changes in mass or the restoring force affect oscillatory motion.

Answer 1

The Correct Option is C

To find the time period of oscillation for the floating test tube, we consider it as a simple harmonic oscillator. The restoring force is provided by the buoyancy which depends on the volume of water displaced. The effective spring constant \( k \) for the system can be related to the density of water, the cross-sectional area of the tube, and the acceleration due to gravity. Given that the cross-sectional area \( A = 10 \, {cm}^2 = 0.001 \, {m}^2 \) and the gravitational force \( g = 10 \, {m/s}^2 \), the spring constant can be estimated by considering the weight of the displaced water when the tube is depressed slightly. \[ k = \rho g A \] Where \( \rho \) is the density of water (approximately \(1000 \, {kg/m}^3\)). Substituting the known values: \[ k = 1000 \times 10 \times 0.001 = 10 \, {N/m} \] The mass \( m \) of the tube including the mercury is \(6 \, {g} + 10 \, {g} = 16 \, {g} = 0.016 \, {kg}\). The time period \( T \) of oscillation for a simple harmonic oscillator is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] \[ T = 2\pi \sqrt{\frac{0.016}{10}} \approx 0.25 \, {s} \]