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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Fluid Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Fluid Mechanics

A cylindrical object of outer diameter 10 cm , height 20 cm , and...

A cylindrical object of outer diameter 10 cm10 \text{ cm}, height 20 cm20 \text{ cm}, and density 8000 kg m−38000 \text{ kg m}^{-3} is held by a vertical spring of spring constant 500 N m−1500 \text{ N m}^{-1}. The object is initially in equilibrium, half dipped in water. If it is slightly depressed and released, calculate the time period of the resulting oscillations.

Options

  1. A1.86 s1.86 \text{ s}
  2. B0.75 s0.75 \text{ s}
  3. C1.25 s1.25 \text{ s}
  4. D0.93 s0.93 \text{ s}

Correct answer

D. 0.93 s0.93 \text{ s}

Step-by-step solution

Mass of the cylinder, m=πr2hρm = \pi r^2 h \rho

m=π(0.05)2(0.2)×8000=4π kgm = \pi (0.05)^2 (0.2) \times 8000 = 4\pi \text{ kg}

When the cylinder is depressed by a small distance xx, the restoring force is provided by the spring and the additional buoyant force.

F=−kx−(πr2x)ρwgF = -kx - (\pi r^2 x) \rho_w g

F=−(k+πr2ρwg)xF = -(k + \pi r^2 \rho_w g)x

Effective spring constant, keff=k+πr2ρwgk_{eff} = k + \pi r^2 \rho_w g

Using g=10 m s−2g = 10 \text{ m s}^{-2}:

keff=500+π(0.05)2(1000)(10)=500+25π N m−1k_{eff} = 500 + \pi (0.05)^2 (1000)(10) = 500 + 25\pi \text{ N m}^{-1}

Time period, T=2πmkeffT = 2\pi \sqrt{\dfrac{m}{k_{eff}}}

T=2π4π500+25π=4ππ500+25πT = 2\pi \sqrt{\dfrac{4\pi}{500 + 25\pi}} = 4\pi \sqrt{\dfrac{\pi}{500 + 25\pi}}

T≈12.56×3.14578.5≈12.56×0.0737≈0.93 sT \approx 12.56 \times \sqrt{\dfrac{3.14}{578.5}} \approx 12.56 \times 0.0737 \approx 0.93 \text{ s}

Answer: 0.93 s0.93 \text{ s}
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