Q104 / 180 · AIIMS 2014 · Physics

A bottle weighing $220 \mathrm{~g}$ and of area of crosssection $50 \mathrm{~cm}^2$, and height $4 \mathrm{~cm}$ oscillates on the surface of water in vertical position. Its frequency of oscillation is

  • A
    $1.5 \mathrm{~Hz}$
  • Correct answer
    B
    $2.5 \mathrm{~Hz}$
  • C
    $3.5 \mathrm{~Hz}$
  • D
    $4.5 \mathrm{~Hz}$

Solution

Let h be the depth of bottle in water then $\mathrm{A} h \rho g=m g$ or $h=\frac{m}{A \rho}=\frac{200}{50 \times 1}=4 \mathrm{~cm}$. Now time period, $T=2 \pi \sqrt{\frac{h}{g}}$ or $v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{g}{h}}$ $=\frac{7}{2 \times 22} \sqrt{\frac{980}{4}}=2.5 \mathrm{~Hz}$

Hence, the correct answer is option: 'B'.

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