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Q. A bottle weighing $220 \,g$ and of area of cross-section $50\,cm^2$, and height $4\,cm$ oscillates on the surface of water in vertical position. Its frequency of oscillation is

JIPMEROscillations

Solution:

Let h be the depth of bottle in water then
$Ah\rho g = mg or h =\frac{m}{A\rho}=\frac{200}{50\times1}=4 cm.$
Now time period,
$T=2\pi\sqrt{\frac{h}{g}} or \upsilon=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{g}{h}}$
$\quad =\frac{7}{2\times22} \sqrt{\frac{980}{4}}=2.5$ Hz