In a thin vertical U-tube of very long arms is some mercury. Period of small oscillations of mercury is 2 s, if in one arm , 100grams of water is poured, the period of small oscillations becomes 3 s, neglect surface tension then mass of the mercury is 16P grams. Then P value is

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answer is 5.

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Detailed Solution

Since cross section is uniform, mass of both the liquids must be proportional to length occupied by them now $m_{H g} = \frac{m_{H_{2} O} \times T_{H g}^{2}}{T_{H g + H_{2} O}^{2} - T_{H g}^{2}} = 80 g = 16 P \Rightarrow P = 5$

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