The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10s. If the metallic bob is immersed in water, the new time period becomes 5 $\sqrt{x}$ s. The value of x will be ____.

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answer is 5.

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Detailed Solution

${mg}^{'} = mg - F_{B}$

$\begin{array}{l} g^{1} = \frac{mg - F_{B}}{m} = \frac{mg - m_{w} g}{m} = \frac{ρ_{B} Vg - ρ_{w} Vg}{ρ_{B} V} (∵ m = ρv) \\ = (\frac{ρ_{B} - ρ_{w}}{ρ_{B}}) g = \frac{5 - 1}{5} \times g = \frac{4}{5} g \end{array}$

$\frac{T^{'}}{T} = \sqrt{\frac{g}{g^{'}}} = \sqrt{\frac{g}{\frac{4 g}{5}}}$

$\Rightarrow T^{'} = T \sqrt{\frac{5}{4}} = \frac{10}{2} \sqrt{5} = 5 \sqrt{5}$

$Compare with 5 \sqrt{x}, we have x = 5$

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