Question 180: Oscillations

Question

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

Accepted Answer

$\mathrm{mg}^{\prime}=\mathrm{mg}-\mathrm{F}_{\mathrm{B}}$
$\mathrm{g}^{\prime}=\frac{\mathrm{mg}-\mathrm{F}_{\mathrm{B}}}{\mathrm{F}_{\mathrm{B}}}$
$=\frac{ho_{\mathrm{B}} \mathrm{Vg}-ho_{\mathrm{w}} \mathrm{Vg}}{ho_{\mathrm{B}} \mathrm{V}}$
$=\left(\frac{ho_{\mathrm{B}}-ho_{\mathrm{w}}}{ho_{\mathrm{B}}}ight) \mathrm{g}$
$=\frac{5-1}{5} 	imes \mathrm{g}$
$=\frac{4}{5} \mathrm{~g}$
We know, $T =2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}$
$\frac{\mathrm{T}^{\prime}}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}^{\prime}}}=\sqrt{\frac{\mathrm{g}}{5} \mathrm{~g}}=\sqrt{\frac{5}{4}}$
$\mathrm{~T}^{\prime}=\mathrm{T} \sqrt{\frac{5}{4}}=\frac{10}{2} \sqrt{5}$
$\mathrm{~T}^{\prime}=5 \sqrt{5}$

Oscillations

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

Integer / Numeric Answer Type