9 text{ kg} of solution is poured into a glas | vedclass

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(A) The mass of the liquid column is $M = 9 	ext{ kg}$.The density of the liquid is $ho = 900 	ext{ kg/m}^3$.The inner diameter of the tube is $D

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(A) The mass of the liquid column is $M = 9 	ext{ kg}$.The density of the liquid is $ho = 900 	ext{ kg/m}^3$.The inner diameter of the tube is $D = 2 \sqrt{\frac{\pi}{5}} 	ext{ m}$, so the radius is $r = \sqrt{\frac{\pi}{5}} 	ext{ m}$.The cross-sectional area of the tube is $A = \pi r^2 = \pi \left(\sqrt{\frac{\pi}{5}}ight)^2 = \frac{\pi^2}{5} 	ext{ m}^2$.The total length $L$ of the liquid column is given by $M = ho A L$.Substituting the values: $9 = 900 	imes \frac{\pi^2}{5} 	imes L$.$9 = 180 \pi^2 L \implies L = \frac{9}{180 \pi^2} = \frac{1}{20 \pi^2} 	ext{ m}$.For a $U$-tube, the time period of oscillation is $T = 2 \pi \sqrt{\frac{L}{2g}}$.Substituting $L = \frac{1}{20 \pi^2}$ and $g = 10 	ext{ m/s}^2$:$T = 2 \pi \sqrt{\frac{1 / (20 \pi^2)}{2 	imes 10}} = 2 \pi \sqrt{\frac{1}{400 \pi^2}} = 2 \pi 	imes \frac{1}{20 \pi} = \frac{2}{20} = 0.1 	ext{ s}$.

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