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9 kg solution is poured into a glass U-tube as shown in the figure below. The tube's inner diameter is 2 sqrt((pi)/(5)) m and the solution oscillates freely up and down about its positon of equilibrium (x - 0). The period of oscillation in seconds is (1 m^(3) of solution has a mass mu = 900 kg, g = 10 m // s^(2) ignor friction and surface tension effects |
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Answer» 0 `:.` RADIUS of tube (r ) `= sqrt((pi)/(5)) m` mass of solution (U) =9 kg we know that, `:'` mass (u) = d.v `:.U = d pi r^(2) L` or ,K `9 xx 900 xx pi xx (sqrt((pi)/(5)))^(2) L` or `9 = 900 xx (pi^(2))/(5) xx L` or, `L = (5)/(100 pi^(2))` Now, `T = 2 pi sqrt((L)/(2g))` Form eq (i) `T = 2 pi sqrt((5)/(200 pi^(2) xx 10))` or, `T = 2 xx sqrt((1)/(400)) = (2)/(20)` or T = 0.1 s |
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