Fluids with Rotational Equllibrium...

A uniform rod A B AB , 4 m 4 m long and weighing 12 k g 12 kg , is supported at end A A (with a string), and a 6 k g 6 kg lead weight at B B (Not shown in the figure). The rod floats as shown in the figure with one-half of its length submerged. The buoyant force on the lead mass is negligible as it is of negligible volume. If Tension in string is T T (in N e w t o n Newton ) and total volume of rod is V V (in m 3 m^3 ).

Find T + V T+V (Numerical values)

D e t a i l s A n d A s s u m p t i o n s Details And Assumptions

Take g = 10 m / s 2 g = 10 m/s^2 Density of liquid = 2 k g / m 3 2 kg/m^3

36 10 24 38

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