A uniform rod $AB, 12 m$ long weighing $24 kg$ , is supported at end B by a flexible light string and a lead weight (of very small size) of $12 kg$ attached at end A.
The rod floats in water with one-half of its length submerged. For this situation, mark out the correct statement. [Take $g = 10 m / s^{2}$ , density of water $= 1000 kg / m^{3}$ ]

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The tension in the string is $12 g$

The volume of the rod is $6.4 \times 10^{- 2} m^{3}$

The point of application of the buoyancy force is passing through C (centre of mass of rod)

The tension in the string is $36 g$

answer is C.

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Detailed Solution

$T + U = w_{1} + w_{2} = 360 N U = \frac{V}{2} ρ_{n} g = (V / 2) (10^{3}) (10) U = 0.5 \times 10^{4} V (ΣMoments) about M = 0 ∴ 120 (\frac{l}{4}) + T (\frac{3 l}{4}) = 240 (\frac{l}{4}) or T = \frac{240 - 120}{3} = 40 N = 4 g Now from Eqs. (i) and (ii), we get V = 6.4 \times 10^{- 2} m^{3}$