Uniform rod $A B, 12 m$ 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 $= 1000 k g / m^{3}$ ]

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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

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

The tension in the string is 40 N

answer is B, C.

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Detailed Solution

$T + U = W_{1} + W_{2} = 360 N$ ……… (i)
$U = \frac{V}{2} ρ_{w} g = (V / 2) (10^{3}) (10)$
Or $U = 0.5 \times 10^{4} V$ …….. (ii)
$(Σ M o m e n t s)$ 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)
$V = 6.4 \times 10^{- 2} m^{3}$