A uniform rod AB, 4m long and weighing 12 kg, is supported at end A, with a 6 kg lead weight at B. The rod floats as shown in figure with one-half of its length submerged. The buoyant force on the lead mass is negligible as it is of negligible volume. Find the tension in the cord and the total volume of the rod

Options

A. The tension in the string is 36g
B. The tension in the string is 12g
C. the volume of the rod is `6.4xx10^(-2)m^(3)`
D. The point of application of the buoyancy force is passing through C (centre of mass of rod)

Answer

C

Explanation

`T+U=w_(1)+w_(2)=360N` …(i)
`U=V/2 rho_(w)g=(V//2)(10^3)(10)`
or, `U=0.5xx10^(4)V` …(ii)
`(Sigma` moments ) about `M =0`
`:. (120)(l/4)+T((3l)/(4))=240(l/4)`
or, `T=(240-120)/(3)`
`=40N=4g`
Now from Eqs. (i) and (ii) , we get
`V=6.4xx10^(-2)m^(3)`.