A rod of mass `m = 2kg`, length `l = 1m`, has uniform cross-section area `A = (8)/(3)xx10^(-3)m^(2)` but its density is non-uniform. It is hinged about one end kept in water of density `rho_(w) = 10^(3)kg//m^(3)`. At the equilibirum, the rod becomes horizontal. Then `(g=10 m//s^(2))`. Choose the correct option(s) :

Options

A. Centre of mass of the rod is at a distance of `(2)/(3)m` from the hingeed end
B. Centre of mass of the rod is at a distance of `(3)/(4)m` from the hinged end
C. Force exerted by the rod on the hinge support is `(20)/(3)N`
D. It we displaced the rod slightly by retatinf downwards, it will oscillate

Answer

A, C

Explanation

`B = rholv_("sub") g = 10^(3) xx ((8)/(3) xx 10^(-3)) xx f xx 10`
`B = (80)/(3)N`

Balaning torque about the hinge point,
`((80)/(3)) xx (1)/(2) = (20)(x)`
`(A,B) , x = (2)/(3)m`
Applying horce balance,
`R + 20 = (80)/(3)`
(C) `R + 20 = (80)/(3)`
(D) If we displace the rod, still torquqe will remain balanced.