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 hinged 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 displace the rod slightly by rotating downwards, it will oscillate

Answer

A, C

Explanation

A rof of ………….
`B = rho_(l)V_(sub)g=10^(3)xx((8)/(3)xx10^(-3))xx1xx10`
`B = (80)/(3)N`

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