A uniform solid sphere of radius R is in equilibrium inside a liquid whose density varies with depth from free surface as `rho=rho_(0)(1+h/(h_(0)))` where h is depth from free surface. Density of sphere `sigma` will be :

Options

A. `sigma=rho_(0)(1+d/(2h_(0)))`
B. `sigma=rho_(0)(1-d/(2h_(0)))`
C. `sigma=rho_(0)(1+(2d)/(h_(0)))`
D. `sigma=rho_(0)(1+d/(h_(0)))`

Explanation

`dB=pi(R^(2)-y^(2))dy rho_(0)(1+(d-y)/(h_(0)))g`
`dB=(pirho_(0)g)/(h_(0))(R^(2)-y^(2))(h_(0)+d-y) dy`
`=(pirho_(0)g)/(h_(0))[R^(2) (h_(0)+d)dy-R^(2)ydy -(h_(0)+d)y^(2)dy+y^(3)dy]`
`B=int_(y=-R)^(+R) dB=(pirho_(0)g)/(h_(0)) (R^(2)(h_(0)+d)y-(R^(2)y^(2))/2-(h_(0)+d)(y^(3))/3+(y^(4))/4)_(-R)^(+R)`

`B=(pirho_(0)g)/(h_(0))[(h_(0)+d)R^(2)(2R)-((h_(0)+d))/3(2R^(3))]=(pirho_(0)g)/(h_(0))[4/3(h_(0)+d)R^(3)]`
`=4/3 piR^(3)g (rho_(0))/(h_(0))(h_(0)+d)=4/3 piR^(3) g sigma rArr sigma=(rho_(0))/(h_(0))(h_(0)+d)`
`sigma =rho_(0)(1+d/(h_(0)))`

Alternate solution
`sigmav g=int[rho_(0)(1+(d-x)/(h_(0)))dvg+rho_(0)(1+(d+x)/(h_(0)))dvg]`
`sigmav=2rho_(0)(1+d/(h_(0)))int_(0)^(v//2)dV=rho_(0)v(1+d/(v_(0)))`
`rArr sigma=rho_(0)(1+d/(h_(0)))`