A hemispherical bowl just floats without sinking in a liquid of density `1.2xx10^(3)kg//m^(3)`. If outer diameter and the density of the bowl are `1m` and `2 xx 10^(4) kg//m^(3)` respectively, then the inner diameter of bowl will be

Options

A. `0.94 m`
B. `0.97 m`
C. `0.98 m`
D. `0.99 m`

Answer

C

Explanation

Weight of the bowl `=mg = Vrhog =(4)/(3) pi [((D)/(2))^(3)-((d)/(2))^(3)]rhog`
where D is the outer diameter, d is the inner diameter and `rho` is the density of bowl
Weight of the liquid by the bowl `= Vsigmag = (4)/(3) pi((D)/(2))^(3) sigma g`
where sigma is the density of the liquid
` (4)/(3)pi((D)/(2))^(3) sigmag = (4)/(3)pi [((D)/(2))^(3)-((d)/(2))^(3)]rhog implies ((1)/(2))^(3) xx 1.2 xx 10^(3) = [((1)/(2))^(3)-((d)/(2))^(3)] 2xx10^(4)`
By solving we get `d = 0.98 m`