Question

Please solve q 167?

Answer 1

`0.98` m

Explanation:

Let us denote the density of the liquid and the material of the bowl by `rho_l` and `rho_b`, respectively. If the outer and inner radii of the hemispherical bowl are `R` and `r`, respectively, then (since the bowl is just floating)

• the weight of the bowl `= 2/3 pi (R^3-r^3) rho_b g`
• the buoyant force on the ball `= 2/3 pi R^3 rho_l g`

Hence

`2/3 pi R^3 rho_l g= 2/3 pi (R^3-r^3) rho_b g implies`

`1-r^3/R^3 = rho_l/rho_b = (1.2times 10^3)/(2times 10^4) = 0.06 implies`

`(r/R)^3 = 0.94 implies r/R = root{3}{0.94}~~ 0.98`