Please solve q 167?

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Please solve q 167?

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Answer 1 · 2018-04-05T07:12:55

$0.98$ $0.98$ m

Explanation:

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

the weight of the bowl $= \frac{2}{3} π (R^{3} - r^{3}) ρ_{b} g$ $= 2/3 pi (R^3-r^3) rho_b g$
the buoyant force on the ball $= \frac{2}{3} π R^{3} ρ_{l} g$ $= 2/3 pi R^3 rho_l g$

Hence

$\frac{2}{3} π R^{3} ρ_{l} g = \frac{2}{3} π (R^{3} - r^{3}) ρ_{b} g \Rightarrow$ $2/3 pi R^3 rho_l g= 2/3 pi (R^3-r^3) rho_b g implies$

$1 - \frac{r^{3}}{R^{3}} = \frac{ρ_{l}}{ρ_{b}} = \frac{1.2 \times 10^{3}}{2 \times 10^{4}} = 0.06 \Rightarrow$ $1-r^3/R^3 = rho_l/rho_b = (1.2times 10^3)/(2times 10^4) = 0.06 implies$

${(\frac{r}{R})}^{3} = 0.94 \Rightarrow \frac{r}{R} = \sqrt[3]{0.94} \approx 0.98$ $(r/R)^3 = 0.94 implies r/R = root{3}{0.94}~~ 0.98$

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Please solve q 167?

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