A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 10^3kg/m^3. If outer diameter and the density of the bowl are

Q: A hemispherical bowl just floats without sinking in a liquid of density $1.2 × 10^3kg/m^3$. If outer diameter and the density of the bowl are $1 m$ and $2 × 10^4 kg/m^3$ respectively, then the inner diameter of the bowl will be........ $m$

c (c)Weight of the bowl $= mg$ = $V\rho g$$ = \frac{4}{3}\pi \left[ {{{\left( {\frac{D}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]\rho g$ where D = Outer diameter , d = Inner diameter $\rho $ = Density of bowl Weight of the liquid displaced by the bowl $ = V\sigma g$$ = \frac{4}{3}\pi {\left( {\frac{D}{2}} \right)^3}\sigma \,g$ where $\sigma $ is the density of the liquid. For the flotation $\frac{4}{3}\pi {\left( {\frac{D}{2}} \right)^3}\sigma g = \frac{4}{3}\pi \left[ {{{\left( {\frac{D}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]\rho g$ ==> ${\left( {\frac{1}{2}} \right)^3} \times 1.2 \times {10^3} = \left[ {{{\left( {\frac{1}{2}} \right)}^3} - {{\left( {\frac{d}{2}} \right)}^3}} \right]\,2 \times {10^4}$ By solving we get $d = 0.98 m.$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$0.94 $
B$0.97 $
C$0.98 $
D$0.99 $

Diffcult

STD 11 Science
NEET
STD 11 - 9.1. fluid mechanics
Physics

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