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........ $m$
Diffcult
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(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 $d = 0.98 m.$
= $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 $d = 0.98 m.$
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