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A hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$. If outer diameter and the density of the bowl are $1 \mathrm{~m}$ and $2 \times 10^4 \mathrm{~kg} / \mathrm{m}^{\mathrm{3}}$ respectively, then the inner diameter of the bowl will be
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The correct answer is:
$0.98 \mathrm{~m}$
Weight of the bowl $=m g$
$=V \rho g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
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
$\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \sigma g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
$\Rightarrow\left(\frac{1}{2}\right)^3 \times 1.2 \times 10^3=\left[\left(\frac{1}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] 2 \times 10^4$
By solving we get $d=0.98 \mathrm{~m}$.
$=V \rho g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
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
$\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \sigma g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
$\Rightarrow\left(\frac{1}{2}\right)^3 \times 1.2 \times 10^3=\left[\left(\frac{1}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] 2 \times 10^4$
By solving we get $d=0.98 \mathrm{~m}$.
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