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  • A metallic sphere floats in an immiscible mixture of water <img alt="\mathrm{\left(\rho_w=10^3 \mathrm{~kg} / \mathrm{m}^3\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

A metallic sphere floats in an immiscible mixture of water \mathrm{\left(\rho_w=10^3 \mathrm{~kg} / \mathrm{m}^3\right)} and a liquid \mathrm{\left(\rho_L=13.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\right)}  such that its \mathrm{(4 / 5)^{\mathrm{h}}} portion is in water and \mathrm{(1 / 5)^{\text {th }}} portion in the liquid. The density of metal is:

Option: 1

\mathrm{4.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3}


Option: 2

\mathrm{4.0 \times 10^3 \mathrm{~kg} / \mathrm{m}^3}


Option: 3

\mathrm{3.5 \times 10^3 \mathrm{~kg} / \mathrm{m}^3}


Option: 4

\mathrm{1.9 \times 10^3 \mathrm{~kg} / \mathrm{m}^3}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \frac{4}{5} \times\left(\rho-\rho_w\right)=\frac{1}{5} \times\left(\rho_L \rho\right) \\\\ & \Rightarrow \frac{\rho-\rho_w}{\rho_L \rho}=\frac{1}{4} \Rightarrow 5 \rho=\rho_L+4 \rho_w \\\\ & \rho=\frac{13.5+4}{5} \times 10^3 \end{aligned}}

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Nehul

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