A metal sphere floats in an immiscible mixture of water density 103kg-m3 and a liquid of density 13-5-103kg-m3 such that its 45th portion is in water and 15th portion in liquid-The density of metal is

The correct option is A 3.5×10^3kg/m^3 Vρ g=45V×10^3g+15V× 13.5×10^3g ⇒ρ =4+13.45×10^3=3.5×10^3kg/m^3 ...

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Standard XII

Chemistry

Density

A metal spher...

Question

A metal sphere floats in an immiscible mixture of water (density $10^{3} k g / m^{3}$ ) and a liquid of density $(13.5 \times 10^{3} k g / m^{3})$ such that its ${(\frac{4}{5})}^{t h}$ portion is in water and ${(\frac{1}{5})}^{t h}$ portion in liquid.The density of metal is

7.25 \times 10^{3} k g / m^{3}

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3.5 \times 10^{3} k g / m^{3}

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\sqrt{13.5} \times 10^{3} k g / m^{3}

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14.51 k g / m^{3}

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Solution

The correct option is A $3.5 \times 10^{3} k g / m^{3}$
$V ρ g = \frac{4}{5} V \times 10^{3} g + \frac{1}{5} V \times 13.5 \times 10^{3} g$
$\Rightarrow ρ = \frac{4 + 13.4}{5} \times 10^{3} = 3.5 \times 10^{3} k g / m^{3}$

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A metal sphere floats in an immiscible mixture of water (density 103kg/m3) and a liquid of density (13.5×103kg/m3) such that its (45)th portion is in water and (15)th portion in liquid.The density of metal is

The correct option is A 3.5×103kg/m3Vρg=45V×103g+15V×13.5×103g⇒ρ=4+13.45×103=3.5×103kg/m3

A metal sphere floats in an immiscible mixture of water (density $10^{3} k g / m^{3}$ ) and a liquid of density $(13.5 \times 10^{3} k g / m^{3})$ such that its ${(\frac{4}{5})}^{t h}$ portion is in water and ${(\frac{1}{5})}^{t h}$ portion in liquid.The density of metal is

The correct option is A $3.5 \times 10^{3} k g / m^{3}$
$V ρ g = \frac{4}{5} V \times 10^{3} g + \frac{1}{5} V \times 13.5 \times 10^{3} g$
$\Rightarrow ρ = \frac{4 + 13.4}{5} \times 10^{3} = 3.5 \times 10^{3} k g / m^{3}$