A sample of metal weighs 210 g in air,180 g | vedclass

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(D) Let the density of the metal be $\rho$ and the density of the liquid be $\sigma$. Let $V$ be the volume of the sample.According to Archimedes' pri

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Both $(b)$ and $(c)$

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(D) Let the density of the metal be $\rho$ and the density of the liquid be $\sigma$. Let $V$ be the volume of the sample.According to Archimedes' principle,the apparent weight is the actual weight minus the buoyant force.In air: $W_{air} = V \rho g = 210 \ g$ (taking $g$ as a unit factor for weight in grams) ... $(i)$In water: $W_{water} = V(\rho - 1)g = 180 \ g$ ... (ii)In liquid: $W_{liquid} = V(\rho - \sigma)g = 120 \ g$ ... (iii)From $(i)$ and (ii): $V \rho g - V(1)g = 180 \implies 210 - Vg = 180 \implies Vg = 30$.Since $V \rho g = 210$,we have $30 \rho = 210 \implies \rho = 7$.From $(i)$ and (iii): $V \rho g - V \sigma g = 120 \implies 210 - V \sigma g = 120 \implies V \sigma g = 90$.Substituting $Vg = 30$: $30 \sigma = 90 \implies \sigma = 3$.Thus,the relative density of the metal is $7$ and the relative density of the liquid is $3$.

$A$ sample of metal weighs $210 \ g$ in air,$180 \ g$ in water,and $120 \ g$ in a liquid. Then,the relative density $(RD)$ of:

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