The weight of a sphere in air is 50 g. Its w | vedclass

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(A) Let the density of the liquid at $20^{\circ} C$ be $\rho_{1}$ and at $70^{\circ} C$ be $\rho_{2}$.The apparent weight is given by the formula: $W_

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$2$

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(A) Let the density of the liquid at $20^{\circ} C$ be $ho_{1}$ and at $70^{\circ} C$ be $ho_{2}$.The apparent weight is given by the formula: $W_{	ext{apparent}} = W_{	ext{air}} - V ho g$,where $V$ is the volume of the sphere.At $20^{\circ} C$: $40 = 50 - V ho_{1} g \implies V ho_{1} g = 10$.At $70^{\circ} C$: $45 = 50 - V ho_{2} g \implies V ho_{2} g = 5$.Taking the ratio of the two equations:$\frac{V ho_{1} g}{V ho_{2} g} = \frac{10}{5} = \frac{2}{1}$.Therefore,the ratio of the densities is $2 : 1$.

The weight of a sphere in air is $50 \ g$. Its weight is $40 \ g$ in a liquid at temperature $20^{\circ} C$. When the temperature increases to $70^{\circ} C$,its weight becomes $45 \ g$. The ratio of the densities of the liquid at the two given temperatures is: (in $: 1$)

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