JAMB Physics 2024 Paper

Concept:A floating object displaces its own weight of fluid, so the mass of displaced fluid equals the mass of the hydrometer.Explanation:The hydrometer has mass $y\ \mathrm{kg}$ and total volume $V = 2y \times 10^{-5}\ \mathrm{m^3}$.Since $20\%$ of its volume is above the fluid, the submerged volume is $80\%$ of the total volume.$V_{\text{submerged}} = \dfrac{80}{100} \times 2y \times 10^{-5} = 1.6y \times 10^{-5}\ \mathrm{m^3}$.For floating equilibrium, the mass of displaced fluid equals the mass of the hydrometer:$m_{\text{displaced}} = y\ \mathrm{kg}$.The fluid density is given by:$\rho = \dfrac{m_{\text{displaced}}}{V_{\text{submerged}}} = \dfrac{y}{1.6y \times 10^{-5}}$.Cancelling $y$ gives:$\rho = \dfrac{1}{1.6 \times 10^{-5}} = 6.25 \times 10^{4}\ \mathrm{kg\,m^{-3}}$.Answer:The density of the fluid is $6.25 \times 10^{4}\ \mathrm{kg\,m^{-3}}$, which is Option D.