QuestionJuly 22, 2025

Asample has a mass of 2.5 kg When the sample is suspended from the lower end of astring and totally immersed in water, the tension in the string is 11.2 N. What is the smallest density of a liquid in which the sample will float? Remember that rho _(water)=997kg/m^3 1838kg/m^3 1835kg/m^3 1828kg/m^3 1849kg/m^3

Asample has a mass of 2.5 kg When the sample is suspended from the lower end of astring and totally immersed in water, the tension in the string is 11.2 N. What is the smallest density of a liquid in which the sample will float? Remember that rho _(water)=997kg/m^3 1838kg/m^3 1835kg/m^3 1828kg/m^3 1849kg/m^3
Asample has a mass of 2.5 kg When the sample is suspended from the lower end of astring and totally immersed in water, the tension in the string is 11.2 N. What is the smallest density of a liquid in which the sample will float? Remember that rho _(water)=997kg/m^3 1838kg/m^3 1835kg/m^3 1828kg/m^3 1849kg/m^3

Solution
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Answer

\( 1835 \mathrm{~kg} / \mathrm{m}^{3} \) Explanation 1. Calculate the buoyant force in water The tension in the string is 11.2 N, and the weight of the sample is 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 24.5 \, \text{N}. The buoyant force F_b is the difference: F_b = 24.5 \, \text{N} - 11.2 \, \text{N} = 13.3 \, \text{N}. 2. Determine the volume of the sample Using **Archimedes' principle**, F_b = \rho_{\text{water}} \cdot V \cdot g, solve for V: V = \frac{F_b}{\rho_{\text{water}} \cdot g} = \frac{13.3 \, \text{N}}{997 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2} = 0.00137 \, \text{m}^3. 3. Calculate the smallest density of a liquid for floating For the sample to float, its weight must equal the buoyant force in the new liquid: 2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = \rho_{\text{liquid}} \cdot V \cdot g. Solve for \rho_{\text{liquid}}: \rho_{\text{liquid}} = \frac{24.5 \, \text{N}}{0.00137 \, \text{m}^3 \times 9.8 \, \text{m/s}^2} = 1835 \, \text{kg/m}^3.

Explanation

1. Calculate the buoyant force in water<br /> The tension in the string is 11.2 N, and the weight of the sample is $2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 24.5 \, \text{N}$. The buoyant force $F_b$ is the difference: $F_b = 24.5 \, \text{N} - 11.2 \, \text{N} = 13.3 \, \text{N}$.<br />2. Determine the volume of the sample<br /> Using **Archimedes' principle**, $F_b = \rho_{\text{water}} \cdot V \cdot g$, solve for $V$: $V = \frac{F_b}{\rho_{\text{water}} \cdot g} = \frac{13.3 \, \text{N}}{997 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2} = 0.00137 \, \text{m}^3$.<br />3. Calculate the smallest density of a liquid for floating<br /> For the sample to float, its weight must equal the buoyant force in the new liquid: $2.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = \rho_{\text{liquid}} \cdot V \cdot g$. Solve for $\rho_{\text{liquid}}$: $\rho_{\text{liquid}} = \frac{24.5 \, \text{N}}{0.00137 \, \text{m}^3 \times 9.8 \, \text{m/s}^2} = 1835 \, \text{kg/m}^3$.
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