Question:
A sphere made of a material with density $12\text{ kg}\cdot\text{m}^{-3}$ and weighing $100\text{ N}$ in vacuum is immersed in a container of gas. Its weight in gas is $85\text{ N}$. The density of the gas in $\text{kg}\cdot\text{m}^{-3}$ is closest to
A sphere made of a material with density $12\text{ kg}\cdot\text{m}^{-3}$ and weighing $100\text{ N}$ in vacuum is immersed in a container of gas. Its weight in gas is $85\text{ N}$. The density of the gas in $\text{kg}\cdot\text{m}^{-3}$ is closest to
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The ratio of buoyant force to weight in vacuum is equal to the ratio of the fluid density to the object density:
\[ \frac{F_b}{W_{\text{vac}}} = \frac{\rho_{\text{fluid}}}{\rho_{\text{object}}} \]
This direct ratio formula bypasses the need to compute the volume $V$ or $g$.
\[ \frac{F_b}{W_{\text{vac}}} = \frac{\rho_{\text{fluid}}}{\rho_{\text{object}}} \]
This direct ratio formula bypasses the need to compute the volume $V$ or $g$.
Updated On: Jun 16, 2026
- 1.80
- 0.01
- 80.00
- 0.55
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem presents a solid sphere weighed first in a vacuum and then inside a gas.
The apparent loss of weight in the gas is due to the buoyant force exerted by the gas. We need to determine the density of this gas.
Step 2: Key Formula or Approach:
- Archimedes' Principle states that the buoyant force $F_b$ is equal to the weight of the fluid displaced:
\[ F_b = \rho_{\text{fluid}} V g \]
- The weight of the sphere in vacuum is:
\[ W_{\text{vac}} = \rho_{\text{sphere}} V g \]
Step 3: Detailed Explanation:
• The weight in vacuum is $W_{\text{vac}} = 100\text{ N}$.
• The weight in the gas is $W_{\text{gas}} = 85\text{ N}$.
• The buoyant force $F_b$ exerted by the gas is:
\[ F_b = W_{\text{vac}} - W_{\text{gas}} = 100 - 85 = 15\text{ N} \]
• Let us write the expressions for $W_{\text{vac}}$ and $F_b$:
\[ W_{\text{vac}} = \rho_s V g \implies Vg = \frac{W_{\text{vac}}}{\rho_s} \]
\[ F_b = \rho_g V g \]
• Substituting the expression for $Vg$ into the buoyant force formula:
\[ F_b = \rho_g \left( \frac{W_{\text{vac}}}{\rho_s} \right) \]
• Rearranging the equation to solve for the density of the gas $\rho_g$:
\[ \rho_g = \frac{F_b \cdot \rho_s}{W_{\text{vac}}} \]
• Substituting the given numerical values:
\[ \rho_g = \frac{15\text{ N} \times 12\text{ kg/m}^3}{100\text{ N}} \]
\[ \rho_g = \frac{180}{100} = 1.80\text{ kg/m}^3 \]
Step 4: Final Answer:
The density of the gas is closest to 1.80 $\text{kg}\cdot\text{m}^{-3}$.
The problem presents a solid sphere weighed first in a vacuum and then inside a gas.
The apparent loss of weight in the gas is due to the buoyant force exerted by the gas. We need to determine the density of this gas.
Step 2: Key Formula or Approach:
- Archimedes' Principle states that the buoyant force $F_b$ is equal to the weight of the fluid displaced:
\[ F_b = \rho_{\text{fluid}} V g \]
- The weight of the sphere in vacuum is:
\[ W_{\text{vac}} = \rho_{\text{sphere}} V g \]
Step 3: Detailed Explanation:
• The weight in vacuum is $W_{\text{vac}} = 100\text{ N}$.
• The weight in the gas is $W_{\text{gas}} = 85\text{ N}$.
• The buoyant force $F_b$ exerted by the gas is:
\[ F_b = W_{\text{vac}} - W_{\text{gas}} = 100 - 85 = 15\text{ N} \]
• Let us write the expressions for $W_{\text{vac}}$ and $F_b$:
\[ W_{\text{vac}} = \rho_s V g \implies Vg = \frac{W_{\text{vac}}}{\rho_s} \]
\[ F_b = \rho_g V g \]
• Substituting the expression for $Vg$ into the buoyant force formula:
\[ F_b = \rho_g \left( \frac{W_{\text{vac}}}{\rho_s} \right) \]
• Rearranging the equation to solve for the density of the gas $\rho_g$:
\[ \rho_g = \frac{F_b \cdot \rho_s}{W_{\text{vac}}} \]
• Substituting the given numerical values:
\[ \rho_g = \frac{15\text{ N} \times 12\text{ kg/m}^3}{100\text{ N}} \]
\[ \rho_g = \frac{180}{100} = 1.80\text{ kg/m}^3 \]
Step 4: Final Answer:
The density of the gas is closest to 1.80 $\text{kg}\cdot\text{m}^{-3}$.
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