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. 1.80
• B. 0.01
• C. 80.00
• D. 0.55

Hint:

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 $g$.

Answer 1

The Correct Option is A

Step 1: Spot why the weight drops.
In a vacuum the sphere weighs $100$ N, but in gas it reads only $85$ N. The missing $15$ N is the buoyant push of the gas, equal to the weight of gas the sphere shoves aside.

Step 2: Find the buoyant force.
\[ F_b = 100 - 85 = 15\ \text{N}. \]

Step 3: Get the sphere's volume from its true weight.
The real weight is $W = \rho_s V g$, where $\rho_s = 12$. So \[ V = \frac{W}{\rho_s g} = \frac{100}{12 \times 9.8} \approx 0.85\ \text{m}^3. \]

Step 4: Write buoyancy in terms of gas density.
The buoyant force is the displaced gas weight, \[ F_b = \rho_g V g. \]

Step 5: Solve for the gas density.
\[ \rho_g = \frac{F_b}{V g} = \frac{15}{0.85 \times 9.8} \approx 1.8\ \text{kg/m}^3. \]

Step 6: State the answer.
A neat shortcut is $\rho_g = \rho_s \times (F_b / W) = 12 \times (15/100) = 1.8$, which confirms the result. \[ \boxed{\rho_g \approx 1.80\ \text{kg/m}^3} \]