Solution for Question 11
Given Data:
- Length of cylinder, $L = 10 \text{ cm} = 0.1 \text{ m}$
- Density of cylinder, $\rho_c = 0.65 \text{ g/cm}^3$
- Initial Air Density, $\rho_{air} = 1.30 \text{ kg/m}^3 = 0.0013 \text{ g/cm}^3$
- Initial state: Half submerged ($50\%$).
- Final state: Pressure becomes 100 times initial value.
Step 1: Calculate Liquid Density ($\rho_L$)
Initially, the cylinder floats half submerged ($V_{sub} = V/2$). The buoyant force from air is negligible compared to the liquid, but let’s write the exact equation for equilibrium:
$$ \text{Weight} = \text{Buoyancy}_{liquid} + \text{Buoyancy}_{air} $$ $$ \rho_c V g = \rho_L (0.5 V) g + \rho_{air} (0.5 V) g $$ $$ 2 \rho_c = \rho_L + \rho_{air} $$ $$ 2(0.65) = \rho_L + 0.0013 $$ $$ 1.30 = \rho_L + 0.0013 \implies \rho_L \approx 1.30 \text{ g/cm}^3 $$Note: The liquid density is effectively $1.30 \text{ g/cm}^3$.
Step 2: Analyze Final State
Pressure increases 100 times. Assuming air behaves as an ideal gas isothermally, density is proportional to pressure. $$ \rho’_{air} = 100 \times \rho_{air} = 100 \times 1.30 \text{ kg/m}^3 = 130 \text{ kg/m}^3 = 0.13 \text{ g/cm}^3 $$
Let $x$ be the new fraction of the cylinder submerged in the liquid. The fraction in air is $(1-x)$.
$$ \rho_c = \rho_L (x) + \rho’_{air} (1-x) $$ $$ 0.65 = 1.30(x) + 0.13(1-x) $$ $$ 0.65 = 1.30x + 0.13 – 0.13x $$ $$ 0.52 = 1.17x $$ $$ x = \frac{0.52}{1.17} = \frac{52}{117} = \frac{4}{9} \approx 0.444 $$Step 3: Calculate Displacement
The submerged fraction changed from $0.5$ (initial) to $0.444$ (final). Since the submerged portion decreased, the cylinder must have moved upwards.
Change in submerged fraction:
$$ \Delta x = 0.5 – \frac{4}{9} = \frac{1}{2} – \frac{4}{9} = \frac{9-8}{18} = \frac{1}{18} $$The displacement $\Delta h$ is the change in submerged length:
$$ \Delta h = L \times \Delta x = 10 \text{ cm} \times \frac{1}{18} = \frac{10}{18} \text{ cm} \approx 0.555 \text{ cm} $$Rounding to two decimal places, displacement is $0.56 \text{ cm}$.