Question

A cylinder of length 10 cm made of material of density 0.65 g/cm³ floats half submerged in an ideal liquid kept in closed vessel as shown in the figure. Air in the vessel has density 1.30 kg/m³. If pressure of air in the vessel in made 100 times of its initial value, what will you observe?

• A. The cylinder moves upwards.
• B. The cylinder moves downwards.
• C. Displacement of the cylinder is 0.56 cm.
• D. Displacement of the cylinder is 0.60 cm.

Answer 1

Answer: (C)

Displacement of the cylinder is 0.56 cm.

Answer 2

Let $L$ be the length of the cylinder, $A$ be its cross-sectional area, $\rho_c$ be the density of the cylinder, $\rho_l$ be the density of the liquid, and $\rho_{air}$ be the density of the air. The weight of the cylinder is $W = \rho_c A L g$.

Initially, the cylinder is half submerged in the liquid. Let $h_0$ be the initial submerged length in the liquid. So, $h_0 = L/2$. The initial volume of the submerged part in the liquid is $V_{l,0} = A h_0$. The initial volume of the part in the air is $V_{a,0} = A (L-h_0)$.

The initial buoyant force is $F_{B,0} = \rho_l g V_{l,0} + \rho_{air,0} g V_{a,0} = \rho_l g A h_0 + \rho_{air,0} g A (L-h_0)$.

In equilibrium, $W = F_{B,0}$, so $\rho_c A L g = \rho_l g A h_0 + \rho_{air,0} g A (L-h_0)$.

Dividing by $Ag$, we get $\rho_c L = \rho_l h_0 + \rho_{air,0} (L-h_0)$.

Given $L = 10$ cm, $\rho_c = 0.65$ g/cm³, $h_0 = L/2 = 5$ cm.

Air density $\rho_{air,0} = 1.30$ kg/m³ = $1.30 \times 10^{-3}$ g/cm³.

Substituting the values: $0.65 \times 10 = \rho_l \times 5 + 1.30 \times 10^{-3} \times (10-5)$.

$6.5 = 5\rho_l + 1.30 \times 10^{-3} \times 5$.

$6.5 = 5\rho_l + 0.0065$.

$5\rho_l = 6.5 - 0.0065 = 6.4935$.

$\rho_l = \frac{6.4935}{5} = 1.2987$ g/cm³.

The problem states "ideal liquid", which usually implies constant density. The value of $\rho_l = 1.3$ g/cm³ derived earlier by neglecting the air density is close to this value. Let's use $\rho_l = 1.3$ g/cm³ as derived from the initial condition of being half submerged and assuming air density is negligible compared to liquid density in the initial buoyant force calculation, or assume the value given in the problem statement implies $\rho_l = 1.3$ g/cm³ for half submersion when air density is $1.30 \times 10^{-3}$ g/cm³. Let's recheck the initial condition using $\rho_l = 1.3$ g/cm³ and $\rho_{air,0} = 0.0013$ g/cm³.

$h_0 = L \frac{\rho_c - \rho_{air,0}}{\rho_l - \rho_{air,0}} = 10 \times \frac{0.65 - 0.0013}{1.3 - 0.0013} = 10 \times \frac{0.6487}{1.2987} \approx 10 \times 0.5 = 5$ cm.

So using $\rho_l = 1.3$ g/cm³ and $\rho_{air,0} = 0.0013$ g/cm³ is consistent with the half submerged condition.

Now, the pressure of air in the vessel is made 100 times its initial value. Assuming the temperature and volume occupied by air remain approximately constant, the density of air becomes 100 times the initial density.

New air density $\rho_{air,1} = 100 \rho_{air,0} = 100 \times 1.30 \times 10^{-3}$ g/cm³ = $0.130$ g/cm³.

Let $h_1$ be the new submerged length in the liquid. In the new equilibrium, the weight of the cylinder equals the new buoyant force:

$W = F_{B,1} = \rho_l g A h_1 + \rho_{air,1} g A (L-h_1)$.

$\rho_c L = \rho_l h_1 + \rho_{air,1} (L-h_1)$.

$h_1 (\rho_l - \rho_{air,1}) = L (\rho_c - \rho_{air,1})$.

$h_1 = L \frac{\rho_c - \rho_{air,1}}{\rho_l - \rho_{air,1}}$.

Substitute the values: $L = 10$ cm, $\rho_c = 0.65$ g/cm³, $\rho_l = 1.3$ g/cm³, $\rho_{air,1} = 0.130$ g/cm³.

$h_1 = 10 \times \frac{0.65 - 0.130}{1.3 - 0.130} = 10 \times \frac{0.520}{1.170} = 10 \times \frac{52}{117} = 10 \times \frac{4 \times 13}{9 \times 13} = 10 \times \frac{4}{9} = \frac{40}{9}$ cm.

The initial submerged length was $h_0 = 5$ cm. The new submerged length is $h_1 = 40/9$ cm.

Since $h_1 = 40/9 \approx 4.44$ cm is less than $h_0 = 5$ cm, the cylinder moves upwards.

The displacement of the cylinder is the change in the position of its bottom (or top). The bottom of the cylinder was initially at a depth of 5 cm below the surface, and is now at a depth of $40/9$ cm below the surface.

The upward displacement is $h_0 - h_1 = 5 - \frac{40}{9} = \frac{45-40}{9} = \frac{5}{9}$ cm.

$5/9 \approx 0.5555...$ cm.

Comparing with the options:

(a) The cylinder moves upwards. This is correct since the submerged length decreases.

(b) The cylinder moves downwards. This is incorrect.

(c) Displacement of the cylinder is 0.56 cm. $5/9 \approx 0.5556$. Rounded to two decimal places, this is 0.56 cm. This is correct.

(d) Displacement of the cylinder is 0.60 cm. This is incorrect.

Since both (a) and (c) are correct, and (c) provides a specific value for the displacement which is consistent with the calculation, option (c) is the most complete answer. If only one option is to be selected, and a numerical value is provided, it is likely that the question intends for the numerical displacement to be calculated.