Question

A cylinder of mass m and material density ρ hanging from a string is lowered into a vessel of cross - sectional area A containing a liquid of density σ (< ρ) until it is fully immersed. The increase in pressure at the bottom of the vessel is:

• A. Zero
• B. mg/A
• C. mgρ/σA
• D. mσg/ρA

Answer 1

Answer: (D)

mσg/ρA

Answer 2

The correct option is: (D): mσg/ρA.

When the cylinder is fully immersed in the liquid within the vessel, it displaces a volume of liquid equal to its own volume (V). The increase in pressure at the bottom of the vessel is due to the weight of the liquid column above it.

The pressure increase (ΔP) can be calculated using the formula:

ΔP = ρgh

Where:

• ρ is the density of the liquid in the vessel (σ).
• g is the acceleration due to gravity.
• h is the height of the liquid column above the point.

The height h is the height by which the cylinder is submerged in the liquid, and it's also equal to the length of the cylinder (l) as it's fully immersed. The volume of the cylinder (V) can be calculated using the formula:

V = Al

Where:

• A is the cross-sectional area of the cylinder (which is also the cross-sectional area of the vessel).
• l is the length of the cylinder.

Given that the mass of the cylinder is m and the material density is ρ, the volume V is the mass divided by the density:

V = m / ρ

Now, equate V to Al:

m / ρ = Al

Solve for l:

l = (m / ρ) / A

Now, substitute this value of l into the pressure formula:

ΔP = ρgh = σg(m / ρ) / A

This simplifies to:

ΔP = mσg / ρA

So, the increase in pressure at the bottom of the vessel is indeed mσg / ρA, as provided in the answer.