A solid cylinder of length ℓ and cross-sectional area 5 a is immersed such that it floats with its axis vertical at the liquid-liquid interface with length ℓ /4 in the denser liquid as shown in figu

Question

A solid cylinder of length ℓ and cross-sectional area 5 a ​ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length ℓ /4 in the denser liquid as shown in figure. The lower density liquid ( ϱ ) is open to atmosphere having pressure P 0 ​ . The density d of solid cylinder is

TestHub · Accepted Answer

Let the cross-sectional area of the cylinder be $A = \frac{a}{5}$ .

For the cylinder to float in equilibrium, its weight must be balanced by the total buoyant force exerted by the two liquids.

The total length of the cylinder is $ℓ$ . It is given that the length of the cylinder in the denser liquid is $\frac{ℓ}{4}$ . Since the cylinder is completely immersed across the interface, the length of the cylinder in the lighter liquid is:
$ℓ - \frac{ℓ}{4} = \frac{3 ℓ}{4}$

The weight of the solid cylinder is:
$W = Volume \times density \times g = (A ℓ) d g$

The total buoyant force $F_{B}$ is the sum of the weights of the displaced liquids:
$F_{B} = Weight of displaced lighter liquid + Weight of displaced denser liquid$
$F_{B} = A (\frac{3 ℓ}{4}) ϱ g + A (\frac{ℓ}{4}) (3 ϱ) g$

Equating the weight of the cylinder to the buoyant force ( $W = F_{B}$ ):
$A ℓ d g = A (\frac{3 ℓ}{4}) ϱ g + A (\frac{ℓ}{4}) (3 ϱ) g$

Dividing both sides by $A ℓ g$ , we get:
$d = \frac{3}{4} ϱ + \frac{3}{4} ϱ$

$d = \frac{6}{4} ϱ = \frac{3}{2} ϱ$

Answer: $\frac{3}{2} ϱ$

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