A uniform cylindrical rod of length 20\text{ cm} floats vertically in liquid X of density … — JAMB UTME

A uniform cylindrical rod of length $20\text{ cm}$ floats vertically in liquid $X$ of density $800\text{ kg/m}^3$ with $15\text{ cm}$ of its length submerged. When transferred to liquid $Y$ , it floats vertically with $12\text{ cm}$ of its length submerged. What is the density of liquid $Y$ ?

A
$640\text{ kg/m}^3$
$1000\text{ kg/m}^3$ Answer
C
$1333\text{ kg/m}^3$
D
$500\text{ kg/m}^3$

Answer

The density of liquid

Y

1000\text{ kg/m}^3

According to the Law of Flotation, a floating body displaces its own weight of liquid. Therefore,

h_X \rho_X = h_Y \rho_Y

. Substituting

15\text{ cm} \times 800\text{ kg/m}^3 = 12\text{ cm} \times \rho_Y

yields

\rho_Y = 1000\text{ kg/m}^3

Step-by-Step Solution

Apply the Law of Flotation for a floating body of uniform cross-sectional area

A

Weight of rod

W = \text{Upthrust} = A \cdot h_{\text{submerged}} \cdot \rho_{\text{liquid}} \cdot g

A floating body displaces a weight of fluid equal to its own total weight.

Equate the upthrust in liquid

X

to the upthrust in liquid

Y

A \cdot h_X \cdot \rho_X \cdot g = A \cdot h_Y \cdot \rho_Y \cdot g \implies h_X \cdot \rho_X = h_Y \cdot \rho_Y

Since the rod is identical and floating freely in both liquids, its weight

W

remains unchanged.

Substitute the known values (

h_X = 15\text{ cm}

\rho_X = 800\text{ kg/m}^3

h_Y = 12\text{ cm}

) and solve for

\rho_Y

\rho_Y = \frac{h_X \cdot \rho_X}{h_Y} = \frac{15 \times 800}{12} = 1000\text{ kg/m}^3

Rearranging the linear equation yields the density of liquid

Y

Key Concept

Law of Flotation and Hydrometer Principle

Estimated Time:1m 15s

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