A solid block floats in water of density 1000\text{ kg/m}^3 with 60\% of its total volume … — JAMB UTME

A solid block floats in water of density $1000\text{ kg/m}^3$ with $60\%$ of its total volume submerged. When the same block is placed in an unknown liquid $X$ , $80\%$ of its total volume is submerged. What is the density of liquid $X$ ?

$750\text{ kg/m}^3$ Answer
B
$1333\text{ kg/m}^3$
C
$480\text{ kg/m}^3$
D
$800\text{ kg/m}^3$

Answer

The density of liquid

X

750\text{ kg/m}^3

For any floating body, its weight equals the upthrust exerted by the liquid. The upthrust is given by

U = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g

. Since the weight of the block is unchanged,

\rho_{\text{water}} \cdot V_{\text{sub, water}} = \rho_X \cdot V_{\text{sub, X}}

. Substituting the given values:

1000 \times 0.60V = \rho_X \times 0.80V

, yielding

\rho_X = 750\text{ kg/m}^3

Step-by-Step Solution

Apply the Law of Flotation to the block in water

Weight of block

W = \rho_w \cdot V_{\text{sub, water}} \cdot g = 1000 \cdot 0.60V \cdot g = 600 V g

A floating object displaces its own weight of fluid.

Apply the Law of Flotation to the block in liquid X

Weight of block

W = \rho_X \cdot V_{\text{sub, X}} \cdot g = \rho_X \cdot 0.80V \cdot g

The weight of the block remains constant regardless of the fluid.

Equate the two expressions for the weight of the block and solve for

\rho_X

\rho_X \cdot 0.80V \cdot g = 600 V g \implies \rho_X = \frac{600}{0.80} = 750\text{ kg/m}^3

Since both buoyant forces equal the block's weight, set them equal to each other.

Key Concept

Law of Flotation and Archimedes' Principle

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