A metallic sphere weighs 5.0\text{ N} in air. When completely immersed in water, its appar… — JAMB UTME

A metallic sphere weighs $5.0\text{ N}$ in air. When completely immersed in water, its apparent weight is $3.0\text{ N}$ . When completely immersed in an unknown liquid $X$ , its apparent weight is $3.4\text{ N}$ . What is the density of liquid $X$ in $\text{kg/m}^3$ ? (Take the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ ).

Answer: 800 kg/m^3

Answer

The density of liquid

X

800\text{ kg/m}^3

The upthrust in water (

2.0\text{ N}

) gives the volume of the sphere as

2.0 \times 10^{-4}\text{ m}^3

. Using the upthrust in liquid X (

1.6\text{ N}

), the density of liquid X is calculated as

\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3

Step-by-Step Solution

Calculate upthrust in water

U_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}

Upthrust equals the loss in weight of the submerged body in water.

Determine the volume of the metallic sphere

V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3

According to Archimedes' principle, upthrust in water equals the weight of displaced water.

Calculate upthrust in liquid X

U_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}

Loss of weight in liquid X gives the upthrust exerted by liquid X.

Calculate the density of liquid X

\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3

Rearranging

U_X = \rho_X V g

allows solving for the unknown fluid density.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:2m 0s

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