A solid wooden block of density 600\text{ kg/m}^3 and volume 4.0 \times 10^{-3}\text{ m}^3… — JAMB UTME

A solid wooden block of density $600\text{ kg/m}^3$ and volume $4.0 \times 10^{-3}\text{ m}^3$ floats in water of density $1000\text{ kg/m}^3$ . A metal block is placed on top of the wooden block so that the wooden block is just completely submerged while the metal block remains entirely above the water surface. What is the mass of the metal block? (Take $g = 10\text{ m/s}^2$ )

$1.6\text{ kg}$ Answer
B
$2.4\text{ kg}$
C
$4.0\text{ kg}$
D
$6.4\text{ kg}$

Answer

1.6\text{ kg}

When the wooden block is completely submerged, it displaces

4.0 \times 10^{-3}\text{ m}^3

of water, creating an upthrust of

40\text{ N}

. The weight of the wooden block is

24\text{ N}

. For equilibrium, the total downward weight must equal the upthrust (

24\text{ N} + W_{\text{metal}} = 40\text{ N}

), which gives

W_{\text{metal}} = 16\text{ N}

and a corresponding mass of

1.6\text{ kg}

Step-by-Step Solution

Calculate the mass and weight of the wooden block.

m_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{wood}} = 600\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 = 2.4\text{ kg}

, so

W_{\text{wood}} = 2.4 \times 10 = 24\text{ N}

The weight of the wood contributes to the total downward force of the floating system.

Calculate the total upthrust exerted by the water when the wooden block is completely submerged.

U = \rho_{\text{water}} \times V_{\text{wood}} \times g = 1000\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 40\text{ N}

By Archimedes' principle, upthrust equals the weight of the displaced liquid.

Apply the law of flotation to solve for the weight and mass of the metal block.

W_{\text{wood}} + W_{\text{metal}} = U \implies 24\text{ N} + W_{\text{metal}} = 40\text{ N} \implies W_{\text{metal}} = 16\text{ N}

. Thus,

m_{\text{metal}} = \frac{16\text{ N}}{10\text{ m/s}^2} = 1.6\text{ kg}

For the system to float in equilibrium just submerged, total downward weight must equal total upward buoyant force.

Key Concept

Archimedes' Principle and Law of Flotation

Estimated Time:2m 0s

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