A solid sphere of mass 0.40\text{ kg} and relative density 2.5 is held fully submerged in … — JAMB UTME

A solid sphere of mass $0.40\text{ kg}$ and relative density $2.5$ is held fully submerged in a liquid of density $800\text{ kg/m}^3$ by a light string attached to a fixed support. What is the tension in the string? (Take $g = 10\text{ m/s}^2$ )

$2.72\text{ N}$ Answer
B
$3.60\text{ N}$
C
$1.28\text{ N}$
D
$2.40\text{ N}$

Answer

The tension in the string is

2.72\text{ N}

The sphere has a weight of

4.0\text{ N}

acting downward. When fully submerged, it displaces a volume of liquid equal to its own volume (

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

). The displaced liquid of density

800\text{ kg/m}^3

exerts an upward buoyant force (upthrust) of

1.28\text{ N}

. Tension balances the remaining downward force:

T = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

Step-by-Step Solution

Calculate the weight of the sphere

W = mg = 0.40\text{ kg} \times 10\text{ m/s}^2 = 4.0\text{ N}

The force of gravity acting downward on the mass.

Determine the volume of the sphere using its relative density

Density of sphere

\rho_s = 2.5 \times 1000\text{ kg/m}^3 = 2500\text{ kg/m}^3

. Volume

V = \frac{m}{\rho_s} = \frac{0.40}{2500} = 1.6 \times 10^{-4}\text{ m}^3

Relative density is the ratio of the substance's density to the density of water (

1000\text{ kg/m}^3

Calculate the upthrust exerted by the liquid

U = V \cdot \rho_{\text{liquid}} \cdot g = (1.6 \times 10^{-4}) \times 800 \times 10 = 1.28\text{ N}

By Archimedes' principle, upthrust equals the weight of the fluid displaced by the submerged volume.

Calculate tension in the string

T = W - U = 4.0\text{ N} - 1.28\text{ N} = 2.72\text{ N}

For vertical equilibrium of the submerged body, weight acts downward while upthrust and tension act upward.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:1m 30s

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