A spiral spring with a metal extends by 10.5 cm in... | WAEC SSCE - Physics - 2012

Question 1 Report

A spiral spring with a metal extends by 10.5 cm in air. When the metal is fully submerged in water, the spring extends by 6.8 cm. Calculate the relative density of the metal. (Assume Hooke's law is obeyed)

Answer

Relative density from spring extensions

By Hooke's law the extension of the spring is proportional to the force (weight) it supports.

In air the spring supports the full weight of the metal, so its extension is proportional to the weight:

\[ e_1 = 10.5\,\text{cm} \propto W \]

In water the spring supports the apparent weight (weight minus upthrust), so:

\[ e_2 = 6.8\,\text{cm} \propto (W - \text{upthrust}) \]

Therefore the extension due to the upthrust is proportional to

\[ e_1 - e_2 = 10.5 - 6.8 = 3.7\,\text{cm} \]

The upthrust equals the weight of water displaced, so

\[ \text{relative density} = \frac{\text{weight of metal in air}}{\text{weight of equal volume of water}} = \frac{e_1}{e_1 - e_2} \]

\[ = \frac{10.5}{3.7} = 2.84 \]

The relative density of the metal is about 2.84.

Read lesson note on Elasticity (JAMB)

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