A steel wire is suspended vertically from a rigid support When loaded with a weight in air it expands by $L_{a}$ and when the weight is immersed completely in water the extension is reduced to $L_{w}$ Then relative density of the material of the weight is

\frac{L_{a}}{L_{a} - L_{w}}

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\frac{L_{w}}{L_{a}}

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\frac{L_{a}}{L_{w}}

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\frac{L_{w}}{L_{a} - L_{w}}

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Solution

The correct option is A $\frac{L_{a}}{L_{a} - L_{w}}$
From Young's relation,
Stress $\propto$ Strain
$F \propto x$
where $F$ is the force acting causing the wire to stretch, and $x$ is the extension of the wire on application of the force.
Hence $W_{a} = k L_{a}$
and $W_{w} = k L_{w}$
The weight in water is the resultant of weight in air and the buoyant force acting upwards.
Hence $W_{a} - W_{w} = V ρ g$
and $W_{a} = V d g$
$⟹$ Relative density of material of weight= $\frac{d}{ρ}$
$= \frac{W_{a}}{W_{a} - W_{w}}$
$= \frac{L_{a}}{L_{a} - L_{w}}$

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A steel wire is suspended vertically from a rigid support When loaded with a weight in air it expands by La and when the weight is immersed completely in water the extension is reduced to Lw Then relative density of the material of the weight is

A steel wire is suspended vertically from a rigid support When loaded with a weight in air it expands by $L_{a}$ and when the weight is immersed completely in water the extension is reduced to $L_{w}$ Then relative density of the material of the weight is