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

\frac{l_{a}}{l_{w}}

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\frac{l_{a}}{l_{a} - l_{w}}

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\frac{l_{w}}{(l_{a} - l_{w})}

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\frac{l_{w}}{l_{a}}

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Solution

The correct option is B $\frac{l_{a}}{l_{a} - l_{w}}$
Let $V$ be the volume of the load and $ρ$ its relative density
So, $Y = \frac{F L}{A l_{a}} = \frac{V ρ g L}{A l_{a}}$ .....(1)
When the load is immersed in the liquid, then $Y = \frac{F^{'} L}{A l_{w}} = \frac{(V ρ g - V \times 1 \times g) L}{A l_{w}}$ .....(2)
( $∵$ Now net weight= weight-upthrust)
From Eqns (1) and (2), we get
$\frac{ρ}{l_{a}} = \frac{(ρ - 1)}{l_{w}}$
$\Rightarrow ρ = \frac{l_{a}}{(l_{a} - l_{w})}$

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

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