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
, Diffcult
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Let $\mathrm{V}$ be the volume of the $1$ load and $\rho$ its relative density
So, $Y=\frac{F L}{A \ell_{a}}=\frac{V \rho g L}{A \ell_{a}}$ $...(1)$
When the load is immersed in the liquid, then
$Y=\frac{F^{\prime} L}{A \ell_{w}}=\frac{(V \rho g-V \times 1 \times g) L}{A \ell_{w}}$ $...(2)$
(... Now net weight $=$ weight - upthrust) From eqs. $( 1)$ and $(2),$ we get
$\frac{\rho}{\ell_{\mathrm{a}}}=\frac{(\rho-1)}{\ell_{\mathrm{w}}}$ or $\rho=\frac{\ell_{\mathrm{a}}}{\left(\ell_{\mathrm{a}}-\ell_{\mathrm{w}}\right)}$
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