A steel wire is suspended vertically from a r | vedclass

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Question

Let $\mathrm{V}$ be the volume of the $1$ load and $\rho$ its relative density

So, $Y=\frac{F L}{A \ell_{a}}=\frac{V \rho g L}{A \ell_{a}}$ &n

Vedclass · Accepted Answer

$\frac{{{l_a}}}{{{l_a} - {l_w}}}$

Vedclass · Answer

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

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

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