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(B) Let $V$ be the volume of the load and $\rho$ be its relative density.When the weight is in air,the tension in the wire is $T_a = V \rho g$. The ex

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

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(B) Let $V$ be the volume of the load and $\rho$ be its relative density.When the weight is in air,the tension in the wire is $T_a = V \rho g$. The extension is $l_a = \frac{T_a L}{AY} = \frac{V \rho g L}{AY} \quad ...(1)$When the weight is immersed in water,the buoyant force acts upwards. The effective weight (tension) becomes $T_w = V \rho g - V \sigma g$,where $\sigma$ is the density of water. Since $\rho$ is relative density,$\sigma = 1$. So,$T_w = Vg(\rho - 1)$.The extension is $l_w = \frac{T_w L}{AY} = \frac{Vg(\rho - 1)L}{AY} \quad ...(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{l_a}{l_w} = \frac{\rho}{\rho - 1}$$l_a(\rho - 1) = l_w \rho$$l_a \rho - l_a = l_w \rho$$\rho(l_a - l_w) = l_a$$\rho = \frac{l_a}{l_a - l_w}$

$A$ 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 the material of the weight is

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