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

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

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

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

answer is B.

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Detailed Solution

$Y = \frac{F L}{A e}$

In air $Y = \frac{ρ V g L}{A l_{a}}$ (1)

In water $y = \frac{(m g - F_{b})}{A l_{w}} Y = \frac{(ρ V g - V \times 1 \times g) L}{A l_{w}} (2) f r o m 1 a n d 2 e q u$

Solving $δ = \frac{l_{a}}{l_{a} - l_{w}}$

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