A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends by and when the weight is immersed completely in water, the extension is reduced to $l_{w}$ .Then the relative density 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_{2}}{l_{a} - l_{w}}

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

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Solution

The correct option is B $\frac{l_{a}}{l_{a} - l_{w}}$

Case – 1 (in air)
$y = \frac{F L}{A Δ l} = \frac{F L}{A l_{a}} ∴ F_{T h r u s t} = V ρ g$
$= \frac{V ρ g L}{A l_{a}} . . . . . (1)$
Case – 2 when the load is immersed in a liquid,
$W_{n e t} = W_{b} - t r u s t = V ρ g - V \times r h o_{w} \times g = V ρ g - V \times 1 \times g Y = \frac{(V ρ g - V \times 1 \times g) L}{A l_{w}} . . . . . (2)$
Y is same, because material is same (1) = (2),
We get
$ρ = \frac{l_{a}}{l_{a} - l_{w}}$

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