Question

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

• A. $\frac{l_a}{l_w}$
• B. $\frac{l_a}{l_a-l_w}$
• C. $\frac{l_w}{l_a-l_w}$
• D. $\frac{l_w}{l_a}$

Answer 1

The correct option is B $\frac{l_a}{l_a-l_w}$

Let $V$ be the volume of the load and $\rho$ its relative density
So, $Y=\frac{FL}{A{l}_a}=\frac{V\rho gL}{A{l}_a}...\left(1\right)$
When the load is immersed in the liquid, then
$Y=\frac{F^{\prime }L}{A{l}_w}=\frac{(V\rho g-V\times {\rho }_w\times g)L}{A{l}_w}...\left(2\right)$
($\because$ Now net weight = weight - upthrust)
From eqs. (1) and (2), we get
$\frac{\rho }{l_a}=\frac{(\rho -{\rho }_w)}{l_w}$ or $\frac{\rho }{{\rho }_w}=\frac{l_a}{(l_a-l_w)}$