If the ratio of lengths, radii and Young's modulus of steel and brass wires shown in the figure are a, b Brass and c respectively, the ratio between the increase in lengths of brass and steel wires would

$\frac{b^{2} a}{2 c}$

$\frac{b c}{2 a^{2}}$

$\frac{b a^{2}}{2 c}$

$\frac{a}{2 b^{2} c}$

Solution

$\frac{a}{2 b^{2} c}$

Free body diagram of the two blocks are

$\frac{l_{1}}{l_{2}} = a, \frac{r_{1}}{r_{2}} = b, \frac{Y_{1}}{Y_{2}} = c$

Let Young's modulus of steel is Y₁ and of brass is Y₂

Y₁ = $\frac{F_{1} . l_{1}}{A_{1} ∆ l_{1}}$ .... (i)

and Y₂ = $\frac{F_{2} l_{2}}{A_{2} ∆ l_{2}}$ .... (ii)

Dividing Eq. (i) by Eq. (ii), we get

$\frac{Y_{1}}{Y_{2}} = \frac{\frac{F_{1} . l_{1}}{A_{1} . ∆ l_{1}}}{\frac{F_{2} . l_{2}}{A_{2} . ∆ l_{2}}}$

⇒ $\frac{Y_{1}}{Y_{2}} = \frac{F_{1} . A_{2} . l_{1} . ∆ l_{2}}{F_{2} . A_{1} . l_{2} . ∆ l_{1}}$ .....(iii)

Force on steel wire from free body diagram

T = F₁ = 2 g N

Force on brass wire from free body diagram

F₂ = T' = T + 2g

= 4 g N

Now, putting the values of F₁ , F₂ Eq. (iii) we get

$\frac{Y_{1}}{Y_{2}} = (\frac{2 g}{4 g}) . (\frac{π r_{2}^{2}}{π r_{1}^{2}}) . (\frac{l_{1}}{l_{2}}) . (\frac{∆ l_{2}}{∆ l_{1}})$

$c = \frac{1}{2} (\frac{1}{b^{2}}) . a (\frac{∆ l_{2}}{∆ l_{1}})$

⇒ $\frac{∆ l_{1}}{∆ l_{2}} = (\frac{a}{2 b^{2} c})$

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