If the ratio of lengths, radii and Youngs modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be:

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

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\frac{3 a}{2 b^{2} c}

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\frac{3 c}{2 a b^{2}}

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

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Solution

The correct option is B $\frac{3 a}{2 b^{2} c}$
Ref image
Given ; $l_{S} / l_{B} = a, \frac{y_{S}}{y_{B}} = c$
$\frac{r_{S}}{r_{B}} = b$
By definition, young's Modulus = $Y = \frac{l i n e a r s t r e s s}{l i n e a r s t r a i n}$
$\Rightarrow Y = \frac{T / A}{Δ l / l}$
where, T - tension in wire
A - cross sectional area of wire
$Δ l$ - change in length
l - original length
$\Rightarrow Δ l = \frac{T . l}{Y . A}$ __(i)
FBD : Ref. image $T_{3} = m g + T_{B}$ __(ii) Ref. image $T_{B} = 2 m g$ __(iii)
Put (iii) in (ii) : $T_{S} = m g + 2 m g = 3 m g$
$\Rightarrow \frac{T_{S}}{T_{B}} = \frac{3 m g}{2 m g} = \frac{3}{2}$ __(iv)
From (i) :
$\frac{Δ l_{S}}{Δ l_{B}} = \frac{(\frac{T_{S} l_{S}}{Y_{S} A_{S}})}{(\frac{T_{B} l_{B}}{Y_{B} A_{B}})} \begin{matrix} = (\frac{T_{S}}{T_{B}}) . (\frac{l_{S}}{l_{B}}) \frac{Y_{B} A_{B}}{Y_{S} A_{S}} & / f o r a w i r e : A = π r^{2} = \frac{3}{2} \times a \times \frac{1}{c} \times \frac{π r_{B}^{2}}{π r_{S}^{2}} & / f r o m g i v e n r a t i o s \end{matrix}$
$\frac{Δ l_{S}}{Δ l_{B}} = \frac{3 a}{2 c . b^{2}}$
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