Question

Two identical rods of the identical dimension and the young’s modulus ${{Y}_{1}}\text{ and }{{Y}_{2}}$ are joined end to end. The equivalent young’s modulus for the composite rod is:
$\begin{aligned} & A.\text{ }\dfrac{{{Y}_{1}}+{{Y}_{2}}}{2} \\\ & B.\text{ }\dfrac{2{{Y}_{1}}{{Y}_{2}}}{{{Y}_{1}}+{{Y}_{2}}} \\\ & C.\text{ }{{Y}_{1}}+{{Y}_{2}} \\\ & D.\text{ }\dfrac{{{Y}_{1}}{{Y}_{2}}}{{{Y}_{1}}+{{Y}_{2}}} \\\ \end{aligned}$

Answer 1

It is given that dimension of the both rods are equal then we can calculate extensions for both the rods and put the values one equivalent young’s modulus formula

Formula used:
$\begin{aligned} & \sigma =\dfrac{F}{A} \\\ & \varepsilon =\dfrac{\delta }{l} \\\ & Y=\dfrac{\sigma }{\varepsilon } \\\ \end{aligned}$

Complete step by step solution:
It is given that in the question those two rods have the identical dimension; hence their area will be the same.

Now formula for the young’s modulus,
$Y=\dfrac{\sigma }{\varepsilon }$
Where, Y = young’s modulus
σ = stress
ε = strain

If ${{\delta }_{1}}$ is the extension for the rod 1 then,
$\begin{aligned} & \Rightarrow {{Y}_{1}}=\dfrac{F}{A}\times \dfrac{l}{{{\delta }_{1}}} \\\ & \left( \because \sigma =\dfrac{F}{A},\varepsilon =\dfrac{{{\delta }_{1}}}{l} \right) \\\ & \therefore {{\delta }_{1}}=\dfrac{Fl}{A{{Y}_{1}}}....\left( 1 \right) \\\ \end{aligned}$
Where, F = force
l = actual length of the rod
A = area.
Y = young’s modulus

Now similarly if the extension for rod 2 is ${{\delta }_{2}}$then,
$\begin{aligned} & \Rightarrow {{Y}_{1}}=\dfrac{F}{A}\times \dfrac{l}{{{\delta }_{2}}} \\\ & \therefore {{\delta }_{2}}=\dfrac{Fl}{A{{Y}_{1}}}....\left( 2 \right) \\\ \end{aligned}$

Now in equivalent young’s modulus total length will be 2l and the extension will be addition of the both rod’s extension.

Formula for the equivalent young’s modulus,
$Y=\dfrac{F\times 2l}{A\times \left( {{\delta }_{1}}+{{\delta }_{2}} \right)}....\left( 3 \right)$

Now substitute the values of the equation (1) and the equation (2) in the equation (3)
$\begin{aligned} & \Rightarrow \dfrac{2Fl}{A\times \left( \dfrac{Fl}{A{{Y}_{1}}}+\dfrac{Fl}{A{{Y}_{2}}} \right)} \\\ & \Rightarrow \dfrac{2Fl}{\dfrac{A\times Fl}{A}\left( \dfrac{1}{{{Y}_{1}}}+\dfrac{1}{{{Y}_{2}}} \right)} \\\ & \Rightarrow \dfrac{D}{\left( \dfrac{{{Y}_{1}}+{{Y}_{2}}}{{{Y}_{1}}{{Y}_{2}}} \right)} \\\ & \therefore Y=\dfrac{2{{Y}_{1}}{{Y}_{2}}}{{{Y}_{1}}+{{Y}_{2}}} \\\ \end{aligned}$

Hence, the option (B) is correct.

Note:
As an approach to solve this question first we derive a formula for the extension for the both the rods. Then we put the values of the extensions of the two rods in one equivalent young’s modulus formula, which gives us, correct formula for equivalent young’s modulus.