Question

Three bars having length l, 2l and 3l and area of cross-section A, 2A and 3A are joined rigidly end to end. Compound rod is subjected to a stretching force F. The increase in the length of the rod is (Young’s modulus of the material is Y and bars are massless)

$A.\,\,\dfrac{13Fl}{2AY}$
$B.\,\,\dfrac{3Fl}{AY}$
$C.\,\,\dfrac{9Fl}{2AY}$
$D.\,\,\dfrac{13Fl}{AY}$

Answer 1

The formula used to calculate Young’s modulus of material should be used to solve this problem. The Young’s modulus is the ratio of the product of force and length by the product of area and the change in length.

Formula used:
$Y=\dfrac{FL}{A\Delta L}$

Complete step by step answer:
From given, we have the data,
The length of the first bar = l
The area of the first bar = A
The length of the first bar = 2l
The area of the first bar = 2A
The length of the first bar = 3l
The area of the first bar = 3A
The stretching force = F
Young’s modulus of material = Y

The formula for calculating Young’s modulus of the material is given as follows.
$Y=\dfrac{FL}{A\Delta L}$
Where F is the stretching force, L is the length of the material, A is the cross-sectional area of the material and $\Delta L$ is the change in the length of material.
$\Rightarrow \Delta L=\dfrac{FL}{AY}$
The increase in the length of a rod is the sum of the increase in the length of 3 bars, as the rod is a compound rod of these 3 bars.
Firstly, compute the increase in the length of all the 3 bars separately.
The increase in the length of the first bar is calculated as follows.
The length of the first bar = l
The area of the first bar = A
$\Delta (l)=\dfrac{Fl}{AY}$
The increase in the length of the second bar is calculated as follows.
The length of second bar = 2l
The area of the second bar = 2A

& \Delta (2l)=\dfrac{F(2l)}{(2A)Y} \\\ & \Rightarrow \Delta (2l)=\dfrac{Fl}{AY} \\\ \end{aligned}$$ The increase in the length of the third bar is calculated as follows. The length of third bar = 3l The area of the third bar = 3A $$\begin{aligned} & \Delta (3l)=\dfrac{F(3l)}{(3A)Y} \\\ & \Rightarrow \Delta (3l)=\dfrac{Fl}{AY} \\\ \end{aligned}$$ The total increase in length of the rod is calculated as follows. $$\begin{aligned} & \therefore {{(\Delta l)}_{total}}=\Delta (l)+\Delta (2l)+\Delta (3l) \\\ & \Rightarrow {{(\Delta l)}_{total}}=\dfrac{Fl}{AY}+\dfrac{Fl}{AY}+\dfrac{Fl}{AY} \\\ & \Rightarrow {{(\Delta l)}_{total}}=\dfrac{3Fl}{AY} \\\ \end{aligned}$$ As the increase in length of rod formed by joining 3 rods of length l, 2l and 3l having area A, 2A and 3A is $$\dfrac{3Fl}{AY}$$, thus, the option (B) is correct. **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem. The uniformity in the case of units of the parameters should be maintained.