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# A Brass rod of length $2m$ and cross sectional area $2c{{m}^{2}}$ is attached end-to-end to a steel rod of length L and cross sectional area $1c{{m}^{2}}$. The compound rod is subjected to equal and opposite pulls of magnitude $5\times {{10}^{4}}N$ at its ends. If the elongations of the two rods are equal, the length of the steel rod (L) is (${{\gamma }_{Brass}}=1\times {{10}^{11}}N{{m}^{-2}}$ and ${{\gamma }_{Steel}}=2\times {{10}^{11}}N{{m}^{-2}}$)(A). $1.5m$ (B). $1.8m$(C). $1m$(D). $2m$

Last updated date: 16th Jun 2024
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Hint: A rod made of connecting a brass and a steel rod experience force from its ends due to which it undergoes a strain. The spring constant depends on the young’s modulus, area of cross section and the length while according to the hooke's law, force is the product of spring constant and change in length. Using equations from both relations, we can calculate the length of steel rod.
Formulas used:
$k=\dfrac{\gamma A}{l}$
$F=k\Delta l$

Given, a brass rod has length $2m$ and cross sectional area $2c{{m}^{2}}$and the young’s modulus is ${{\gamma }_{Brass}}=1\times {{10}^{11}}N{{m}^{-2}}$. Therefore, the spring constant of Brass is given by-
$k=\dfrac{\gamma A}{l}$ - (1)
Here, $k$ is the spring constant
$\gamma$ is the young’s modulus
$A$ is area of cross section
$l$ is the length
Substituting values for Brass in eq (1), we get,
$k=\dfrac{1\times {{10}^{11}}\times 2\times {{10}^{-4}}}{2}={{10}^{7}}N{{m}^{-1}}$
For steel rod, length is L and the cross sectional area is $1c{{m}^{2}}$ and the young’s modulus is ${{\gamma }_{Steel}}=2\times {{10}^{11}}N{{m}^{-2}}$
Substituting values for steel in eq (1), we get,
$k'=\dfrac{2\times {{10}^{11}}\times 1\times {{10}^{-4}}}{L}=\dfrac{2\times {{10}^{7}}}{L}N{{m}^{-1}}$
According to Hooke's law,
$F=k\Delta l$ - (2)
Here, $F$ is the force applied
$k$ is the spring constant
$\Delta l$ is the change in length
The change in length for both roods is equal as is the force acting on both rods. From eq (2), the spring constant of both rods is also equal, therefore,
\begin{align} & {{10}^{7}}=\dfrac{2\times {{10}^{7}}}{L} \\ & \therefore L=2m \\ \end{align}
Therefore, the length of the steel rod is $2m$. Hence, the correct option is (D).

Note:
The young’s modulus of a material is the measure of elongation or compression before it reaches the elastic limit. The force is negative of the product of spring constant and changes in length because it is the restoring force which develops in the body opposite to the force applied.