Answer
Verified
410.7k+ views
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:
$k=\dfrac{\gamma A}{l}$
$F=k\Delta l$
Complete answer:
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.
Formulas used:
$k=\dfrac{\gamma A}{l}$
$F=k\Delta l$
Complete answer:
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE