
There is some change in length when a 33000N tensile force is applied on a steel rod of area of cross-section ${{10}^{-3}}{{m}^{2}}$. The change of temperature required to produce the same elongation, if the steel rod is heated is (The modulus of elasticity is $3\times {{10}^{11}}N/{{m}^{2}}$and the coefficient of linear equation if steel is $1.1\times {{10}^{-5}}{{/}^{0}}C$
A. \[20^{0}C\]
B. ${{15}^{0}}C$
C. ${{10}^{0}}C$
D. ${{0}^{0}}C$
Answer
562.2k+ views
Hint:We are given that a force is being exerted, we know force can produce some extension in length and that can be found out using Young’s Modulus. We also know that change in temperature can also result in extension. So, we can find out the extension in the two cases and since the change is the same, we can equate both to find out the change in temperature.
Complete step by step answer:
We know Young’s modulus is given by \[Y=\dfrac{Fl}{A\Delta l}\], where F is the force applied, l is the original length and A is the area. So, elongation produced can be given as \[\Delta l=\dfrac{Fl}{AY}\]--(1)
Also, elongation in length due to change of temperature is given by $\Delta l=l\alpha \Delta T$, where $\Delta l $is the change in the length and $\alpha $ is the coefficient of linear expansion.
Since, the change in length is equal in both the case, equating this with the equation (1) we get,
$\dfrac{Fl}{AY}=l\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY}=\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY\alpha }=\Delta T$
Putting the values, we get,
$\Delta T=\dfrac{33000}{{{10}^{-3}}\times 3\times {{10}^{11}}\times 1.1\times {{10}^{-5}}}$
$\therefore\Delta T={{10}^{0}}C$
So, the correct option is C.
Note:Young’s modulus is dependent on the nature of the material and it does not depend on other parameters such as length or area of the cross-section or the temperature.Here the area was uniform and thus we did not have to take into account deformation of shape. If we are given a problem in which two wires of different materials, then we have to use two sets of young’s moduli for each of them.
Complete step by step answer:
We know Young’s modulus is given by \[Y=\dfrac{Fl}{A\Delta l}\], where F is the force applied, l is the original length and A is the area. So, elongation produced can be given as \[\Delta l=\dfrac{Fl}{AY}\]--(1)
Also, elongation in length due to change of temperature is given by $\Delta l=l\alpha \Delta T$, where $\Delta l $is the change in the length and $\alpha $ is the coefficient of linear expansion.
Since, the change in length is equal in both the case, equating this with the equation (1) we get,
$\dfrac{Fl}{AY}=l\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY}=\alpha \Delta T$
$\Rightarrow\dfrac{F}{AY\alpha }=\Delta T$
Putting the values, we get,
$\Delta T=\dfrac{33000}{{{10}^{-3}}\times 3\times {{10}^{11}}\times 1.1\times {{10}^{-5}}}$
$\therefore\Delta T={{10}^{0}}C$
So, the correct option is C.
Note:Young’s modulus is dependent on the nature of the material and it does not depend on other parameters such as length or area of the cross-section or the temperature.Here the area was uniform and thus we did not have to take into account deformation of shape. If we are given a problem in which two wires of different materials, then we have to use two sets of young’s moduli for each of them.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a labelled diagram of the human heart and label class 11 biology CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

