Answer
413.4k+ views
Hint : To solve this question, we need to use the formula for the Young’s modulus of a string in terms of its geometrical parameters. Then, putting the values given in the question, we can get the required value of the Young’s modulus of the material wire.
Formula used: The formula which has been used to solve this question is given by
$ Y = \dfrac{{Fl}}{{A\Delta l}} $ , here $ Y $ is the young’s modulus of a string of length $ l $ and area of cross section $ A $ , $ F $ is the force applied on it due to which its length gets changed by $ \Delta l $ .
Complete step by step answer
We know that the Young’s modulus for a wire can be written as
$ Y = \dfrac{{Fl}}{{A\Delta l}} $ (1)
According to the question, the length of the wire is $ l = 10m $ , the area of cross section of the wire is $ A = 2m{m^2} = 2 \times {10^{ - 6}}{m^2} $ , the load on the wire is equal to the weight of $ 1kg $ mass. So the force becomes
$ F = 1g $
Substituting $ g = 9.8m{s^{ - 2}} $ , we get
$ F = 9.8N $
Also, the elongation of the wire is given to be of $ 0.001\% $ . So the elongation is given by
$ \Lambda l = \dfrac{{0.001}}{{100}} \times l $
As $ l = 10m $ , so we have
$ \Delta l = \dfrac{{0.001}}{{100}} \times 10 $
$ \Rightarrow \Delta l = {10^{ - 4}}m $
Substituting these values in (1) we get
$ Y = \dfrac{{9.8 \times 10}}{{2 \times {{10}^{ - 6}} \times {{10}^{ - 4}}}} $
On solving we get
$ Y = 4.9 \times {10^{11}}N/{m^2} $
Thus the young’s modulus of the material of the wire is equal to $ 4.9 \times {10^{11}}N/{m^2} $ .
Note
We should not forget to convert the value of the area given in this question into the Si units. It is given in square millimeters. So it has to be converted into square meters. Also, in this question, the value of the length is not needed. This is because the elongation is given in the form of the length, so solving the length would get cancelled.
Formula used: The formula which has been used to solve this question is given by
$ Y = \dfrac{{Fl}}{{A\Delta l}} $ , here $ Y $ is the young’s modulus of a string of length $ l $ and area of cross section $ A $ , $ F $ is the force applied on it due to which its length gets changed by $ \Delta l $ .
Complete step by step answer
We know that the Young’s modulus for a wire can be written as
$ Y = \dfrac{{Fl}}{{A\Delta l}} $ (1)
According to the question, the length of the wire is $ l = 10m $ , the area of cross section of the wire is $ A = 2m{m^2} = 2 \times {10^{ - 6}}{m^2} $ , the load on the wire is equal to the weight of $ 1kg $ mass. So the force becomes
$ F = 1g $
Substituting $ g = 9.8m{s^{ - 2}} $ , we get
$ F = 9.8N $
Also, the elongation of the wire is given to be of $ 0.001\% $ . So the elongation is given by
$ \Lambda l = \dfrac{{0.001}}{{100}} \times l $
As $ l = 10m $ , so we have
$ \Delta l = \dfrac{{0.001}}{{100}} \times 10 $
$ \Rightarrow \Delta l = {10^{ - 4}}m $
Substituting these values in (1) we get
$ Y = \dfrac{{9.8 \times 10}}{{2 \times {{10}^{ - 6}} \times {{10}^{ - 4}}}} $
On solving we get
$ Y = 4.9 \times {10^{11}}N/{m^2} $
Thus the young’s modulus of the material of the wire is equal to $ 4.9 \times {10^{11}}N/{m^2} $ .
Note
We should not forget to convert the value of the area given in this question into the Si units. It is given in square millimeters. So it has to be converted into square meters. Also, in this question, the value of the length is not needed. This is because the elongation is given in the form of the length, so solving the length would get cancelled.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)