A wire of length $ 10m $ and area of cross section $ 2sq{\text{. }}mm $ is stretched by a load of $ 1kg $ weight. An elongation of $ 0.001\% $ is produced. The Young’s modulus of the material of the wire is
(A) $ 49 \times {10^{11}}N/{m^2} $
(B) $ 4.9 \times {10^{11}}N/{m^2} $
(C) $ 9.8 \times {10^{11}}N/{m^2} $
(D) $ 98 \times {10^{11}}N/{m^2} $
Answer
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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.
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