Question

An iron wire of length $ 4m $ and diameter $ 2mm $ is loaded with a weight of $ 8kg $ , if the young’s modulus $ 'Y' $ for iron is $ 2 \times {10^{11}}N{m^{ - 2}} $ then the increase in length of the wire is
(A) $ 0.2mm $
(B) $ 0.5mm $
(C) $ 2mm $
(D) $ 1mm $

Answer 1

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 increase in length of the 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)
We know that the area of cross section of a wire is given by
 $ A = \pi {r^2} $
Where $ r $ is the radius of the wire. Now, as the radius is equal to half the diameter, that is, $ r = \dfrac{D}{2} $ , so the area of cross section becomes
 $ A = \pi {\left( {\dfrac{D}{2}} \right)^2} $
 $ \Rightarrow A = \dfrac{{\pi {D^2}}}{4} $ (2)
Substituting (2) in (1) we have
 $ Y = \dfrac{{4Fl}}{{\pi {D^2}\Delta l}} $
So the increase in the length of the wire is given by
 $ \Delta l = \dfrac{{4Fl}}{{\pi {D^2}Y}} $ (3)
According to the question, the length of the wire is $ l = 4m $ , the young’s modulus of the wire is $ Y = 2 \times {10^{11}}N{m^{ - 2}} $ the diameter of the wire is $ D = 2mm = 2 \times {10^{ - 3}}m $ .
Also the load on the wire is equal to the weight of $ 8kg $ mass. So the force becomes
 $ F = 8g $
Substituting $ g = 9.8m{s^{ - 2}} $ , we get
 $ F = 78.4N $
Substituting these values in (3) we get
 $ \Delta l = \dfrac{{4 \times 78.4 \times 4}}{{\pi {{\left( {2 \times {{10}^{ - 3}}} \right)}^2} \times 2 \times {{10}^{11}}}} $
On solving we get
 $ \Delta l = 49.9 \times {10^{ - 5}}m $
We know that $ 1m = 1000mm $ . So we get
 $ \Delta l = 49.9 \times {10^{ - 5}} \times 1000mm $
 $ \Rightarrow \Delta l = 0.499mm \approx 0.5mm $
Thus, the increase in the length of the wire is equal to $ 0.5mm $ .
Hence, the correct answer is option B.

Note
We should not forget to convert the values of the quantities given in the question into their respective SI units. In this question the length of the wire was given in millimeters, which is not an SI unit. So it was supposed to be converted into meters.