Question

A student performs an experiment to determine the Young’s modulus of a wire exactly $ 2m $ long by Searle’s method. In a particular reading, he measures the extension in the length of the wire to be $ 0.8mm $ with an uncertainty of $ \pm 0.05mm $ at a load of exactly $ 1.0kg $ . He also measures the diameter of the wire to be $ 0.4mm $ with an uncertainty of $ \pm 0.01mm $ . The young’s Modulus obtained from the reading is:
(Given Young’s modulus of a wire is given by $ Y = \dfrac{{Fl}}{{Ae}} $ , where $ F $ is the force applied on the wire, $ l $ is the length of the wire, $ A $ is the area of cross section of the wire and $ e $ is the elongation produced in the wire due to the load. Take $ g = 9.8m/{s^2} $ )
(A) $ (2 \pm 0.3) \times {10^{11}}N/{m^2} $
(B) $ (2 \pm 0.2) \times {10^{11}}N/{m^2} $
(C) $ (2 \pm 0.1) \times {10^{11}}N/{m^2} $
(D) $ (2 \pm 0.5) \times {10^{11}}N/{m^2} $

Answer 1

Hint: We know that the young’s modulus could be found out by, $ Y = \dfrac{{Fl}}{{Ae}} $ . But we also know that the area can be given by, $ A = \pi {R^2} = \pi {\left( {\dfrac{d}{2}} \right)^2} = \pi \left( {\dfrac{d}{4}} \right) $ . So we need to substitute this value in the formula for Young’s modulus. Then substituting the values given in the question we will get the answer.

Formulas used We will use the formula $ Y = \dfrac{{Fl}}{{Ae}} $ where $ F $ is the force applied, $ l $ is the length of the body, $ A $ is the area, and $ e $ is the elongation produced.
We will also use the formula $ A = \pi {R^2} = \pi {\left( {\dfrac{d}{2}} \right)^2} = \pi \left( {\dfrac{{{d^2}}}{4}} \right) $ where $ A $ is the area, $ R $ is the radius and $ d $ is the diameter.

Complete Step by Step Solution
The young’s Modulus can be found out by, $ Y = \dfrac{{Fl}}{{Ae}} $ but we also know that $ A = \pi {R^2} = \pi \left( {\dfrac{{{d^2}}}{4}} \right) $ , Thus substituting $ A $ in $ Y $ , We can get , $ Y = \dfrac{{Fl}}{{\pi \left( {\dfrac{{{d^2}}}{4}} \right)e}} $ .
We also know from the problem the values of diameter of the wire, $ d = 0.4mm $ and its uncertainty to be $ \Delta d = \pm 0.01mm $ , Also the extension in the wire when a load of mass $ m = 1.0kg $ is connected will be , $ e = 0.8mm $ and its uncertainty will be $ \Delta e = \pm 0.05mm $ .
We also know that the force is given by, $ F = mg $ and $ g = 9.8m/{s^2} $ .
 $ Y = \dfrac{{1.0 \times 9.8 \times 2}}{{\pi \left( {\dfrac{{{{0.4}^2}}}{4}} \right) \times 0.8}} $
Solving for $ Y $ we get, $ Y = 194.96 \times {10^9}N/{m^2} \simeq 2 \times {10^{11}}N/{m^2} $
Similarly, we know to find the accuracy of $ Y $ ,
 $ \dfrac{{\Delta Y}}{Y} = - \dfrac{{2\Delta D}}{D} - \dfrac{{\Delta e}}{e} $
Substituting the values, we know,
 $ \dfrac{{\Delta Y}}{Y} = - \dfrac{{2(0.01)}}{{(0.4)}} - \dfrac{{0.05}}{{0.8}} $
Solving the equation, we get,
 $ \dfrac{{\Delta Y}}{Y} = - 0.1125 $
We now know that $ Y \simeq 2 \times {10^{11}}N/{m^2} $ so,
 $ \Delta Y = - 0.1125 \times Y = - 0.1125 \times 2 \times {10^{11}} $
 $ \Rightarrow \Delta Y = 0.225 \times {10^{11}} $
Thus, the value of $ Y $ will be $ Y = \left( {2 \pm 0.2} \right) \times {10^{11}}N/{m^2} $
So the correct answer will be option B.

Note
Here we can see that the terms $ F $ and $ l $ have been omitted out of the accuracy expression, this is because in the above problem we know that length of wire is $ 2m $ exactly (hence no change ) and the force is also fixed.