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Question: In the Searle’s method to determine the young’s modulus of a wire, a steel of length \(156\;{\rm{cm}...

In the Searle’s method to determine the young’s modulus of a wire, a steel of length 156  cm156\;{\rm{cm}} and diameter 0.054  cm0.054\;{\rm{cm}} is taken as experimental wire. The average increase in length for 1.5  kg  wt1.5\;{\rm{kg}}\;{\rm{wt}} is found to be 0.050  cm0.050\;{\rm{cm}}. Then the young’s modulus of the wire is
(A) 3.002×1011  N/m23.002 \times {10^{11}}\;N/{m^2}
(B) 1.002×1011  N/m21.002 \times {10^{11}}\;N/{m^2}
(C) 2.002×1011  N/m22.002 \times {10^{11}}\;N/{m^2}
(D) 2.5×1011  N/m22.5 \times {10^{11}}\;N/{m^2}

Explanation

Solution

In this question, you need to be clear with the concept and formulas of stress, strain, and Young’s modulus. First, you need to calculate stress and strain by substituting the values in their formulas. Later you need to substitute the values in the formula of Young’s modulus, in order to get the answer.

Complete step by step answer:
Given Data
The initial length of the steel wire is 156 cm.
The diameter of the steel wire is 0.054 cm.
The change in length of the steel wire is ΔL=0.050  cm\Delta L = 0.050\;{\rm{cm}}.
The mass of the steel wire is 1.5 kg wt.

We know that the acceleration due to gravity is g=10  m/s2g = 10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}.
We can be easily calculate the radius of the steel wire as,
r=d2   ⟹  r=0.054  cm×10−2  m1  cm2   ⟹  r=2.7×10−4  m r = \dfrac{d}{2}\\\ \implies r = \dfrac{{0.054\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}{2}\\\ \implies r = 2.7 \times {10^{ - 4}}\;{\rm{m}}

We know that the expression for the area of cross-section of the wire is given as,
A=πr2A = \pi {r^2}.

We now substitute the values in above expression,
\begin{array} A = \pi \times {\left( {2.7 \times {{10}^{ - 4}}\;{\rm{m}}} \right)^2}\\\ A = 2.28 \times {10^{ - 7}}\;{{\rm{m}}^{\rm{2}}} \end{array}

We can calculate the stress in the steel wire as,
σ=mgA   ⟹  σ=1.5  kg  wt×10  m/s22.28×10−7  m2   ⟹  σ=6.57×107  N/m2 \sigma = \dfrac{{mg}}{A}\\\ \implies \sigma = \dfrac{{1.5\;{\rm{kg}}\;{\rm{wt}} \times 10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{2.28 \times {{10}^{ - 7}}\;{{\rm{m}}^2}}}\\\ \implies \sigma = 6.57 \times {10^7}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}

Further, we can calculate the value of strain in the steel wire as,
ε=ΔLL   ⟹  ε=0.050  cm×10−2  m1  cm156  cm×10−2  m1  cm   ⟹  ε=3.20×10−4 \varepsilon = \dfrac{{\Delta L}}{L}\\\ \implies \varepsilon = \dfrac{{0.050\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}{{156\;{\rm{cm}} \times \dfrac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}\\\ \implies \varepsilon = 3.20 \times {10^{ - 4}}

As we know that the expression for the Young’s modulus is given as,
Y=σεY = \dfrac{\sigma }{\varepsilon }

Now we substitute the values in the above expression,
\begin{array} Y = \dfrac{{6.57 \times {{10}^7}\;{\rm{N/}}{{\rm{m}}^2}}}{{3.20 \times {{10}^{ - 4}}}}\\\ Y \approx 2.002 \times {10^{11}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}} \end{array}

Therefore, the Young’s modulus of the steel wire is 2.002×1011  N/m22.002 \times {10^{11}}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}.

So, the correct answer is “Option C”.

Note:
You need to focus on the calculation part, as you can go wrong while calculating. Moreover, you can make the mistake of not calculating the radius of the wire and substituting the diameter directly. An alternative method that we can use is by substituting all the value in the formula of Young’s modulus directly.