$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$
Where $F$ is the force applied, $L$ is the length, $D$ is the diameter and $\Delta L$ is the extension of the wire respectively.
As each wire is made up of same material therefore their $Young's$ modulus is same for each wire.
Foe all the four wires, $Y,F\,(=tension)$ are the same.
$\therefore \Delta L \propto \frac{L}{{{D^2}}}$
$In\,\,\left( a \right)\,\,\frac{L}{{{D^2}}} = \frac{{200\,cm}}{{{{\left( {0.2\,cm} \right)}^2}}} = 5 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( b \right)\,\,\,\frac{L}{{{D^2}}} = \frac{{300\,cm}}{{{{\left( {0.3\,cm} \right)}^2}}} = 3.3 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( c \right)\,\,\frac{L}{{{D^2}}} = \frac{{50\,cm}}{{{{\left( {0.05\,cm} \right)}^2}}} = 20 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( d \right)\,\,\frac{L}{{{D^2}}} = \frac{{100\,cm}}{{{{\left( {0.1\,cm} \right)}^2}}} = 10 \times {10^3}\,\,c{m^{ - 1}}$
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