$Y = \frac{{FL}}{{A\Delta L}} = \frac{{4FL}}{{\pi {D^2}\Delta L}}\,\,or\,\,\Delta L = \frac{{4FL}}{{\pi {D^2}Y}}$
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.
$\therefore \Delta L \propto \frac{L}{{{D^2}}}$
$In\,\,\left( a \right)\,\,\frac{L}{{{D^2}}} = \frac{{200\,cm}}{{{{\left( {0.2\,cm} \right)}^2}}} = 5 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( b \right)\,\,\,\frac{L}{{{D^2}}} = \frac{{300\,cm}}{{{{\left( {0.3\,cm} \right)}^2}}} = 3.3 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( c \right)\,\,\frac{L}{{{D^2}}} = \frac{{50\,cm}}{{{{\left( {0.05\,cm} \right)}^2}}} = 20 \times {10^3}\,c{m^{ - 1}}$
$In\,\,\left( d \right)\,\,\frac{L}{{{D^2}}} = \frac{{100\,cm}}{{{{\left( {0.1\,cm} \right)}^2}}} = 10 \times {10^3}\,\,c{m^{ - 1}}$



$Assertion$ $(A)$ : In Vernier calliper if positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
$Reason$ $(R)$ : The zero error in Vernier Calliper might have happened due to manufacturing defect or due to rough handling.
In the light of the above statements, choose the correct answer from the options given below :

