A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $P \times 10^{11} \, \text{Nm}^{-2}$, where the value of $P$ is: (Take $g = 3\pi \, \text{m/s}^2$)

<p>Let’s break down the solution step by step:</p> <p>First, note the given values:</p> <p><p>Length of the wire, $L = 3 \, \text{m}$</p></p> <p><p>Radius of the wire, $r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$</p></p> <p><p>Extension of the wire, $\Delta L = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m}$</p></p> <p><p>Mass attached, $m = 50 \, \text{kg}$</p></p> <p><p>Acceleration due to gravity, $g = 3\pi \, \text{m/s}^2$</p></p> <p><p>The force (weight) acting on the wire is:</p> <p>$F = mg = 50 \times 3\pi = 150\pi \, \text{N}$</p></p> <p><p>Next, calculate the cross-sectional area, $A,$ of the wire:</p> <p>$$A = \pi r^2 = \pi (3 \times 10^{-3})^2 = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6} \, \text{m}^2$$</p></p> <p><p>Young’s modulus, $E,$ is given by:</p> <p>$E = \frac{FL}{A \Delta L}$</p></p> <p><p>Substitute the values into the formula:</p> <p>$E = \frac{(150\pi) \times 3}{(9\pi \times 10^{-6}) \times (1 \times 10^{-4})}$</p></p> <p><p>Simplify step by step:</p></p> <p><p>Multiply in the numerator:</p> <p>$150\pi \times 3 = 450\pi$</p></p> <p><p>Multiply in the denominator:</p> <p>$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$</p></p> <p><p>Cancel the common factor $\pi$ in the numerator and denominator:</p> <p>$E = \frac{450}{9 \times 10^{-10}}$</p></p> <p><p>Divide 450 by 9:</p> <p>$$ E = \frac{450}{9} \times \frac{1}{10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11} \, \text{Nm}^{-2} $$</p></p> <p><p>The problem states that Young's modulus can be expressed as $P \times 10^{11} \, \text{Nm}^{-2}.$ Here we have:</p> <p>$P = 5$</p></p> <p>Thus, the correct option is:</p> <p>Option D (5).</p>