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 is $$P \times 10^{11}$$ Nm$$^{-2}$$, where P is : (Take $$g = 3\pi$$ m/s$$^2$$)

The extension in a wire under a load is related to Young’s modulus $$Y$$ by the formula
$$Y = \frac{F\,L}{A\,\Delta L}$$
where
  $$F$$ = stretching force,
  $$L$$ = original length of the wire,
  $$A$$ = cross-sectional area, and
  $$\Delta L$$ = extension produced.

First, find each quantity from the data given:

Force
The load is a mass of $$50\ \text{kg}$$, so the force is its weight:
$$F = m g = 50 \times 3\pi = 150\pi\ \text{N}$$

Original length
$$L = 3\ \text{m}$$

Extension
$$\Delta L = 0.1\ \text{mm} = 0.1 \times 10^{-3}\ \text{m} = 1 \times 10^{-4}\ \text{m}$$

Cross-sectional area
The wire has radius $$r = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$$.
Area $$A = \pi r^{2} = \pi \left(3 \times 10^{-3}\right)^{2} = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6}\ \text{m}^{2}$$

Substitute into the formula for $$Y$$
$$Y = \frac{150\pi \times 3}{9\pi \times 10^{-6} \times 1 \times 10^{-4}}$$

Simplify step by step:

• Multiply force and length:
$$150\pi \times 3 = 450\pi$$

• Multiply area and extension:
$$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$$

• Divide:
$$Y = \frac{450\pi}{9\pi \times 10^{-10}} = \frac{450}{9} \times 10^{10} = 50 \times 10^{10}\ \text{N m}^{-2}$$

• Write in the required power-of-ten form:
$$50 \times 10^{10} = 5 \times 10^{11}\ \text{N m}^{-2}$$

Hence the Young’s modulus is $$P \times 10^{11}\ \text{N m}^{-2}$$ with $$P = 5$$.

Therefore, the correct option is Option A.