In an experiment to determine the Young's modulus of wire of a length exactly $$1 \text{ m}$$, the extension in the length of the wire is measured as $$0.4 \text{ mm}$$ with an uncertainty of $$\pm 0.02 \text{ mm}$$ when a load of $$1 \text{ kg}$$ is applied. The diameter of the wire is measured as $$0.4 \text{ mm}$$ with an uncertainty of $$\pm 0.01 \text{ mm}$$. The error in the measurement of Young's modulus $$(\Delta Y)$$ is found to be $$x \times 10^{10} \text{ N m}^{-2}$$. The value of $$x$$ is ______.

Young's modulus is given by:

$$Y = \dfrac{FL}{\pi r^2 \Delta L} = \dfrac{4FL}{\pi d^2 \Delta L}$$

We start by substituting the values $$F = mg = 1 \times 9.8 = 9.8 \text{ N}$$, $$L = 1 \text{ m}$$, $$d = 0.4 \text{ mm} = 4 \times 10^{-4} \text{ m}$$, and $$\Delta L = 0.4 \text{ mm} = 4 \times 10^{-4} \text{ m}$$ into the expression for Y:

$$Y = \dfrac{4 \times 9.8 \times 1}{\pi \times (4 \times 10^{-4})^2 \times 4 \times 10^{-4}}$$

This gives

$$Y = \dfrac{39.2}{\pi \times 16 \times 10^{-8} \times 4 \times 10^{-4}} = \dfrac{39.2}{\pi \times 6.4 \times 10^{-11}}$$

Therefore,

$$Y = \dfrac{39.2}{2.0106 \times 10^{-10}} = 1.95 \times 10^{11} \text{ N m}^{-2}$$

Next, the fractional error in Y is expressed as

$$\dfrac{\Delta Y}{Y} = \dfrac{\Delta F}{F} + \dfrac{\Delta L}{L} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta(\Delta L)}{\Delta L}$$

Since F and L have no uncertainty mentioned, this reduces to

$$\dfrac{\Delta Y}{Y} = 2 \times \dfrac{0.01}{0.4} + \dfrac{0.02}{0.4} = 0.05 + 0.05 = 0.10$$

Hence, the absolute error in Y becomes

$$\Delta Y = 0.10 \times 1.95 \times 10^{11} = 1.95 \times 10^{10} \approx 2 \times 10^{10} \text{ N m}^{-2}$$

Therefore, the value of $$x$$ is 2.