In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1 m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

Young's modulus is given by $$Y = \frac{FL}{\pi r^2 \Delta L}$$, where $$F$$ is the applied force, $$L$$ is the original length, $$r$$ is the radius, and $$\Delta L$$ is the elongation.

The fractional error in $$Y$$ is obtained by adding the fractional errors of each measured quantity. Since $$r$$ appears squared, its fractional error is doubled: $$\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta(\Delta L)}{\Delta L}$$.

The force is $$F = mg$$, so $$\frac{\Delta F}{F} = \frac{\Delta m}{m} = \frac{1 \text{ g}}{1000 \text{ g}} = 0.001$$.

The fractional error in length is $$\frac{\Delta L}{L} = \frac{1 \text{ mm}}{1000 \text{ mm}} = 0.001$$.

The fractional error in radius is $$\frac{\Delta r}{r} = \frac{0.001 \text{ cm}}{0.2 \text{ cm}} = 0.005$$, so $$2\frac{\Delta r}{r} = 0.01$$.

The fractional error in elongation is $$\frac{\Delta(\Delta L)}{\Delta L} = \frac{0.001 \text{ cm}}{0.5 \text{ cm}} = 0.002$$.

Adding all contributions: $$\frac{\Delta Y}{Y} = 0.001 + 0.001 + 0.01 + 0.002 = 0.014 = 1.4\%$$.

Therefore, the fractional error in the value of Young's Modulus is $$1.4\%$$.