Young's modulus is determined by the equation given by $$Y = 49000 \frac{m}{l} \frac{dyn}{cm^2}$$ where $$M$$ is the mass and $$l$$ is the extension of wire used in the experiment. Now error in Young modulus $$(Y)$$ is estimated by taking data from $$M - l$$ plot in graph paper. The smallest scale divisions are $$5 \text{ g}$$ and $$0.02 \text{ cm}$$ along load axis and extension axis respectively. If the value of $$M$$ and $$l$$ are $$500 \text{ g}$$ and $$2 \text{ cm}$$ respectively then percentage error of $$Y$$ is :

We need to find the percentage error in Young's modulus $$Y = 49000\frac{M}{l}$$.

For $$Y = k\frac{M}{l}$$, taking the logarithm gives $$\ln Y = \ln k + \ln M - \ln l$$, and differentiating yields $$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l}$$.

From the graph paper, the smallest scale division along the load axis is 5 g, so $$\Delta M = 5$$ g, and along the extension axis it is 0.02 cm, giving $$\Delta l = 0.02$$ cm.

$$\frac{\Delta Y}{Y} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 = 2\%$$

The correct answer is Option (2): 2%.