1. The given relation is

$Y = 49000\,\frac{M}{l}\; \frac{\text{dyne}}{\text{cm}^2}$

So, ignoring the constant, Young's modulus depends on

$Y \propto \frac{M}{l}$

2. Hence the fractional error in $Y$ is

$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l}$

because for quotient/product, relative errors add.

3. From the graph paper scales:

• Smallest division along load axis $= 5\,\text{g}$
• Smallest division along extension axis $= 0.02\,\text{cm}$

Taking the maximum reading error as half of the smallest division,

$\Delta M = \frac{5}{2} = 2.5\,\text{g}$ $\Delta l = \frac{0.02}{2} = 0.01\,\text{cm}$

4. Given:

$M = 500\,\text{g}, \qquad l = 2\,\text{cm}$

So,

$\frac{\Delta M}{M} = \frac{2.5}{500} = 0.005$

$\frac{\Delta l}{l} = \frac{0.01}{2} = 0.005$

Therefore,

$\frac{\Delta Y}{Y} = 0.005 + 0.005 = 0.01$

5. Converting into percentage:

$\%\text{ error in }Y = 0.01 \times 100 = 1\%$

6. Comparison with options:

• A: $2\%$
• B: $0.02\%$
• C: $0.5\%$
• D: $0.2\%$

None matches $1\%$.

So the physically correct result is:

$\boxed{1\%}$

The stored answer $2\%$ would arise only if full least count were used as absolute error instead of half least count:

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

But standard experimental reading error from graph paper is usually taken as half the smallest division.