Elasticity And Young's Modulus

As stated in NCERT Class 11, Chapter 9 (Mechanical Properties of Solids), Young's modulus is defined as the ratio of tensile stress to longitudinal strain within the elastic limit, $Y=\frac{FL}{A\Delta L}$. Here stress measures the internal restoring force per unit area while strain measures fractional change in length, and their ratio is a material constant. Substituting the given values: $F=100\text{N}$, $L=2\text{m}$, $A=1{\text{mm}}^2=1\times 10^{-6}{\text{m}}^2$, and $\Delta L=1\text{mm}=1\times 10^{-3}\text{m}$. Thus $Y=\frac{100\times 2}{1\times 10^{-6}\times 1\times 10^{-3}}=\frac{200}{1\times 10^{-9}}=2\times 10^{11}{\text{N/m}}^2$. It is important to convert millimetre quantities to metres before substituting, since mixing units is the most common source of error in such problems. The option $1\times 10^{11}$ is wrong because it doubles the area incorrectly. The option $4\times 10^{11}$ wrongly halves the extension. The option $5\times 10^{10}$ arises from mistakenly multiplying the area and extension before dividing. A plausibility check confirms the answer matches the known Young's modulus of steel (~$2\times 10^{11}{\text{N/m}}^2$), which is far larger than that of rubber or copper, reflecting steel's high stiffness, so the result is dimensionally and physically reasonable. The large numerical value is expected because the strain produced here is extremely small for such a substantial load.