The Young's modulus and Poisson's ratio of a material are respectively Y and $\sigma$. The force required to decrease the area of cross-section of a wire made of this material by $\Delta \mathrm{A}$ is

• A

  $\frac{\mathrm{Y} \Delta \mathrm{A}}{4 \sigma}$
• B

  $\frac{2 \mathrm{Y} \Delta \mathrm{A}}{\sigma}$
• Correct answer
  C

  $\frac{\mathrm{Y} \Delta \mathrm{A}}{2 \sigma}$
• D

  $\frac{\mathrm{Y} \Delta \mathrm{A}}{\sigma}$

Hence, the correct answer is option: 'C'.