Two wires A and B have the same length and area of cross section. But Young’s modulus of A is two times the Young’s modulus of B. Then the ratio of force constant of A to that of B is

The correct option is 2.

Understanding the Force Constant of a Wire:
When a wire is stretched by a force, it behaves like a spring. According to Hooke's Law, the restoring force is proportional to the extension:
$F=k\Delta L$
where F is the applied force, $\Delta L$ is the extension, and k is the force constant (spring constant) of the wire.

Relating Force Constant to Young's Modulus:
Young's modulus (Y) is defined as the ratio of tensile stress to tensile strain:
$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$
where A is the area of cross-section and L is the original length of the wire.

Rearranging the formula to solve for force F:
$F=\left(\frac{YA}{L}\right)\Delta L$

By comparing this equation with Hooke's Law ($F=k\Delta L$), we get the expression for the force constant:
$k=\frac{YA}{L}$

Finding the Ratio:
For two wires A and B with the same length (L) and area of cross-section (A), the force constant is directly proportional to the Young's modulus:
$k\propto Y$

Therefore, the ratio of the force constant of A to that of B is:
$\frac{k_A}{k_B}=\frac{Y_A}{Y_B}$

Since the Young's modulus of A is twice that of B ($Y_A=2Y_B$):
$\frac{k_A}{k_B}=2$

Thus, the ratio of the force constant of A to that of B is 2.