Q:

About Springs Nature

Why when we coil some straight wires (may be metallic or nonmetallic) then it gives a restoring force (some what like spring)?Why when the wire was straight it doesn't give that appreciable amount of restoring force?
Does in this case also Force is proportional to length the coiled wire extended. If yes then consider giving the ratio of Force to length extended (i.e. Effective spring constant) in terms of young's modulus(Y) and other required terms.

A:

The deformation mode for a straight wire is stretching, which is a relatively stiff mode of deformation.  The spring stiffness for a straight wire is given by:
$$k=\frac{EA}{L}$$
where $E$ is Young's modulus, $A$ is the cross-sectional area, and $L$ is the length of the wire.  For a coiled spring the deformation is dominated by torsion of the wire, which is a relatively compliant mode of deformation.  For a coiled spring the stiffness is approximately:
$$k=\frac{G a^4}{4 n R^3}$$
where $a$ is the radius of the wire, $R$ is the radius of the coil, $G$ is the shear modulus, and $n$ is the number of coils.
EDIT: To address the question of where the second formula comes from, I think the following explanation is the simplest I can come up with.  Let's say the axial force applied to the spring is $P$ and it is applied at the center of the coils. The displacement of the spring is $\Delta$.  The stiffness of the spring is then $k=P/\Delta$. The torque that is carried at any cross-section of the wire is $T=PR$, where $R$ is the radius of the coils. Now, during the process of loading the spring the force $P$ does work and the coils store strain energy.  If we equate the work done by the force during the loading process to the strain energy stored in the coils, then we should be able to derive the formula.
$$\frac{1}{2} P\Delta=2n\pi R\int_A \frac{\tau^2}{2 G} dA$$
where again $n$ is the number of coils, $G$ is the shear modulus, and $A$ is the cross-section of the wire. Now, for a wire, the shear stress can be related to the torque as,
$$\tau=\frac{Tr}{I_p}$$
where $r$ is the radial position within the cross-section, and $I_p=\pi a^4/2$ is the polar moment of inertia of the cross-section that has a wire radius of $a$. So, our integral then becomes:
$$2n\pi R\left(\frac{8\pi T^2}{2G\pi^2 a^8}\int_{0}^{a} r^3 dr\right)=\frac{2nRT^2}{Ga^4}=\frac{2nR^3P^2}{Ga^4}=\frac{P^2}{2k}$$
Solving for $k$ gives:
$$k=\frac{Ga^4}{4nR^3}$$

