Q.Consider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they are stretched by the same load then the stress on Y is

STRESS is the internal restoring force per unit area that develops inside a body when a deforming force acts on it: $\sigma=\dfrac{F}{A}$. Even a deformation too small to see with the naked eye still produces this internal force, because the relative positions of the body's atoms or molecules have shifted from equilibrium. The SI unit of stress is N m$^{-2}$, also called the pascal (Pa), and its dimensional formula is $[ML^{-1}T^{-2}]$; stress is, in general, a TENSOR quantity, since the internal force on a small surface element can have components both normal and tangential to that surface, and these components can differ depending on the orientation chosen for the surface -- this is exactly why stress is NOT simply a scalar, unlike quantities such as pressure, viscosity, or surface tension. Resolving the internal force F on a cross-sectional element $\Delta A$ into a normal component $F_n$ and a tangential component $F_t$ gives two basic kinds of stress: LONGITUDINAL (normal) STRESS, $\sigma_n=F_n/\Delta A$, further split into TENSILE STRESS (the two sides of $\Delta A$ pulled apart, as in a stretched wire) and COMPRESSIVE STRESS (the two sides pushed together, as in a loaded pillar); and SHEARING (tangential) STRESS, $\sigma_t=F_t/\Delta A$. A third kind, VOLUME STRESS, arises when a body is squeezed uniformly from every direction (as when fully immersed in a fluid at pressure P), and is numerically equal to that pressure. STRAIN is the resulting fractional change in the body's size: for a rod of natural length L stretched to $L+\Delta L$, $\varepsilon=\Delta L/L$ -- a pure ratio, so strain is dimensionless and has no unit. Strain is classified to match stress: LONGITUDINAL STRAIN ($\varepsilon_l=\Delta L/L$, split into tensile and compressive strain), SHEARING STRAIN ($\varepsilon_s=x/h=\tan\theta\approx\theta$ for a cuboid of height h whose top face is displaced sideways by x under a tangential force -- this equals the angle of shear itself for small deformations), and VOLUME STRAIN ($\varepsilon_v=\Delta V/V$). Radius and area enter stress calculations through $A=\pi r^2$, so for the SAME applied load, a wire of LARGER cross-sectional area experiences PROPORTIONALLY LESS stress -- a thicker wire is always under less stress than a thinner one carrying the identical load, since the same force is spread over a bigger area.