Q.State the SI unit and dimensions of stress.

The coefficient of elasticity of a body depends only on the nature/material of the substance, not on its size, shape, or dimensions.

Modulus of elasticity = stress/strain. Since both stress (force per unit area) and strain (fractional deformation) are already normalized quantities (independent of the actual size of the sample), their ratio — the modulus — comes out to be a fixed number characteristic of the material itself (e.g. steel always has the same Young's modulus, ~2×10¹¹ Pa, regardless of whether the sample is a thin wire or a thick rod). This is what allows elastic moduli to be tabulated as material properties.

Hooke's law (stress proportional to strain) defines the modulus of elasticity as the constant of proportionality.

Hooke's law states that, within the elastic limit, stress is directly proportional to strain: stress = E x strain, where E is called the modulus of elasticity (Young's modulus for tensile/compressive stress, bulk modulus for volume stress, or shear modulus for shearing stress, depending on the type of deformation). Stress, strain, and elastic limit are themselves separately defined quantities/concepts that appear IN Hooke's law, but the law itself is what establishes and defines the modulus of elasticity as their ratio.

The elastic limit is the maximum deforming force (or stress) up to which a body shows perfectly elastic behaviour — beyond it, the deformation becomes permanent (plastic).

When a deforming force is applied to a solid body, it deforms (changes shape/size), and if the force is removed, an elastic body returns to its original shape. This holds true only up to a certain limiting value of the force/stress, called the elastic limit. Beyond the elastic limit, the material develops a permanent set — it does not fully return to its original dimensions even after the force is removed; the body has been permanently, or plastically, deformed.

Regaining the original shape/size after a deforming force is removed is the property of elasticity.

When an external (deforming) force is applied to a body, it changes the body's shape or size — this is deformation. Elasticity is the property by virtue of which a body opposes this deformation and, once the deforming force is removed, tends to regain its original shape and size (perfectly, for a perfectly elastic body, e.g. an idealised steel spring within its elastic limit). This is distinct from inertia, which is a body's tendency to resist any change in its state of rest or motion (a property related to mass, not to deformation and recovery of shape).

Shearing strain = (relative displacement of the faces) / (distance between the faces) = Δx / L, a dimensionless angle-like quantity describing shape distortion without volume change.

When equal and opposite tangential forces act on the two opposite faces of a body (like the flat ends of a cylinder), the body deforms by having one face shift sideways relative to the other, while the length between them stays essentially unchanged. If Δx is this relative (tangential) displacement and L is the perpendicular distance between the two faces, the shearing strain is defined as:

shearing strain = Δx / L = tan θ ≈ θ (for small angles)

where θ is the angle of shear. This is distinct from longitudinal strain (Δl/l along the direction of the applied force) and volume strain (ΔV/V). Since the question describes exactly this ratio (relative displacement of faces to the length between them), it describes shearing strain.

Strain = (change in length)/(original length); stretching a wire to double its length gives strain = (2L-L)/L = 1.

Longitudinal strain is defined as

Strain = change in length / original length = (Delta L) / L

If the wire is stretched to double its original length L, the new length is 2L, so

Delta L = 2L - L = L

Strain = (Delta L)/L = L/L = 1

Strain is a dimensionless ratio, so it has no units -- a strain of 1 means the wire's length increased by 100% of its original length.

Pressure, viscosity (coefficient of viscosity), and surface tension are all scalar quantities in this course — each is specified completely by a magnitude alone. Stress, in contrast, depends on the plane/surface on which it is evaluated (normal stress vs shear stress in different directions), so it is not a simple scalar.

Check each option:

1. Pressure — force per unit area acting normally, has magnitude only (no direction associated with the quantity itself); scalar.
2. Viscosity (coefficient of viscosity, η) — a material property relating shear stress to velocity gradient; specified by magnitude alone; scalar.
3. Stress — internal restoring force per unit area within a deformed body; its value depends on the orientation of the internal surface considered (it can be normal stress or shear/tangential stress on different planes at the very same point), so a full description needs more than a single number — it is a tensor quantity, not a scalar.
4. Surface tension — force per unit length acting along the surface of a liquid, described fully by magnitude alone; scalar. So the one quantity here that is NOT a (simple) scalar is stress.

Longitudinal strain is the fractional change in length of a body produced by a force applied along its length: strain = deltaL / L.

When a deforming force is applied along the length of a rod or wire (either stretching it or compressing it), it produces a change in length without a significant change in the cross-sectional shape. This type of strain, restricted to length alone, is called longitudinal strain.

If L is the original length of the wire and deltaL is the change in length produced by the applied force, then:

Longitudinal strain = deltaL / L

Since it is a ratio of two lengths, longitudinal strain has no units and no dimensions.

Stress $=$ Force/Area, and Area $\propto r^2$; with the same load, the thinner wire carries higher stress.

Let the radius of wire B be $r$; then the radius of wire A is $2r$ (given).

Cross-sectional areas:

$A_A=\pi(2r)^2=4\pi r^2,\qquad A_B=\pi r^2$

Both wires are stretched by the same load $F$. Stress is Force/Area:

$\text{Stress}_A=\frac{F}{4\pi r^2},\qquad \text{Stress}_B=\frac{F}{\pi r^2}$

Ratio:

$\frac{\text{Stress}_B}{\text{Stress}_A}=\frac{F/(\pi r^2)}{F/(4\pi r^2)}=4$

So the stress on B is four times the stress on A (B, being thinner, bears a much higher stress for the same load).