Q.How should a force be applied on a body to produce shearing 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, …

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 separate …

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 materi …

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 tend …

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)

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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

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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. …

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:

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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: …