JEE Corner — Advanced Elasticity

What JEE Adds to This Chapter

Sections 1 to 11 built elasticity properly, and that build is complete for the Board syllabus. What JEE adds is not harder physics — it is the same three moduli — but a family of set-ups where the quantity you need is not constant along the object, so you cannot simply write $\Delta L = \frac{FL}{AY}$ and stop.

A rod that is thicker at one end than the other. A wire hanging under its own weight, where the tension at the top is not the tension at the bottom. A rod spun about one end, where the tension dies away to nothing at the far tip. A rigid bar resting on three wires that must all end up on one straight line. A composite rod squeezed between two walls that will not move. A mass dropped onto a wire instead of lowered onto it.

Every one of them is answered the same way: stop treating the object as a single thing, slice it, and add up the slices.

Most of what follows sits outside the rationalised syllabus body text, but JEE Main and JEE Advanced ask it every year, so it is developed here from first principles.

The nine things this section teaches

#	Skill	Why it earns marks
1	Elongation of a non-uniform bar by integration — tapering, conical, varying width	The examiner picks a shape with no formula. You build one
2	A heavy wire under its own weight, and the cone that is exactly one third of it	The tension varies, so the strain varies
3	A rod spun about an end or a centre	Tension is largest at the axis and zero at the tip
4	Series and parallel wires, treated as springs of constant $k = \frac{YA}{L}$	Two lines instead of two pages
5	The rigid bar on several wires — force balance, torque balance, compatibility	Three equations, always the same three
6	Thermal stress in a composite rod and in a bimetallic strip	The one place two materials must be solved together
7	Strain energy, energy released on fracture, and the dropped mass	The extension comes out at exactly twice the static one
8	Extension that changes with the state of motion — lifts, vertical circles	The wire does not change; the tension does
9	Torsion in depth — the couple by integration, hollow shafts, the torsional pendulum	$R^4$ makes this the most sensitive formula in the chapter

Running underneath all nine is a single habit: before writing any formula, ask whether the force, the area, or both, are the same everywhere along the object. If either varies, you integrate. If two bodies share one constraint, you write a compatibility equation. Nothing else in this section is difficult.

Conventions, fixed now

Symbols. $\sigma$ denotes Poisson's ratio throughout. Stress is written $\frac{F}{A}$ wherever possible, and where a symbol is unavoidable it is $\sigma_L$, $\sigma_s$ and $\sigma_h$ for longitudinal, shearing and hydraulic stress. Strain is $\varepsilon$. The moduli are $Y$ (Young's), $G$ (shear) and $B$ (bulk), with compressibility $k = \frac{1}{B}$. Elsewhere $\sigma$ often denotes stress, with $\nu$ or $\mu$ for Poisson's ratio — check the convention before copying a formula out of anything.

Careful: $k$ does double duty in this section, as the compressibility $\frac{1}{B}$ and as the force constant $\frac{YA}{L}$ of a wire treated as a spring. They never appear in the same problem, and every use below says which one it is.

Constants. Unless a problem states otherwise: $Y_{\text{steel}} = 2.0\times10^{11}$ Pa, $Y_{\text{copper}} = 1.2\times10^{11}$ Pa, $Y_{\text{brass}} = 0.91\times10^{11}$ Pa, $Y_{\text{aluminium}} = 0.70\times10^{11}$ Pa, $G_{\text{steel}} = 0.84\times10^{11}$ Pa, $B_{\text{steel}} = 1.6\times10^{11}$ Pa, $\rho_{\text{steel}} = 7800$ kg/m$^3$, $\alpha_{\text{steel}} = 1.2\times10^{-5}$ per degree, $\alpha_{\text{copper}} = 1.7\times10^{-5}$ per degree, $\alpha_{\text{brass}} = 1.9\times10^{-5}$ per degree. Every problem states which constants it used, and no problem mixes $g = 9.8$ with $g = 10$.

Key Point — the master equation of this whole section: $$\Delta L = \int_0^{L}\frac{F(x)\,dx}{A(x)\,Y}$$ Everything else is a special case. Constant $F$ and constant $A$ collapses it to $\frac{FL}{AY}$. A varying area gives you the tapering-bar family. A varying force gives you the self-weight and rotating-rod family. Write this line first and you can never be lost.

[Exam Tip] Three questions, asked before any algebra, decide almost every problem below. Is the force the same at every cross-section? Is the area the same at every cross-section? Do two or more bodies have to fit together at the end? A "no" to the first two means integrate; a "yes" to the third means write a compatibility equation. Answer all three and the method is chosen before you have written a symbol.

Non-Uniform Bars: Slice, Stretch, Add

The formula $\Delta L = \frac{FL}{AY}$ has one silent assumption buried in it: that every cross-section of the bar carries the same force over the same area, and therefore suffers the same strain. Break either half of that and the formula is simply wrong.

The method, in four lines

1. Take an element of length $dx$ at a distance $x$ from a stated end. It is short enough that the force and the area are constant across it.
2. Write the force $F(x)$ across that element, and the area $A(x)$ of that element. This is the only step where thinking is required.
3. Stretch the element: $d(\Delta L) = \frac{F(x)\,dx}{A(x)\,Y}$, which is $\frac{FL}{AY}$ applied to a bar of length $dx$.
4. Integrate from one end to the other.

Case 1: the area varies, the force does not

A rod tapering from radius $a$ to radius $b$. Pulled by an axial force $F$ at both ends, so every section carries the same $F$. Taking $x$ from the end of radius $a$, the radius at $x$ is $a + \frac{(b-a)x}{L}$, so $A(x) = \pi\left(a + \frac{(b-a)x}{L}\right)^2$ and

$$\Delta L = \int_0^{L}\frac{F\,dx}{\pi\left(a+\frac{(b-a)x}{L}\right)^2 Y} = \boxed{\ \frac{FL}{\pi a b Y}\ }$$

Look hard at that answer. The area doing the work is $\pi a b$ — the geometric mean of the two end areas, since $\sqrt{\pi a^2 \cdot \pi b^2} = \pi a b$ — and not $\pi\left(\frac{a+b}{2}\right)^2$, the area of a uniform bar of the mean radius. Whenever $a \ne b$ the geometric mean is the smaller of the two, so the real bar always stretches more than that mean-radius bar. Put $a = b$ and the result collapses to $\frac{FL}{\pi a^2 Y}$, as any correct answer must.

A bar of rectangular section, constant thickness $t$, width growing linearly from $b_1$ to $b_2$:

$$\Delta L = \int_0^{L}\frac{F\,dx}{t\left(b_1+\frac{(b_2-b_1)x}{L}\right)Y} = \boxed{\ \frac{FL}{Y\,t\,(b_2-b_1)}\ln\frac{b_2}{b_1}\ }$$

A logarithm, because a linear width puts $x$ in the denominator to the first power rather than the second. Notice that the two tapering results look nothing alike — you cannot guess one from the other, and the examiner knows it.

Case 2: the force varies, the area does not

A uniform bar hanging under its own weight. Measure $x$ down from the support. The section at $x$ has to hold up everything below it, a length $L - x$ of bar:

$$F(x) = \rho A g (L - x) \qquad\Longrightarrow\qquad \Delta L = \int_0^{L}\frac{\rho g (L-x)}{Y}\,dx = \boxed{\ \frac{\rho g L^2}{2Y} = \frac{WL}{2AY}\ }$$

where $W = \rho A L g$ is the total weight. That second form is the memorable one: the elongation is what you would get by hanging half the bar's weight from the free end of a weightless bar. Not the whole weight — half of it, because the top of the bar carries everything and the bottom carries nothing.

A solid cone hung from its base, apex pointing down. Now both vary, and they very nearly cancel. Measure $x$ up from the apex; the radius there is $\frac{Rx}{L}$, so the area is $\pi\frac{R^2x^2}{L^2}$ and the weight hanging below that section is the weight of the little cone of height $x$, namely $\frac{1}{3}\pi\frac{R^2x^2}{L^2}x\rho g$. The stress at $x$ is their quotient:

$$\frac{F(x)}{A(x)} = \frac{\rho g x}{3} \qquad\Longrightarrow\qquad \Delta L = \int_0^{L}\frac{\rho g x}{3Y}\,dx = \boxed{\ \frac{\rho g L^2}{6Y}\ }$$

Exactly one third of the uniform bar's answer, and completely independent of how fat the cone is. Both facts are asked directly.

Case 3: a rod that is spinning

A rod of mass $m$, length $L$, area $A$ and density $\rho$ rotates in a horizontal plane about a vertical axis through one end, at angular speed $\omega$. Gravity plays no part; what stretches the rod is that every element needs a centripetal force, and only the material closer to the axis can supply it.

The section at distance $x$ from the axis must supply the centripetal force for everything beyond it:

$$T(x) = \int_x^{L}\rho A\,\omega^2 s\,ds = \frac{\rho A \omega^2}{2}\left(L^2-x^2\right)$$

Read three things off it immediately:

• $T$ is largest at the axis, where it equals $\frac{\rho A\omega^2L^2}{2} = \frac{m\omega^2L}{2}$ — the same as if the whole mass sat at the midpoint, which is exactly where the centre of mass is.
• $T$ is zero at the free tip. Nothing lies beyond it to be pulled round.
• The rod will break at the axis, not at the tip.

Integrating gives the elongation:

$$\Delta L = \int_0^{L}\frac{T(x)\,dx}{AY} = \frac{\rho\omega^2}{2Y}\left(L^3-\frac{L^3}{3}\right) = \boxed{\ \frac{\rho\,\omega^2L^3}{3Y} = \frac{m\omega^2L^2}{3AY}\ }$$

Spin the same rod about its centre instead and the same method gives $T_{\max} = \frac{\rho A\omega^2L^2}{8}$ at the middle and

$$\Delta L = \frac{\rho\,\omega^2L^3}{12Y}$$

one quarter of the about-an-end value. Half the length on each side, and the cube in $L^3$ does the rest.

The five results, collected

Situation	Elongation
uniform bar, axial load $F$	$\dfrac{FL}{AY}$
rod tapering between radii $a$ and $b$, axial load $F$	$\dfrac{FL}{\pi a b Y}$
bar of thickness $t$, width $b_1 \to b_2$, axial load $F$	$\dfrac{FL}{Yt(b_2-b_1)}\ln\dfrac{b_2}{b_1}$
uniform bar under its own weight	$\dfrac{\rho g L^2}{2Y} = \dfrac{WL}{2AY}$
solid cone hung from its base	$\dfrac{\rho g L^2}{6Y}$
rod spun about one end	$\dfrac{\rho\,\omega^2L^3}{3Y}$
rod spun about its centre	$\dfrac{\rho\,\omega^2L^3}{12Y}$

Key Point: In every one of these the stress is what varies from point to point, and the elongation is the running total of the local strain. If you can write $\frac{F(x)}{A(x)}$ correctly, the integration is ordinary school calculus. The marks are in step 2, not step 4.

[Exam Tip] Check every non-uniform result by collapsing it. Put $a = b$ in the taper and you must get $\frac{FL}{\pi a^2Y}$. Put $\omega = 0$ in the spinning rod and you must get zero. Set the cone's answer beside the bar's and the ratio must be exactly $\frac{1}{3}$. A result that fails one of these checks is wrong, and you will find out in ten seconds rather than at the end of the paper.

Composite Wires and the Rigid Bar

A wire of Young's modulus $Y$, area $A$ and length $L$ obeys $F = \frac{YA}{L}\,\Delta L$, which is Hooke's law for a spring with force constant

$$k = \frac{YA}{L}$$

Once you have written that line, every wire-combination problem in this chapter becomes a spring problem, and you already know how springs combine.

Series and parallel, in one table

	Series (end to end)	Parallel (side by side)
what is shared	the tension $T$	the extension $\Delta L$
what adds	the extensions	the tensions
combination	$\dfrac{1}{k} = \dfrac{1}{k_1}+\dfrac{1}{k_2}$	$k = k_1+k_2$
the softer wire	stretches more	carries less load
stress	equal only if the areas are equal	equal only if the $\frac{Y}{L}$ values are equal

The two rows in bold are the whole thing. In series the same tension runs right through, so the wire with the smaller $\frac{YA}{L}$ takes the bigger share of the stretch. In parallel everything is forced to stretch by the same amount, so the stiffer wire takes the bigger share of the load — load divides in the ratio of the $k$ values.

The rigid bar: three equations, always the same three

Here is the arrangement the paper actually sets. A light rigid bar hangs horizontally from two or more vertical wires, and a load is hung somewhere along it. Find the tensions.

You have more unknowns than the two equations of statics can supply, so you need a third relation, and it comes from geometry:

1. Force balance. $\displaystyle\sum T_i = W$.
2. Torque balance about any point on the bar. $\displaystyle\sum T_i x_i = W x_{\text{load}}$.
3. Compatibility. The bar is rigid, so after everything has stretched it is still straight. The extensions of the wires, plotted against position along the bar, must lie on a straight line: $$e(x) = e_0 + m x, \qquad e_i = \frac{T_i}{k_i}$$

For two wires that third condition is just "the two ends are joined by a straight bar", which is automatic; two equations suffice. For three or more it bites, and it is what makes the problem solvable.

Key Point: For three equally spaced wires the compatibility condition has a form worth memorising: $$e_1 + e_3 = 2e_2$$ the middle extension is the average of the outer two. With identical wires that becomes $T_1 + T_3 = 2T_2$ directly.

Two wires, one horizontal rod — the classic

A light rod hangs from a wire at each end. Where must the load go?

The question has two different answers, and you must read which one is being asked.

• For the rod to stay horizontal, the two extensions must be equal: $$\frac{T_1L_1}{A_1Y_1} = \frac{T_2L_2}{A_2Y_2} \qquad\Longrightarrow\qquad \frac{T_1}{T_2} = \frac{A_1Y_1/L_1}{A_2Y_2/L_2} = \frac{k_1}{k_2}$$
• For the two wires to be under equal stress, the tensions must be in the ratio of the areas: $$\frac{T_1}{A_1} = \frac{T_2}{A_2} \qquad\Longrightarrow\qquad \frac{T_1}{T_2} = \frac{A_1}{A_2}$$

Then torque balance about one end converts a tension ratio into a distance. With the wires a distance $d$ apart and the load $x$ from wire 1, $T_1 x = T_2 (d - x)$ gives

$$\boxed{\ \frac{x}{d-x} = \frac{T_2}{T_1}\ }$$

Note the inversion: the load sits closer to the wire that carries the larger tension. Get that backwards and the answer is the reflection of the right one, which is exactly what the distractors are built from.

An honest warning about "equal strain"

"Equal extension" and "equal strain" are the same condition only when the wires have the same length. If the two wires are of different lengths, equal strain means $\frac{\Delta L_1}{L_1} = \frac{\Delta L_2}{L_2}$, whereas a rigid horizontal bar demands $\Delta L_1 = \Delta L_2$. Read the wording, and if the problem says "the rod remains horizontal", it is the extensions that are equal, not the strains.

[Exam Tip] Whenever a problem gives you more wires than equations of statics, stop and count. Two wires: statics is enough. Three wires: you need compatibility, and the shortcut $e_1 + e_3 = 2e_2$ finishes it in one line. Four or more equally spaced: the extensions still lie on a straight line, so write $e_i = e_0 + m x_i$ with two unknowns, and solve force balance and torque balance for $e_0$ and $m$.

Thermal Stress When Two Materials Share One Constraint

Section 9 established the single-material result: a rod clamped between rigid walls and heated through $\Delta T$ develops a stress

$$\frac{F}{A} = Y\alpha\,\Delta T$$

independent of both its length and its cross-section. JEE takes the next step and puts two different materials under one constraint, where neither the stress nor the strain is the same in the two of them.

Two rods end to end between rigid walls

A rod of material 1 (length $L_1$, area $A_1$, modulus $Y_1$, expansion coefficient $\alpha_1$) is joined end to end with a rod of material 2, and the pair is clamped between walls that do not move. Heat the whole assembly through $\Delta T$.

Step 1 — how much would they grow if free? $$\text{free expansion} = \left(\alpha_1 L_1 + \alpha_2 L_2\right)\Delta T$$

Step 2 — what force squeezes exactly that much back out? The rods are in series, so the same compressive force $F$ runs through both, and the two compressions add: $$\text{compression} = \frac{FL_1}{A_1Y_1}+\frac{FL_2}{A_2Y_2}$$

Step 3 — set them equal, because the total length is unchanged:

$$\boxed{\ F = \frac{\left(\alpha_1L_1+\alpha_2L_2\right)\Delta T}{\dfrac{L_1}{A_1Y_1}+\dfrac{L_2}{A_2Y_2}}\ }$$

Two things follow at once, and both are examined.

• The force is the same in both rods, so if the areas are equal the stress is the same in both — even though the two materials are quite different. Series, not parallel.
• The strains are not equal. The junction between the two rods does not stay put: it moves, by $$\delta = \alpha_1L_1\Delta T - \frac{FL_1}{A_1Y_1}$$ measured from its original position. Compute it from either side and you must get the same answer with the opposite sign convention — a free check worth thirty seconds.

A rod inside a tube: the parallel version

Now weld a rod of material 1 along the axis of a tube of material 2, joining them only at the two end caps, and heat the pair. Nothing external holds them; they hold each other.

Both must finish at the same length, so if $\alpha_2 > \alpha_1$ the tube wants to be longer, is prevented, and ends up in compression, while the rod is dragged out into tension. There are no external forces, so the two internal forces are equal and opposite, $F_1 = F_2 = F$:

$$\alpha_1\Delta T + \frac{F}{A_1Y_1} \;=\; \alpha_2\Delta T - \frac{F}{A_2Y_2} \qquad\Longrightarrow\qquad \boxed{\ F = \frac{\left(\alpha_2-\alpha_1\right)\Delta T}{\dfrac{1}{A_1Y_1}+\dfrac{1}{A_2Y_2}}\ }$$

Notice how much simpler this is: only the difference of the expansion coefficients appears, and the lengths have cancelled entirely.

Key Point: Both problems are the same two-step idea — let it expand freely, then apply the force that restores the constraint. What differs is the constraint. Two rods between walls: the total length is fixed, the rods are in series, the force is shared. Rod inside a tube: the two lengths must be equal to each other, the members are in parallel, the forces are equal and opposite.

The bimetallic strip

Rivet two strips of different metals face to face, each of thickness $t$, and heat them. The strip with the larger $\alpha$ wants to be longer; it cannot slide, so the pair has no option but to bend, with the high-$\alpha$ metal on the outside of the curve.

Let the assembly bend into an arc of radius $R$ measured to the interface, subtending an angle $\theta$. The centre line of each strip sits a distance $\frac{t}{2}$ from the interface, so

$$\left(R+\frac{t}{2}\right)\theta = L\left(1+\alpha_2\Delta T\right), \qquad \left(R-\frac{t}{2}\right)\theta = L\left(1+\alpha_1\Delta T\right)$$

Divide one by the other and use $\frac{1+\alpha_2\Delta T}{1+\alpha_1\Delta T}\approx 1+(\alpha_2-\alpha_1)\Delta T$:

$$\frac{R+\frac{t}{2}}{R-\frac{t}{2}} \approx 1 + (\alpha_2-\alpha_1)\Delta T \qquad\Longrightarrow\qquad \boxed{\ R \approx \frac{t}{\left(\alpha_2-\alpha_1\right)\Delta T}\ }$$

with $t$ the thickness of one strip. Three consequences, all of them tested:

• $R$ does not depend on the length of the strip. A longer strip curls through a bigger angle, not a tighter curve.
• $R \propto t$: thin strips curl tightly. That is why a thermostat strip is thin.
• $R \propto \frac{1}{\Delta T}$, so the curvature $\frac{1}{R}$ is proportional to the temperature rise. That linearity is precisely what makes the device a usable thermometer and switch.

For a strip clamped at one end, the free tip swings sideways by $R(1-\cos\theta)$, which for a small angle is very nearly $\frac{L^2}{2R}$ — and therefore grows as the square of the length.

This treatment ignores the elastic stresses the two strips exert on each other, which shift the neutral surface slightly; a full analysis brings the two Young's moduli in as well. For the ratios and the order of magnitude that objective questions want, the geometric result above is the one to use, and it is the one every paper expects.

[Exam Tip] In every thermal problem in this section, write the free expansion on one line and the elastic compression on the next, then equate. Do not try to write a single formula from memory — the composite results look similar enough to each other that recalling the wrong one is easy, while the two-step route cannot go wrong.

When the Load Is Moving: Lifts, Vertical Circles and a Dropped Mass

A wire does not know what is happening at the far end of it. All it feels is the tension, and its extension is $\Delta L = \frac{TL}{AY}$ whatever that tension came from. So every "elasticity meets dynamics" problem is really two problems bolted together: find the tension from Newton's laws, then feed it to the wire.

The easy half: tension from the motion

Situation	Tension in the wire
mass $m$ hanging at rest	$mg$
lift accelerating up with $a$	$m(g+a)$
lift accelerating down with $a$	$m(g-a)$
lift in free fall	$0$ — the wire returns to its natural length
mass in a horizontal circle, wire nearly horizontal	$m\omega^2 L$
vertical circle, at the lowest point	$\dfrac{mv^2}{L}+mg$
vertical circle, wire horizontal	$\dfrac{mv^2}{L}$
vertical circle, at the top	$\dfrac{mv^2}{L}-mg$

For a stone that just completes a vertical circle, $v_{\text{top}}^2 = gL$, and energy conservation gives $v_{\text{low}}^2 = v_{\text{top}}^2+4gL = 5gL$. So

$$T_{\text{top}} = 0, \qquad T_{\text{horizontal}} = 3mg, \qquad T_{\text{lowest}} = 6mg$$

and since the extension is proportional to the tension, the wire at the bottom of the swing is stretched six times as far as it would be if the stone simply hung still. That factor of six is asked constantly, and it is the reason a rope that holds a static load can still snap when the load starts swinging.

A subtlety worth one line. In a horizontal circle the radius is $L + \Delta L$, not $L$, so strictly $T = m\omega^2(L+\Delta L)$ and $\Delta L = \frac{TL}{AY}$ must be solved together, giving $\Delta L = \frac{m\omega^2L^2}{AY - m\omega^2 L}$. In every realistic case $AY \gg m\omega^2L$ and the correction is invisible; quote the simple answer, and mention the exact one only if the question makes a point of it.

The hard half: a mass dropped instead of lowered

Now the classic. A wire hangs from a ceiling at its natural length. A mass $m$ is attached to the lower end and released from rest. How far does the wire stretch?

The wrong answer is $x_0 = \frac{mg}{k}$, with $k = \frac{YA}{L}$. That is the extension you get if you lower the mass gently, supporting it all the way down until the wire alone can hold it. Released from rest, the mass arrives at $x_0$ moving, and keeps going.

Use energy. At the lowest point the mass is momentarily at rest, so all the work gravity has done has gone into the wire:

$$mgx = \frac{1}{2}kx^2 \qquad\Longrightarrow\qquad \boxed{\ x_{\max} = \frac{2mg}{k} = 2x_0\ }$$

Key Point: A suddenly applied load produces twice the extension, and therefore twice the stress, of the same load applied gradually. A rope that is safe when a weight is lowered onto it can break when the same weight is dropped onto it from no height at all.

What actually happens next is simple harmonic motion: the mass oscillates about the static position $x_0$ with amplitude $x_0$, so it runs between $0$ and $2x_0$. At the lowest point the wire pulls up with $kx_{\max} = 2mg$ while gravity pulls down with $mg$, so the net force is $mg$ upwards and the mass is thrown back. It stops permanently at $x_0$ only when damping has removed the energy.

And the missing half of the energy. Lower the mass gently through $x_0$ and gravity does $mgx_0$ of work while the wire stores only $\frac{1}{2}kx_0^2 = \frac{1}{2}mgx_0$. The other half was not destroyed and did not go into the wire: your hand took it out, doing negative work as it let the mass down. Section 8 set this up; here is where it pays.

Dropped from a height $h$

Let the mass fall a height $h$ before it starts stretching the wire. Gravity now does work over $h + x$:

$$mg(h+x) = \frac{1}{2}kx^2 \qquad\Longrightarrow\qquad \boxed{\ x = x_0\left(1+\sqrt{1+\frac{2h}{x_0}}\right)\ }$$

Two limits confirm it. Put $h = 0$ and you get $2x_0$, the sudden-release answer. Let $h$ grow large and $x \to \sqrt{\frac{2mgh}{k}}$, which is just $mgh = \frac{1}{2}kx^2$ — all the fall energy stored in the wire, the drop dominating everything else.

The stress goes up in exactly the same ratio, and it goes up fast. Since $x_0$ is typically a fraction of a millimetre, dropping the load through even a few millimetres multiplies the peak stress by five or ten. This is why a crane is never allowed to let a load fall onto a slack cable, and why lifting slings are rated for "shock loads" separately.

Strain energy and what a fracture releases

The energy the wire holds is the area under its own force-extension line:

$$U = \frac{1}{2}F\,\Delta L = \frac{1}{2}\frac{YA(\Delta L)^2}{L} = \frac{F^2L}{2AY}, \qquad u = \frac{U}{\text{volume}} = \frac{1}{2}\frac{F}{A}\varepsilon = \frac{1}{2}Y\varepsilon^2 = \frac{\left(F/A\right)^2}{2Y}$$

At the instant a cable snaps, all of that is released at once. A steel cable of length 20 m and cross-section 3.0 cm$^2$ carrying a stress of $5.0\times10^{8}$ Pa, with $Y = 2.0\times10^{11}$ Pa, holds

$$u = \frac{\left(5.0\times10^{8}\right)^2}{2\left(2.0\times10^{11}\right)} = 6.25\times10^{5}\ \text{J/m}^3, \qquad U = 6.25\times10^{5}\times\left(20\times3.0\times10^{-4}\right) = 3750\ \text{J}$$

Nearly four kilojoules, dumped in a few milliseconds, which is why a parting cable whips. Notice that the energy per cubic metre depends on the stress and the modulus alone — not on the length, not on the area. The length and area only decide how much material there is to hold it.

[Exam Tip] The single most expensive slip in this block is writing $U = F\,\Delta L$. The force grows from zero to $F$ as the wire stretches, so the work is the area of a triangle, not a rectangle, and the factor $\frac{1}{2}$ is not optional. If a question offers you both $F\Delta L$ and $\frac{1}{2}F\Delta L$ among the options, that is the entire question.

Torsion in Depth

Section 5 introduced torsion: twisting is shear wrapped around an axis, the shearing angle at radius $r$ is $\theta = \frac{r\phi}{L}$, and the restoring couple is proportional to the twist. This block takes it as far as JEE goes.

The couple, built by integration

Take a thin annulus of radius $r$ and thickness $dr$ in the cross-section of a shaft of length $L$ twisted through $\phi$. Its shearing strain is $\frac{r\phi}{L}$, so it carries a shearing stress $G\frac{r\phi}{L}$ over an area $2\pi r\,dr$. That force acts at a distance $r$ from the axis, so its moment is the stress times the area times $r$:

$$\tau = \int_{R_1}^{R_2} G\frac{r\phi}{L}\left(2\pi r\,dr\right)r = \frac{2\pi G\phi}{L}\int_{R_1}^{R_2} r^3\,dr = \frac{\pi G\phi}{2L}\left(R_2^4-R_1^4\right)$$

Writing $J$ for the polar second moment of area of the section makes every result compact:

$$\boxed{\ \tau = C\phi, \qquad C = \frac{GJ}{L}, \qquad J_{\text{solid}} = \frac{\pi R^4}{2}, \qquad J_{\text{hollow}} = \frac{\pi}{2}\left(R_2^4-R_1^4\right)\ }$$

$C$ is the torsional rigidity — the couple per radian of twist, in N m per radian.

The stress distribution, and why hollow beats solid

The shearing stress at radius $r$ is

$$\sigma_s(r) = G\frac{r\phi}{L} = \frac{\tau r}{J}$$

Linear in $r$: zero on the axis, maximum at the surface. The material near the axis is barely working, yet it weighs just as much as the material near the rim. Remove it and you lose very little stiffness while losing a lot of weight.

Make the comparison properly. Take a hollow shaft of inner radius $R_1$ and outer radius $2R_1$, and a solid shaft of the same material, same length and same mass — so the same cross-sectional area, $\pi R^2 = \pi\left(4R_1^2-R_1^2\right)$, giving $R = \sqrt{3}\,R_1$. Then

$$\frac{C_{\text{hollow}}}{C_{\text{solid}}} = \frac{\left(2R_1\right)^4-R_1^4}{\left(\sqrt{3}R_1\right)^4} = \frac{15R_1^4}{9R_1^4} = \frac{5}{3}$$

Sixty-seven per cent stiffer for exactly the same weight, and it carries a given torque at a lower peak stress too. This is why drive shafts, bicycle frames, scaffolding poles and aircraft spars are tubes.

The $R^4$ is the headline

$C \propto \frac{R^4}{L}$. Nothing else in this chapter is anywhere near as sensitive to a dimension.

Change	Effect on $C$	Effect on $\phi$ for a given couple
double the radius	$\times 16$	$\div 16$
double the length	$\div 2$	$\times 2$
5% error in measuring $R$	22% error	22% error
switch steel to copper ($G$ halves)	$\div 2$	$\times 2$

The torsional pendulum

Hang a body of moment of inertia $I$ from a wire of torsional rigidity $C$ and twist it. The restoring couple is $-C\phi$, so

$$I\frac{d^2\phi}{dt^2} = -C\phi \qquad\Longrightarrow\qquad \boxed{\ T = 2\pi\sqrt{\frac{I}{C}} = 2\pi\sqrt{\frac{2IL}{\pi G R^4}}\ }$$

This is angular simple harmonic motion, the exact analogue of a mass on a spring with $I$ in place of $m$ and $C$ in place of $k$. Two consequences:

• The period does not depend on the amplitude, provided the wire stays elastic.
• $T \propto \frac{1}{R^2}$. Double the wire's radius and the period drops to a quarter of what it was.

Invert it and you have a laboratory method for measuring $G$: time the oscillations, get $C = \frac{4\pi^2I}{T^2}$, and then $G = \frac{2CL}{\pi R^4}$.

Energy, and combinations

Twisting stores energy exactly as stretching does, with $C$ playing the part of $k$:

$$W = \int_0^{\phi}C\phi'\,d\phi' = \frac{1}{2}C\phi^2$$

and the same factor of $\frac{1}{2}$ trap lies waiting. Wires combine in torsion exactly as springs do:

• In series (one hung below the other, the same couple through both): $\dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}$, and the twists add.
• In parallel (both attached to the same disc, both twisted by the same angle): $C = C_1 + C_2$, and the couples add.

Key Point: Torsion is governed by $G$ and never by $Y$, because twisting is shear. Any answer to a torsion problem containing $Y$ is wrong unless the problem gave you $Y$ and $\sigma$ and expected you to get $G$ from $Y = 2G(1+\sigma)$ first.

[Exam Tip] Most torsion questions are ratio questions in disguise. Write $C \propto \frac{GR^4}{L}$ and $T \propto \sqrt{\frac{I}{C}}$, cancel everything the two cases share, and you are finished without ever evaluating $\pi$ or $G$. Only reach for numbers when the question asks for an actual couple, period or stress.

The Four Constants, Used in Both Directions — and the Traps

Four constants, two independent

For a homogeneous isotropic solid there are only two independent elastic constants. Any two of $Y$, $G$, $B$ and $\sigma$ determine the other two, through relations that Section 7 derived:

$$Y = 3B\left(1-2\sigma\right), \qquad Y = 2G\left(1+\sigma\right), \qquad \sigma = \frac{3B-2G}{2\left(3B+G\right)}, \qquad \frac{9}{Y} = \frac{3}{G}+\frac{1}{B}$$

JEE uses them in both directions, and the direction decides which one to pick.

You are given	You want	Use
$Y$ and $\sigma$	$G$	$G = \dfrac{Y}{2(1+\sigma)}$
$Y$ and $\sigma$	$B$	$B = \dfrac{Y}{3(1-2\sigma)}$
$Y$ and $G$	$\sigma$	$\sigma = \dfrac{Y}{2G}-1$
$Y$ and $B$	$\sigma$	$\sigma = \dfrac{1}{2}\left(1-\dfrac{Y}{3B}\right)$
$G$ and $B$	$\sigma$	$\sigma = \dfrac{3B-2G}{2(3B+G)}$
$G$ and $B$	$Y$	$Y = \dfrac{9GB}{3B+G}$

Three limiting cases worth knowing cold

If	then	meaning
$\sigma = 0.5$	$B \to \infty$ and $Y = 3G$	volume exactly conserved: incompressible
$\sigma = 0.25$	$Y = 2.5G$ and $Y = 1.5B$, so $G = 0.4Y$, $B = \dfrac{2Y}{3}$	the "Poisson solid", a common exam choice
$\sigma = 0$	$Y = 2G = 3B$	no lateral contraction at all (cork is close)

The bounds are $-1 \le \sigma \le 0.5$. A value above $0.5$ would make $B$ negative, meaning the solid expands when you squeeze it, and no ordinary material does that.

Being honest about real data

Test the relations on genuine tabulated values, taking $Y$, $G$ and $B$ from the same source for each metal and asking whether they actually fit:

Metal	$Y$ (Pa)	$G$ (Pa)	$B$ (Pa)	$\sigma$ from $Y,G$	$\sigma$ from $Y,B$	$Y$ rebuilt from $G,B$
Steel	$2.0\times10^{11}$	$0.84\times10^{11}$	$1.6\times10^{11}$	$0.19$	$0.29$	$2.14\times10^{11}$, 7% high
Copper	$1.2\times10^{11}$	$0.42\times10^{11}$	$1.4\times10^{11}$	$0.43$	$0.36$	$1.15\times10^{11}$, 5% low
Aluminium	$0.70\times10^{11}$	$0.25\times10^{11}$	$0.72\times10^{11}$	$0.40$	$0.34$	$0.67\times10^{11}$, 4% low

The relations tested on real tabulated moduli. Agreement is good, not perfect.

The two routes to $\sigma$ disagree by five to seven hundredths, and $Y$ rebuilt from $G$ and $B$ misses the tabulated $Y$ by four to seven per cent. That is not an arithmetic error. Real metals are polycrystalline and not perfectly isotropic, the tabulated values are averages over different samples and different measurement techniques, and the relations assume a perfectly isotropic continuum. In an exam you use the relations as exact, because the question intends them to be; in a laboratory you would not expect better than five to ten per cent.

The four traps of this chapter, priced

1. The wrong area for shear. For longitudinal stress the area is the section perpendicular to the force. For shearing stress it is the face the force acts along — the face parallel to the force. A block $0.40$ m long, $0.10$ m deep and $0.20$ m tall, sheared along the top: the correct area is $0.40\times0.10 = 0.04$ m$^2$, not the perpendicular section $0.10\times0.20 = 0.02$ m$^2$. Using the second doubles the answer, and the error factor changes with the block's proportions, so you cannot recognise it from the size of the number. Name the loaded face in writing before you compute.

2. Forgetting the $\frac{1}{2}$ in the energy. The restoring force grows from zero to $F$, so the work is $\frac{1}{2}F\Delta L$. Both $F\Delta L$ and $\frac{1}{2}F\Delta L$ will be among the options.

3. Treating breaking stress as if it depended on length. Breaking stress is a material property, measured in pascal, and a wire cut in half has exactly the same breaking stress and therefore carries exactly the same breaking load. What does depend on length is the maximum length a wire can hang under its own weight, $L_{\max} = \frac{\sigma_{\text{break}}}{\rho g}$ — that is a different question with a different answer.

4. Reaching for the wrong modulus. Stretch or compress along one direction: $Y$. Slide or twist: $G$. Squeeze from every side at once: $B$. A twisted shaft never uses $Y$; a wire under a hanging load never uses $G$.

Key Point — the four sentences to carry out of this section:

1. If the force or the area varies along the body, integrate; the master equation is $\Delta L = \int \frac{F(x)\,dx}{A(x)Y}$.
2. A wire is a spring with $k = \frac{YA}{L}$; wires in series share the tension, wires in parallel share the extension.
3. A rigid bar keeps its wires' extensions on a straight line — that is the third equation you were missing.
4. A load dropped rather than lowered doubles the extension and doubles the stress.

[Exam Tip] In this chapter almost every wrong answer is one of five things: the wrong modulus, the wrong area, a missing $\frac{1}{2}$, a radius used where a diameter was given, or an SI slip on mm$^2$ ($1$ mm$^2 = 10^{-6}$ m$^2$, not $10^{-3}$). Before you commit an answer, run those five names down the margin. It takes eight seconds and it is the highest-yield habit in the whole chapter.

Solved Examples, Part 1: Integration and Compatibility

Values used in this section, unless a problem states otherwise: $Y_{\text{steel}} = 2.0\times10^{11}$ Pa, $Y_{\text{copper}} = 1.2\times10^{11}$ Pa, $Y_{\text{aluminium}} = 0.70\times10^{11}$ Pa, $G_{\text{steel}} = 0.84\times10^{11}$ Pa, $\rho_{\text{steel}} = 7800$ kg/m$^3$, $g = 9.8$ m/s$^2$. Every constant is restated inside the solution that uses it.

Example 1: A rod that tapers along its length

A steel rod 1.0 m long tapers uniformly from a diameter of 20 mm at one end to 10 mm at the other. It is pulled by an axial force of 20 kN. Find its elongation, and compare it with what a uniform rod of the mean radius would give. Take $Y_{\text{steel}} = 2.0\times10^{11}$ Pa.

Solution:

1. Convert the diameters to radii, and be careful here — this is where half the marks are lost. $$a = 10\ \text{mm} = 0.010\ \text{m}, \qquad b = 5\ \text{mm} = 0.005\ \text{m}$$

2. Set up the integral. The force is the same at every section (an axial pull), the area is not. At a distance $x$ from the thick end the radius is $a + \frac{(b-a)x}{L}$, so $$\Delta L = \int_0^{L}\frac{F\,dx}{\pi\left(a+\frac{(b-a)x}{L}\right)^{2}Y}$$

3. Do the integral. The substitution $u = a+\frac{(b-a)x}{L}$ turns it into $\int u^{-2}du$, and it comes out as $$\Delta L = \frac{FL}{\pi a b Y}$$

4. Substitute. $$\Delta L = \frac{\left(20\times10^{3}\right)\left(1.0\right)}{\pi\left(0.010\right)\left(0.005\right)\left(2.0\times10^{11}\right)} = \frac{2.0\times10^{4}}{3.1416\times10^{7}} = 6.37\times10^{-4}\ \text{m}$$

5. The comparison. A uniform rod of the mean radius $7.5$ mm has $A = \pi\left(0.0075\right)^2 = 1.767\times10^{-4}$ m$^2$, giving $$\Delta L = \frac{FL}{AY} = \frac{2.0\times10^{4}}{\left(1.767\times10^{-4}\right)\left(2.0\times10^{11}\right)} = 5.66\times10^{-4}\ \text{m}$$

Final Answer: The tapered rod stretches $6.37\times10^{-4}$ m, that is 0.637 mm. The mean-radius estimate gives 0.566 mm. Measured against the exact value, as every error in this chapter is, that estimate is 11.1% low: $\frac{0.637-0.566}{0.637} = 0.111$. (Turned round the other way, the tapered rod stretches 12.5% more than the mean-radius rod — same two numbers, different denominator, so always say which one you divided by.)

Takeaway: $\pi a b$ is the geometric mean of the two end areas, and the geometric mean of two unequal numbers is always less than the arithmetic mean. So a real tapered bar always stretches more than the "average thickness" shortcut predicts, and the thinner end is doing far more than its share of the stretching.

Example 2: The bar whose width grows linearly

An aluminium bar 0.60 m long has a constant thickness of 10 mm, while its width increases uniformly from 20 mm at one end to 40 mm at the other. It carries an axial pull of 12 kN. Find its elongation. Take $Y_{\text{aluminium}} = 0.70\times10^{11}$ Pa.

Solution:

1. Area at distance $x$ from the narrow end, with $t = 0.010$ m the thickness: $$A(x) = t\left(b_1+\frac{\left(b_2-b_1\right)x}{L}\right)$$

2. Integrate. With the area linear in $x$ rather than quadratic, the integral is a logarithm: $$\Delta L = \int_0^{L}\frac{F\,dx}{A(x)Y} = \frac{FL}{Y\,t\,\left(b_2-b_1\right)}\ln\frac{b_2}{b_1}$$

3. Substitute, with $b_2-b_1 = 0.020$ m and $\frac{b_2}{b_1}=2$: $$\Delta L = \frac{\left(12\times10^{3}\right)\left(0.60\right)}{\left(0.70\times10^{11}\right)\left(0.010\right)\left(0.020\right)}\ln 2 = \left(5.143\times10^{-4}\right)\left(0.6931\right)$$ $$\Delta L = 3.57\times10^{-4}\ \text{m}$$

4. Sanity check against the mean width, 30 mm: $A = 3.0\times10^{-4}$ m$^2$ and $\Delta L = 3.43\times10^{-4}$ m — only 3.8% below the exact value.

Final Answer: $\Delta L = 3.57\times10^{-4}$ m, that is 0.357 mm.

Takeaway: Compare this with Example 1. Both bars taper, both are pulled axially — and the answers have completely different forms, one a simple product and one a logarithm, because a circular taper makes the area go as $x^2$ while a width taper makes it go as $x$. You cannot guess a non-uniform result; you have to set the integral up. Notice also how much smaller the correction is here (3.8%) than there (11.1%): a linear area varies far more gently than a quadratic one.

Example 3: A cone hanging under its own weight

A solid steel cone of height 3.0 m hangs with its circular base fixed to a ceiling and its apex pointing straight down. Find the elongation produced by its own weight, and compare it with that of a uniform steel bar of the same length. Take $\rho_{\text{steel}} = 7800$ kg/m$^3$, $Y = 2.0\times10^{11}$ Pa, $g = 9.8$ m/s$^2$.

Solution:

1. Measure $x$ up from the apex. If the base radius is $R$, the radius at height $x$ is $\frac{Rx}{L}$, so $$A(x) = \pi\frac{R^2x^2}{L^2}$$

2. The weight below that section is the weight of the little cone of height $x$: $$W(x) = \rho g\cdot\frac{1}{3}\pi\frac{R^2x^2}{L^2}\cdot x$$

3. The stress at $x$ — and watch $R$ and $L$ cancel completely: $$\frac{W(x)}{A(x)} = \frac{\rho g x}{3}$$

4. Integrate the strain: $$\Delta L = \int_0^{L}\frac{\rho g x}{3Y}\,dx = \frac{\rho g L^2}{6Y}$$

5. Substitute: $$\Delta L = \frac{\left(7800\right)\left(9.8\right)\left(3.0\right)^{2}}{6\left(2.0\times10^{11}\right)} = \frac{6.880\times10^{5}}{1.2\times10^{12}} = 5.73\times10^{-7}\ \text{m}$$

6. The uniform bar of the same length gives $\frac{\rho gL^2}{2Y} = 1.72\times10^{-6}$ m.

Final Answer: The cone stretches $5.73\times10^{-7}$ m, that is 0.573 micrometre — exactly one third of the uniform bar's $1.72\times10^{-6}$ m.

Takeaway: Two things fell out that a student never expects. First, the base radius cancelled: a fat cone and a thin cone of the same height stretch by the same amount. Second, the factor is exactly $\frac{1}{3}$, and it comes from the $\frac{1}{3}$ in the volume of a cone. Both are asked directly, so it is worth carrying $\frac{\rho gL^2}{6Y}$ alongside $\frac{\rho gL^2}{2Y}$.

Example 4: Spinning a rod about one end

A steel rod of length 1.2 m and uniform cross-section 4.0 mm$^2$ is rotated in a horizontal plane about a vertical axis through one end, at 300 revolutions per minute. Find the maximum tension, the maximum stress, and the total elongation. Take $\rho_{\text{steel}} = 7800$ kg/m$^3$ and $Y = 2.0\times10^{11}$ Pa.

Solution:

1. Angular speed and mass: $$\omega = \frac{2\pi\left(300\right)}{60} = 31.42\ \text{rad/s}, \qquad m = \rho A L = \left(7800\right)\left(4.0\times10^{-6}\right)\left(1.2\right) = 0.03744\ \text{kg}$$

2. Tension at distance $x$ from the axis. The section there must supply the centripetal force for everything beyond it: $$T(x) = \int_x^{L}\rho A\omega^2 s\,ds = \frac{\rho A\omega^2}{2}\left(L^2-x^2\right)$$

3. Maximum tension, at the axis where $x = 0$: $$T_{\max} = \frac{\rho A\omega^2L^2}{2} = \frac{m\omega^2L}{2} = \frac{\left(0.03744\right)\left(31.42\right)^{2}\left(1.2\right)}{2} = 22.2\ \text{N}$$

4. Maximum stress: $$\frac{T_{\max}}{A} = \frac{22.17}{4.0\times10^{-6}} = 5.54\times10^{6}\ \text{Pa}$$

5. Elongation: $$\Delta L = \int_0^{L}\frac{T(x)}{AY}dx = \frac{\rho\omega^2}{2Y}\left(L^3-\frac{L^3}{3}\right) = \frac{\rho\omega^2L^3}{3Y}$$ $$\Delta L = \frac{\left(7800\right)\left(987.0\right)\left(1.728\right)}{3\left(2.0\times10^{11}\right)} = 2.22\times10^{-5}\ \text{m}$$

Final Answer: $T_{\max} = 22.2$ N at the axis, peak stress $5.54\times10^{6}$ Pa, elongation $2.22\times10^{-5}$ m, that is 0.022 mm.

Takeaway: $T_{\max} = \frac{m\omega^2L}{2}$ is the same as putting the whole mass at the midpoint — which is where the centre of mass is. That is a genuine shortcut for the maximum tension, and it is not a shortcut for the elongation, which needs the whole profile and gives $\frac{\rho\omega^2L^3}{3Y}$, not $\frac{m\omega^2L^2}{2AY}$. Spin the same rod about its centre instead and the elongation drops to one quarter of this.

Example 5: Three identical wires, and a load that is not in the middle

A light rigid horizontal bar hangs from three identical vertical steel wires attached at positions 0, 0.60 m and 1.20 m along it. Each wire is 2.0 m long with a cross-section of 1.0 mm$^2$ and $Y = 2.0\times10^{11}$ Pa. A load of 600 N is hung at 0.30 m from the first wire. Find the tension in each wire and the extension of each.

Solution:

1. The spring constant of each wire: $$k = \frac{YA}{L} = \frac{\left(2.0\times10^{11}\right)\left(1.0\times10^{-6}\right)}{2.0} = 1.0\times10^{5}\ \text{N/m}$$

2. Count the unknowns. Three tensions, but only two equations of statics. The third relation is the rigidity of the bar: after stretching, it is still straight, so the extensions lie on a straight line, $e(x) = e_0 + mx$. Since the wires are identical, $T_i = k\,e(x_i)$.

3. Force balance. $T_1+T_2+T_3 = 600$, that is $k\left(3e_0+3\left(0.60\right)m\right) = 600$.

4. Torque balance about the first wire, with the load 0.30 m along: $$T_2\left(0.60\right)+T_3\left(1.20\right) = 600\left(0.30\right) = 180$$

5. Solve the pair. Eliminating $e_0$ gives $m = -\frac{W}{4kd}$ with $d = 0.60$ m, and then $e_0 = \frac{7W}{12k}$. Feeding those back: $$T_1 = \frac{7W}{12} = 350\ \text{N}, \qquad T_2 = \frac{W}{3} = 200\ \text{N}, \qquad T_3 = \frac{W}{12} = 50\ \text{N}$$

6. Check both equations. Sum $= 350+200+50 = 600$ N. Torque $= 200\left(0.60\right)+50\left(1.20\right) = 120+60 = 180$ N m. Both hold.

7. Extensions, from $e = \frac{T}{k}$: $$e_1 = 3.5\ \text{mm}, \qquad e_2 = 2.0\ \text{mm}, \qquad e_3 = 0.5\ \text{mm}$$

Final Answer: $T_1 = 350$ N, $T_2 = 200$ N, $T_3 = 50$ N, with extensions 3.5 mm, 2.0 mm and 0.5 mm.

Takeaway: Look at the three extensions: 3.5, 2.0, 0.5 — falling in equal steps of 1.5 mm. That is the compatibility condition made visible, $e_1+e_3 = 2e_2$, and it is the fastest way to check your own answer. Note also that the far wire is still in tension, not compression; if the load had been outside the wires, that would not have been true, and a wire cannot push.

Example 6: Where must the load hang for the rod to stay horizontal?

A light rigid rod 1.0 m long hangs horizontally from a steel wire at one end and a copper wire at the other. Both wires are 2.0 m long; the steel wire has a cross-section of 1.0 mm$^2$ and the copper wire 2.0 mm$^2$. Take $Y_{\text{steel}} = 2.0\times10^{11}$ Pa and $Y_{\text{copper}} = 1.2\times10^{11}$ Pa. Where must a load be hung (a) for the rod to stay horizontal, and (b) for the two wires to be under equal stress?

Solution:

(a) Horizontal — equal extensions.

1. The spring constants are $$k_{\text{steel}} = \frac{\left(2.0\times10^{11}\right)\left(1.0\times10^{-6}\right)}{2.0} = 1.0\times10^{5}\ \text{N/m}, \qquad k_{\text{copper}} = \frac{\left(1.2\times10^{11}\right)\left(2.0\times10^{-6}\right)}{2.0} = 1.2\times10^{5}\ \text{N/m}$$

2. Equal extensions means the tensions are in the ratio of the spring constants: $$\frac{T_{\text{steel}}}{T_{\text{copper}}} = \frac{k_{\text{steel}}}{k_{\text{copper}}} = \frac{1.0}{1.2}$$

3. Torque balance about the steel end, with the load $x$ from the steel wire: $$T_{\text{steel}}\,x = T_{\text{copper}}\left(1.0-x\right) \qquad\Longrightarrow\qquad \frac{x}{1.0-x} = \frac{T_{\text{copper}}}{T_{\text{steel}}} = 1.2$$ $$x = \frac{1.2}{2.2} = 0.545\ \text{m from the steel wire}$$

(b) Equal stress.

4. Equal stress means $\frac{T_{\text{steel}}}{A_{\text{steel}}} = \frac{T_{\text{copper}}}{A_{\text{copper}}}$, so the tensions are in the ratio of the areas, $1 : 2$.

5. Torque balance again: $$\frac{x}{1.0-x} = \frac{T_{\text{copper}}}{T_{\text{steel}}} = 2 \qquad\Longrightarrow\qquad x = \frac{2}{3} = 0.667\ \text{m from the steel wire}$$

Final Answer: (a) 0.545 m from the steel wire; (b) 0.667 m from the steel wire. Two different places.

Takeaway: Two conditions, two answers, and 12 cm between them. "Stays horizontal" is about extensions; "equal stress" is about areas. The examiner will phrase one of them and offer the other among the options, every time. Note the inversion in the torque step: the load sits nearer the wire carrying the larger tension, which is why $\frac{x}{d-x} = \frac{T_2}{T_1}$ and not the other way up.