Steel wire of length L at 40°C is suspended from the ceiling and then a mass m is hung from its free

Q: Why do we equate the two changes in length?

The problem states that the wire regains its *original* length $L$. This means the amount it stretched downwards due to the weight is exactly equal to the amount it shrank upwards due to the cooling.

Q: Why doesn't the original length $L$ matter?

Both the mechanical extension and the thermal contraction are directly proportional to the original length $L$. When you equate them, $L$ appears on both sides and beautifully cancels out. The required mass depends only on the material properties and the temperature change!

Q: What value of $g$ should I use?

In JEE problems, unless specified otherwise, using $g = 9.8 \text{ m/s}^2$ or $g = 10 \text{ m/s}^2$ might give slightly different decimals. Here, using $g = 9.8$ gives $3.2$ kg, and $g = 10$ gives $3.14$ kg. Both round to the nearest integer, which is $3$.

Q: Does the order of events matter?

Not for the final state! Whether you cool it first and then hang the mass, or hang the mass and then cool it, the final equilibrium length depends only on the final temperature and the final suspended mass.

Q: Why did we use the formula for linear expansion instead of volume expansion?

Because we are only concerned with the change in the 1D length of the wire. The changes in the cross-sectional area due to temperature are negligible and don't affect the length constraint given in the problem.

Question

Steel wire of length $L$ at $40^{\circ}	ext{C}$ is suspended from the ceiling and then a mass $m$ is hung from its free end. The wire is cooled down from $40^{\circ}	ext{C}$ to $30^{\circ}	ext{C}$ to regain its original length $L$. The coefficient of linear thermal expansion of the steel is $10^{-5}/^{\circ}	ext{C}$, Young's modulus of steel is $10^{11}	ext{ N/m}^2$ and radius of the wire is $1	ext{ mm}$. Assume that $L \gg 	ext{diameter of the wire}$. Then the value of $m$ in kg is nearly.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the delicate balance of opposing forces. In this problem, we witness a fascinating tug-of-war between mechanical stress and thermal contraction. Let's dive into the mechanics of this steel wire!

### The Setup: A Tug of War Between Forces

Imagine a steel wire of length $L$ suspended from a ceiling. Initially, it rests at a comfortable $40^{\circ}	ext{C}$.

When we hang a mass $m$ from its free end, gravity pulls it downwards, causing the wire to stretch. But then, we cool the wire down to $30^{\circ}	ext{C}$. Cooling causes the atoms in the steel lattice to pull closer together, shrinking the wire.

The problem presents a perfect equilibrium: the wire regains its exact original length $L$. This means the downward mechanical stretch is perfectly neutralized by the upward thermal contraction.

### The Mechanical Stretch

First, let's quantify the mechanical stretch. When a force $F = mg$ is applied to a wire of cross-sectional area $A = \pi r^2$, it experiences longitudinal stress. According to Hooke's Law, the extension $\Delta l_1$ is given by:

$$ \Delta l_1 = \frac{FL}{AY} = \frac{mgL}{\pi r^2 Y} $$

This is our first key piece of the puzzle. It tells us exactly how much the wire lengthens due to the hanging mass.

### The Thermal Contraction

Next, we cool the wire. The change in temperature is $\Delta 	heta = 40^{\circ}	ext{C} - 30^{\circ}	ext{C} = 10^{\circ}	ext{C}$. The thermal contraction $\Delta l_2$ depends on the original length, the coefficient of linear expansion $\alpha$, and the temperature drop:

$$ \Delta l_2 = L \alpha \Delta 	heta $$

This equation dictates how much the wire shrinks as it loses thermal energy.

### The Grand Equalizer

Since the wire regains its original length $L$, the mechanical extension must be exactly equal to the thermal contraction. We equate the two expressions:

$$ \frac{mgL}{\pi r^2 Y} = L \alpha \Delta 	heta $$

Notice something beautiful here? The original length $L$ appears on both sides of the equation! We can cancel it out. This reveals a profound physical truth: **the mass required to maintain the original length is completely independent of how long the wire initially was!**

Rearranging the formula to solve for the mass $m$, we get our master equation:

$$ m = \frac{\pi r^2 Y \alpha \Delta 	heta}{g} $$

### The Final Calculation

Now, it is time to plug in the numbers. We must be careful to convert all units to standard SI units. The radius $r = 1 	ext{ mm} = 10^{-3} 	ext{ m}$.

Substituting the given values into our master equation:

$$ m = \frac{\pi (10^{-3})^2 (10^{11}) (10^{-5}) (10)}{9.8} $$

Let's simplify the powers of $10$: $(10^{-6}) 	imes (10^{11}) 	imes (10^{-5}) 	imes 10 = 10^1 = 10$.

$$ m = \frac{10\pi}{9.8} \approx \frac{31.4159}{9.8} \approx 3.205 	ext{ kg} $$

The question asks for the value of $m$ in kg *nearly*. The closest integer to $3.205$ is $3$.

Therefore, the mass hung from the wire is approximately **$3 	ext{ kg}$**.

Enter Numerical Value:

Visualized Solution

$L, T_{1} = 4 0^{\circ} C$

$Δ l_{1} = \frac{F L}{A Y}$

$Δ l_{2} = Lα Δ θ$

$Δ l_{1} = Δ l_{2}$

$\frac{m g L}{π r ^{2} Y} = Lα Δ θ$

$m = \frac{π r ^{2} Y α Δ θ}{g}$

$m = \frac{π ( 1 0 ^{- 3} ) ^{2} ( 1 0 ^{11} ) ( 1 0 ^{- 5} ) ( 10 )}{9.8}$

$m \approx 3 kg$

The Sigma Insight: Thermal Expansion

The Setup

The Mechanical Stretch

The Thermal Contraction

The Grand Equalizer

The Final Calculation

Similar Questions

The pressure that has to be applied to the ends of a steel wire of length $10 cm$ to keep its length constant when its temperature is raised by $10 0^{\circ} C$ is (For steel, Young's modulus is $2 \times 1 0^{11} Nm^{- 2}$ and coefficient of thermal expansion is $1.1 \times 1 0^{- 5} K^{- 1}$ )

When the temperature of a metal wire is increased from $0^{\circ} C$ to $1 0^{\circ} C$ , its length increases by $0.02%$ . The percentage change in its mass density will be closest to

A metal rod of Young's modulus $Y$ and coefficient of thermal expansion $α$ is held at its two ends such that its length remains invariant. If its temperature is raised by $t^{\circ} C$ , the linear stress developed in it is

Two rods $A$ and $B$ of identical dimensions are at temperature $3 0^{\circ} C$ . If $A$ is heated upto $18 0^{\circ} C$ and $B$ upto $T^{\circ} C$ , then new lengths are the same. If the ratio of the coefficients of linear expansion of $A$ and $B$ is $4 : 3$ , then the value of $T$ is

Two different wires having lengths $L_{1}$ and $L_{2}$ and respective temperature coefficients of linear expansion $α_{1}$ and $α_{2}$ , are joined end-to-end. Then the effective temperature coefficient of linear expansion is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L,T1​=40∘C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl1​=AYFL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl2​=LαΔθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl1​=Δl2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } πr2YmgL​=LαΔθ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=gπr2YαΔθ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=9.8π(10−3)2(1011)(10−5)(10)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m≈3 kg