A block of weight 100 N is suspended by copper and steel wires of same cross sectional area 0.5 cm^2

Q: Why do we take the angles with the horizontal instead of the vertical?

The problem explicitly states that the angles subtended by the wires are 'with the ceiling', which is horizontal. Always read the reference line for angles carefully!

Q: Can we solve this using Lami's Theorem?

Absolutely! Lami's Theorem is a great shortcut here. The angle between $T_S$ and $W$ is $90^\circ + 60^\circ = 150^\circ$, and between $T_C$ and $W$ is $90^\circ + 30^\circ = 120^\circ$. The angle between $T_S$ and $T_C$ is $180^\circ - 60^\circ - 30^\circ = 90^\circ$. Applying $\frac{T_S}{\sin 120^\circ} = \frac{T_C}{\sin 150^\circ} = \frac{100}{\sin 90^\circ}$ will give you the exact same tensions much faster!

Q: Why did the area $0.5\text{ cm}^2$ not affect the final answer?

Because we were asked for a *ratio* of elongations, and both wires had the *same* cross-sectional area. When you divide the two elongation expressions, the area term $A$ cancels out completely. The specific value $0.5\text{ cm}^2$ is just extra information meant to distract you.

Q: What if the wires had different areas?

If the areas were different, say $A_C$ and $A_S$, they wouldn't cancel out. You would simply include the inverse ratio of their areas $\left(\frac{A_S}{A_C}\right)$ in your final multiplication step.

Q: Is it necessary to convert Young's Modulus to different units?

No, as long as you are consistent. Here, Young's Modulus is in $\text{N/m}^2$, lengths are in meters, and forces are in Newtons. Everything is in standard SI units, so no conversions are needed. Even if they weren't, taking a ratio often cancels out the need for unit conversions if both quantities share the same units.

Question

A block of weight $100	ext{ N}$ is suspended by copper and steel wires of same cross sectional area $0.5	ext{ cm}^2$ and, length $\sqrt{3}	ext{ m}$ and $1	ext{ m}$, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are $30^\circ$ and $60^\circ$, respectively. If elongation in copper wire is $(\Delta \ell_C)$ and elongation in steel wire is $(\Delta \ell_S)$, then the ratio $\frac{\Delta \ell_C}{\Delta \ell_S}$ is _______. [Young's modulus for copper and steel are $1 	imes 10^{11}	ext{ N/m}^2$ and $2 	imes 10^{11}	ext{ N/m}^2$ respectively]

Sigma Percentile · Accepted Answer

The beauty of physics often lies in how different domains seamlessly intertwine. In this classic JEE Advanced problem, we are presented with a system that perfectly marries **Newtonian Mechanics** (specifically, static equilibrium) with the **Properties of Matter** (elasticity and Hooke's Law).

At first glance, the problem might look like a standard elasticity question, but the real challenge is unlocking the forces hidden within the wires. Let's break down the journey step-by-step.

### Analyzing the Setup

Imagine you are standing exactly at the knot where the block is suspended. What forces do you feel?

First, there is the undeniable downward pull of gravity. The block has a weight of $W = 100	ext{ N}$, acting straight down. To counteract this and keep the knot from falling, the two wires must pull upwards and outwards.

The steel wire pulls up and to the left with a tension $T_S$, making an angle of $60^\circ$ with the horizontal ceiling. By alternate interior angles, it also makes a $60^\circ$ angle with the horizontal at the knot. Similarly, the copper wire pulls up and to the right with a tension $T_C$, making a $30^\circ$ angle with the horizontal.

Because the block is perfectly still, the system is in **static equilibrium**. This means the net force in any direction must be exactly zero.

### The Master Equations of Equilibrium

To solve for the unknown tensions, we must resolve the forces into their horizontal ($x$) and vertical ($y$) components.

**1. Horizontal Equilibrium ($\sum F_x = 0$):**
The leftward pull must exactly balance the rightward pull. 
$$T_S \cos 60^\circ = T_C \cos 30^\circ$$

Substituting the trigonometric values:
$$T_S \left(\frac{1}{2}ight) = T_C \left(\frac{\sqrt{3}}{2}ight)$$
$$T_S = \sqrt{3} T_C$$

This is a beautiful, simple relationship. It tells us that the steel wire is carrying significantly more tension than the copper wire.

**2. Vertical Equilibrium ($\sum F_y = 0$):**
The combined upward pull of both wires must support the entire $100	ext{ N}$ weight.
$$T_S \sin 60^\circ + T_C \sin 30^\circ = 100$$

Substituting the sine values:
$$T_S \left(\frac{\sqrt{3}}{2}ight) + T_C \left(\frac{1}{2}ight) = 100$$

### Solving for the Tensions

Now, we substitute our horizontal relationship ($T_S = \sqrt{3} T_C$) into the vertical equation:
$$(\sqrt{3} T_C) \left(\frac{\sqrt{3}}{2}ight) + \frac{T_C}{2} = 100$$
$$\frac{3 T_C}{2} + \frac{T_C}{2} = 100$$
$$\frac{4 T_C}{2} = 100$$
$$2 T_C = 100 \implies T_C = 50	ext{ N}$$

With $T_C$ found, $T_S$ is simply:
$$T_S = 50\sqrt{3}	ext{ N}$$

We have successfully cracked the mechanics part of the problem!

### Hooke's Law and Elongation

Now we transition to elasticity. We need to find the ratio of the elongations, $\frac{\Delta \ell_C}{\Delta \ell_S}$.

Recall the definition of Young's Modulus ($Y$):
$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta \ell / \ell}$$

Rearranging this to solve for elongation ($\Delta \ell$):
$$\Delta \ell = \frac{F \ell}{A Y}$$

Let's set up the ratio for the copper and steel wires:
$$\frac{\Delta \ell_C}{\Delta \ell_S} = \frac{\frac{T_C \ell_C}{A_C Y_C}}{\frac{T_S \ell_S}{A_S Y_S}}$$

We can group the terms logically:
$$\frac{\Delta \ell_C}{\Delta \ell_S} = \left(\frac{T_C}{T_S}ight) \left(\frac{\ell_C}{\ell_S}ight) \left(\frac{A_S}{A_C}ight) \left(\frac{Y_S}{Y_C}ight)$$

The problem generously states that both wires have the **same cross-sectional area** ($0.5	ext{ cm}^2$). This means the area ratio $\left(\frac{A_S}{A_C}ight)$ is simply $1$. The specific value of $0.5	ext{ cm}^2$ is a classic distractor!

### The Final Calculation

Let's plug in all our known ratios:
- Tension ratio: $\frac{T_C}{T_S} = \frac{50}{50\sqrt{3}} = \frac{1}{\sqrt{3}}$
- Length ratio: $\frac{\ell_C}{\ell_S} = \frac{\sqrt{3}}{1} = \sqrt{3}$
- Young's Modulus ratio: $\frac{Y_S}{Y_C} = \frac{2 	imes 10^{11}}{1 	imes 10^{11}} = 2$

Multiplying them all together:
$$\frac{\Delta \ell_C}{\Delta \ell_S} = \left(\frac{1}{\sqrt{3}}ight) 	imes (\sqrt{3}) 	imes (1) 	imes (2)$$

The $\sqrt{3}$ terms cancel out perfectly, leaving us with a highly satisfying, clean integer:
$$\frac{\Delta \ell_C}{\Delta \ell_S} = 2$$

### Conclusion

This problem is a masterclass in structured thinking. By isolating the mechanics from the elasticity, we avoided getting overwhelmed by variables. Always remember to look for ratios—they often cause messy terms (like the $\sqrt{3}$ and the cross-sectional area) to vanish elegantly!

Enter Numerical Value:

Visualized Solution

$Free Body Diagram$

$Equilibrium of Forces$

$Horizontal Equilibrium$

$Relation between Tensions$

$Vertical Equilibrium$

$Solving for Tensions$

$Hooke’s Law for Elongation$

$Ratio of Elongations$

$Final Calculation$

$Conclusion & Extensions$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equations of Equilibrium

Solving for the Tensions

Hooke's Law and Elongation

The Final Calculation

Conclusion

Similar Questions

Two wires of same length and radius are joined end-to-end and loaded. The Young's moduli of the materials of the two wires are $Y_{1}$ and $Y_{2}$ . The combination behaves as a single wire, then its Young's modulus is

Young's moduli of two wires $A$ and $B$ are in the ratio $7 : 4$ . Wire $A$ is $2 m$ long and has radius $R$ . Wire $B$ is $1.5 m$ long and has radius $2 mm$ . If the two wires stretch by the same length for a given load, then the value of $R$ is close to

Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is $1 : 4$ , the ratio of their diameters is

A uniform heavy rod of weight $10 kg ms^{- 2}$ , cross-sectional area $100 cm^{2}$ and length $20 cm$ is hanging from a fixed support. Young's modulus of the material of the rod is $2 \times 1 0^{11} Nm^{- 2}$ . Neglecting the lateral contraction, find the elongation of rod due to its own weight.

A wire elongates by $l$ mm when a load $w$ is hanged from it. If the wire goes over a pulley and two weights $w$ each are hung at the two ends, the elongation of the wire will be (in mm)

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium of Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Horizontal Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation between Tensions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vertical Equilibrium

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Tensions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hooke’s Law for Elongation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Elongations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion & Extensions