A uniform wire of length l of weight w is suspended from the roof with a weight of W at the other end. The stress in the wire at l/3 distance from the top is (W/A + 2/gamma w/A), where A is the cross sectional area of the wire. The value of gamma is _______.

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q49 – JEE-MAIN 2026

← Back

Home Physics Laws of MotionQuestion #210

A uniform wire of length

l

of weight

w

is suspended from the roof with a weight of

W

at the other end. The stress in the wire at

\frac{l}{3}

distance from the top is

(\frac{W}{A} + \frac{2}{γ} \frac{w}{A})

, where

A

is the cross sectional area of the wire. The value of

γ

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Understanding Stress

Stress at any point in a suspended wire is the total force acting on the cross-section at that point divided by the cross-sectional area.

🧠 Full Solution Path

Step 1: Determine the force acting at the specified point.✦ Active

The wire has a total length $l$ and total weight $w$ . The weight per unit length is $\frac{w}{l}$ . We need to find the stress at a distance $x = \frac{l}{3}$ from the top. The length of the wire segment below this point is $l - x = l - \frac{l}{3} = \frac{2 l}{3}$ .

The weight of this segment of the wire is $w^{'} = (\frac{w}{l}) (\frac{2 l}{3}) = \frac{2 w}{3}$ . The total force $F$ acting at this point is the sum of the external weight $W$ and the weight of the wire segment below it.

F = W + w^{'} = W + \frac{2 w}{3}

Step 2: Calculate the stress at the specified point.○ Expand

Stress $σ$ is defined as force per unit area. Given the cross-sectional area $A$ , the stress at $x = \frac{l}{3}$ is:

σ = \frac{F}{A} = \frac{W + \frac{2 w}{3}}{A} = \frac{W}{A} + \frac{2 w}{3 A}

Step 3: Compare with the given stress expression to find

γ

.○ Expand

The problem states that the stress in the wire at $\frac{l}{3}$ distance from the top is $(\frac{W}{A} + \frac{2}{γ} \frac{w}{A})$ . Equating our derived stress with the given expression:

\frac{W}{A} + \frac{2 w}{3 A} = \frac{W}{A} + \frac{2}{γ} \frac{w}{A}

Subtracting $\frac{W}{A}$ from both sides and simplifying:

\frac{2 w}{3 A} = \frac{2}{γ} \frac{w}{A}

Dividing both sides by $\frac{2 w}{A}$ (assuming $w \neq 0$ and $A \neq 0$ ):

\frac{1}{3} = \frac{1}{γ}

Therefore, the value of $γ$ is:

γ = 3

💡 Teacher's Secret Hint

Ensure to correctly identify the portion of the wire whose weight contributes to the force at the specified cross-section.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 49 – JEE-MAIN 2026

Clue 1 of 3: Understanding Stress

🧠 Full Solution Path