Physics Q13 – NEET-UG 2024

Home Physics Mechanical Properties of SolidsQuestion #289

The maximum elongation of a steel wire of

1 m

length if the elastic limit of steel and its Young's modulus, respectively, are

8 \times 10^{8} {N m}^{- 2}

and

2 \times 10^{11} {N m}^{- 2}

, is :

Clue 1 of 3: Elasticity and Hooke's Law

Understand the relationship between stress, strain, and Young's modulus within the elastic limit.

🧠 Full Solution Path

Step 1: Identify the relevant formula✦ Active

The relationship between Young's modulus ( $Y$ ), stress ( $σ$ ), and strain ( $ϵ$ ) is given by Hooke's Law within the elastic limit:

Y = \frac{Stress}{Strain} = \frac{σ}{ϵ}

Strain is defined as the ratio of change in length (elongation, $Δ L$ ) to the original length ( $L$ ):

ϵ = \frac{Δ L}{L}

Step 2: Rearrange the formula and substitute values○ Expand

From the above relations, we can express the elongation $Δ L$ as:

Δ L = \frac{σ \times L}{Y}

Given values are: Elastic limit (maximum stress), $σ = 8 \times 10^{8} {N m}^{- 2}$ Original length, $L = 1 m$ Young's modulus, $Y = 2 \times 10^{11} {N m}^{- 2}$ Substitute these values into the formula:

Δ L = \frac{(8 \times 10^{8} {N m}^{- 2}) \times (1 m)}{2 \times 10^{11} {N m}^{- 2}}

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units in this case) before calculation.

Step 3: Calculate the elongation and convert units○ Expand

Perform the calculation:

Δ L = \frac{8}{2} \times 10^{8 - 11} m = 4 \times 10^{- 3} m

Convert the elongation from meters to millimeters ( $1 m = 1000 mm$ ):

Δ L = 4 \times 10^{- 3} \times 10^{3} mm = 4 mm

💡 Teacher's Secret Hint

Always check the required units for the final answer and perform necessary conversions.

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Physics Question 13 – NEET-UG 2024

Clue 1 of 3: Elasticity and Hooke's Law

🧠 Full Solution Path