Verified Solution • 2025–26 Session moderate

Q8.1: MECHANICAL PROPERTIES OF SOLIDS Solutions — NCERT Physics - Part 2

⚠️ Common Exam Traps & Conceptual Pitfalls

When deriving the ratio of Young's moduli for wires with equal extension and load, students frequently fail to invert the cross-sectional area ratio, incorrectly calculating $A_1/A_2$ instead of $A_2/A_1$. Remember that since $Y \propto 1/A$ for a fixed extension, the area of the second wire must appear in the numerator of the ratio expression.

A steel wire of length 4.7 m and cross-sectional area 3.0 _ 10__ m_ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 _ 10__ m_ under a given load. What is the ratio of the Young's modulus of steel to that of copper?

Step-by-Step Verified Solution

**Answer:**

1.79 : 1

**Step-by-Step Explanation:** Given: Steel wire:

L_1 = 4.7

m,

A_1 = 3.0 \times 10^{-5}

m_. Copper wire:

L_2 = 3.5

m,

A_2 = 4.0 \times 10^{-5}

m_. Both stretch by the same amount

\Delta L

under the same load

F

. Young's modulus:

Y = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}

For steel:

Y_1 = \frac{F L_1}{A_1 \Delta L}

For copper:

Y_2 = \frac{F L_2}{A_2 \Delta L}

Taking the ratio:

\frac{Y_1}{Y_2} = \frac{L_1}{L_2} \times \frac{A_2}{A_1} = \frac{4.7}{3.5} \times \frac{4.0 \times 10^{-5}}{3.0 \times 10^{-5}} = 1.343 \times 1.333 = 1.79

Hence, the ratio of Young's modulus of steel to copper is

1.79 : 1

[citation:1][citation:11].

All Questions in this Chapter

Q8.1 Q8.2 Q8.3 Q8.4 Q8.5 Q8.6 Q8.7 Q8.8 Q8.9 Q8.10 Q8.11 Q8.12 Q8.13 Q8.14 Q8.15 Q8.16

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