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Q.A steel rod has a radius of 10 mm and length of 1 m. A 100 kN force F stretches it along its length. Calculate

(a) stress,
(b) elongation and
(c) strain on the rod, given Young's modulus of steel = 2 x 10^11 N/m^2. OR The average depth of the Indian Ocean is about 3000 m. Calculate the fractional compression (delta V / V) of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 x 10^9 N/m^2. (Take g = 10 m/s^2)
Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2026Subjective· 3mImportance★★★★★
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For the steel rod: stress ≈ 3.18×10⁸ N/m², elongation ≈ 1.59 mm, strain ≈ 1.59×10⁻³.

Given: radius r = 10 mm = 1.0 × 10⁻² m, length L = 1 m, force F = 100 kN = 1.0 × 10⁵ N, Young's modulus Y = 2 × 10¹¹ N/m².

Step 1 — Cross-sectional area:

A = πr² = π × (1.0×10⁻²)² = π × 1.0×10⁻⁴ ≈ 3.1416 × 10⁻⁴ m²

Step 2 — (a) Stress:

Stress = F/A = (1.0×10⁵) / (3.1416×10⁻⁴) ≈ 3.183 × 10⁸ N/m²

Step 3 — (c) Strain (from the definition of Young's modulus, Y = stress/strain): …

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