A wire of length 20 m and area of cross section 1.25 × 10^-4 m^2 is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N).

Question

A wire of length 20 m and area of cross section 1.25 × 10^-4 m^2 is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N).

A wire of length 20 m and area of cross section 1.25 × 10^-4 m² is subjected to a load of 2.5 kg. (1 kg wt = 9.8 N). The elongation produced in wire is 1 × 10^-4 m. Calculate Young’s modulus of the material.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

L = 20 m, A = 1.25 × 10^-4m²,

F = mg = 2.5 × 9.8 N, l = 10^-4 m

Formula Y = \(\frac{FL}{Al}\)

To find: Young’s modulus (Y)

Calculation: From formula,

Y = \(\frac{2.5\times9.8\times20}{1.25\times10^{-4}\times10^{-4}}\) = 3.92 × 10¹⁰Nm^-2

The Young’s modulus of the material is 3.92 × 10¹⁰ Nm^-2.

7 views Answered 2023-02-05 15:29