Question

9.
An aluminium rod has a breaking strain 0.2%.
The minimum cross-sectional area of the rod
in m² in order to support a load of 10^4N is if
Youngs modulus is 7 x 10'Nm-2

1) 1.7 x 10-4 2) 1.7 x 10-3

3) 7.1 x 10-4 4) 1.4 x 10-4​

Answer 1

answer : option (3) 7.1 × 10^-4 m²

it is given that,

percentage strain = ∆l/l × 100 = 0.2%

⇒ strain = ∆l/l = 0.2/100 = 0.002

Young's modulus, Y = 7 × 10^9 N/m²

load , F = 10⁴ N

we know, Young's modulus = stress/strain

⇒Y = (F/A)/(∆l/l) = F/A(∆l/l)

⇒7 × 10^9 = 10⁴/(A × 0.002)

⇒A = 10⁴ × 10³/(7 × 2 × 10^9)

⇒ A = 10^-2/(14) = 0.071 × 10^-2 m²

= 7.1 × 10^-4 m²

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Answer 2

$\bf\large\underline\green{Answer:-}$

3) 7.1 x 10-4