A load of 2 kg produces an extension of 1 mm in a wire of 3 m in length and 1 mm in diameter. The Young's modulus of wire will be

Question

A load of 2 kg produces an extension of 1 mm in a wire of 3 m in length and 1 mm in diameter. The Young's modulus of wire will be (A) 3.25 × 1010 Nm -2 (B) 7.48 × 1012 Nm 2 (C) 7.48 × 1010 Nm -2 (D) 7.48 × 10-10 Nm -2

Tardigrade Admin · Accepted Answer

We know ( text Force × text Length / text Area of cross-section × text elongation )= text Young's Modulus =(F × L/A × Δ L)=Y F=2 × 10 N , A=π ×(1 / 2)2 × 10-6 m 2 , L=3 m Δ L=1 × 10-3 m Substituting values (20 × 3/π × (1)4 × 10-6 × 1 × 10-3)=Y (20 × 3 × 4/3.14 × 10-9)=Y 7.48 × 1010 Nm -2=Y

Q. A load of $2 k g$ produces an extension of $1 mm$ in a wire of $3 m$ in length and $1 mm$ in diameter. The Young's modulus of wire will be

$3.25 \times 1 0^{10} N m^{- 2}$

$7.48 \times 1 0^{12} N m^{2}$

$7.48 \times 1 0^{10} N m^{- 2}$

$7.48 \times 1 0^{- 10} N m^{- 2}$

Q. A load of 2kg produces an extension of 1mm in a wire of 3m in length and 1mm in diameter. The Young's modulus of wire will be

3.25×1010Nm−2

7.48×1012Nm2

7.48×1010Nm−2

7.48×10−10Nm−2

We know Area of cross-section × elongation Force × Length ​= Young’s Modulus =A×ΔLF×L​=Y {F=2×10N,A=π×(1/2)2×10−6m2,<br/>L=3mΔL=1×10−3m} Substituting values π×41​×10−6×1×10−320×3​=Y 3.14×10−920×3×4​=Y 7.48×1010Nm−2=Y

Q. A load of $2 k g$ produces an extension of $1 mm$ in a wire of $3 m$ in length and $1 mm$ in diameter. The Young's modulus of wire will be

$3.25 \times 1 0^{10} N m^{- 2}$

$7.48 \times 1 0^{12} N m^{2}$

$7.48 \times 1 0^{10} N m^{- 2}$

$7.48 \times 1 0^{- 10} N m^{- 2}$