In a tensile test on a metal bar of diameter 0.015 m and length 0.2 m, the relation between the load and elongation within the proportional limit is found to be F=97.2 × 106(Δ L), where F is the load (in N ) and Δ L is the elongation (in m ). The Young's modulus of the material in GPa is

Question

In a tensile test on a metal bar of diameter 0.015 m and length 0.2 m, the relation between the load and elongation within the proportional limit is found to be F=97.2 × 106(Δ L), where F is the load (in N ) and Δ L is the elongation (in m ). The Young's modulus of the material in GPa is (A) 75.5 (B) 85.6 (C) 98.7 (D) 110

Tardigrade Admin · Accepted Answer

Given that, diameter of metal bar (d)=0 ⋅ 015 m length (L) =0 ⋅ 2 m F =97 ⋅ 2 × 106 × Δ L We know that, Y=(F L/A ⋅ Δ L) Y=(97.2 × 106 × 0.2 × 4/3.14 ×(0.015)2) =110063.69 × 106 ≈ 110 GPa

Q. In a tensile test on a metal bar of diameter 0.015m and length 0.2m, the relation between the load and elongation within the proportional limit is found to be F=97.2×106(ΔL), where F is the load (in N ) and ΔL is the elongation (in m ). The Young's modulus of the material in GPa is

75.5

85.6

98.7

110

Given that, diameter of metal bar (d)=0⋅015m length (L)=0⋅2m F=97⋅2×106×ΔL We know that, Y=A⋅ΔLFL​ Y=3.14×(0.015)297.2×106×0.2×4​ =110063.69×106≈110GPa

Given that, diameter of metal bar $(d) = 0 \cdot 015 m$
length $(L) = 0 \cdot 2 m$
$F = 97 \cdot 2 \times 1 0^{6} \times Δ L$
We know that, $Y = \frac{F L}{A \cdot Δ L}$
$Y = \frac{97.2 \times 1 0 ^{6} \times 0.2 \times 4}{3.14 \times ( 0.015 ) ^{2}}$
$= 110063.69 \times 1 0^{6} \approx 110 GP a$