A metal bar with a cross-sectional area of 50mm2 is subjected to a tensile strain of 0.001. What is the tensile force responsible for the strain given that the material has a Young's modulus of 15GNm raise to the power -2

A 0.003N
B 7.5x10 raise to the power 3
C 7.5x 10 raise to the power -10
D 3x10 raise to the power 12

To calculate the tensile force responsible for the given strain, you can use the formula:

Tensile Force = Young's Modulus × Cross-sectional Area × Tensile Strain

Given values:
Cross-sectional Area (A) = 50 mm^2 = 50 × 10^(-6) m^2 (since 1 mm^2 = 10^(-6) m^2)
Tensile Strain (ε) = 0.001
Young's Modulus (Y) = 15 GNm^(-2) = 15 × 10^9 Nm^(-2) (since 1 GN = 10^9 N)

Now, substitute the values into the formula:
Tensile Force = (15 × 10^9 Nm^(-2)) × (50 × 10^(-6) m^2) × (0.001)

Simplifying the equation:
Tensile Force = 7.5 × 10^(-4) × 10^9 N

Rearranging the equation to a suitable form:
Tensile Force = 7.5 × 10^(5 - 4) N
Tensile Force = 7.5 × 10^1 N
Tensile Force = 75 N

Therefore, the tensile force responsible for the given strain is 75 N.

None of the answer options A, B, C, or D matches the correct value of 75 N.

To find the tensile force responsible for the strain, we can use the formula:

Tensile force = Young's modulus * Cross-sectional area * Tensile strain

Given:
- Cross-sectional area (A) = 50mm^2
- Tensile strain (ε) = 0.001
- Young's modulus (Y) = 15GNm^-2 = 15 * 10^9 Nm^-2

Plugging in the values into the formula:

Tensile force = 15GNm^-2 * 50mm^2 * 0.001

Remember to convert mm^2 to m^2 and simplify the units:

1 mm^2 = (1 * 10^-3 m)^2 = 10^-6 m^2

Tensile force = 15 * 10^9 Nm^-2 * 50 * 10^-6 m^2 * 0.001

Simplifying:

Tensile force = 750 N

Therefore, the tensile force responsible for the strain is 750 N, which is option B: 7.5 x 10^2 N.