Final answer:
The energy stored in a stretched wire with given parameters is calculated using the elastic potential energy formula with Young's modulus, resulting in the correct answer, which is option d.) 0.150 J.
Step-by-step explanation:
The student has asked about the energy stored in a wire that has been stretched by a force. The area of cross-section of the wire is 3×10−6m², its original length is 4m, and it has been stretched by 1mm (which is 1×10−3m). The Young's modulus (Y) is given as 2×1011Nm². To find the energy stored in the wire when it is stretched, we use the formula for elastic potential energy stored in a stretched wire which is (1/2) x stress x strain x volume. Stress can be calculated using Hooke's Law (stress = force/area), and strain is given by the change in length over the original length. Since the force is not given, we replace it with (stress x area) and use Young's modulus (stress/strain = Y). After substitution and simplification, the volume (cross-sectional area x original length) is used to find the energy stored.
The correct answer is calculated as follows:
Energy stored = (1/2) x stress x strain x volume
= (1/2) x (Y x strain) x strain x (area x length)
= (1/2) x (2×1011Nm²) x (1×10−3m / 4m) x (3×10−6m² x 4m)
= 0.150 J
Hence, the correct option is (D) 0.150 J.