Answer:
L = 3.35 m
Explanation:
given,
mass(m) = 0.15 Kg
stretch(ΔL) = 6.9 mm
diameter of wire(d) = 0.01 cm
L= ?
Young's modulus (E)= 0.91 x 10¹¹ Pa
[tex]E = \dfrac{stress}{strain}[/tex]
[tex]E = \dfrac{\dfrac{P}{A}}{\dfrac{\Delta l}{L}}[/tex]
[tex]E = \dfrac{PL}{A\Delta L}[/tex]
[tex]L = \dfrac{E A \Delta L}{mg}[/tex]
[tex]L = \dfrac{0.91 \times 10^{11} \pi (0.005)^2 \times 0.0069}{0.15 \times 9.8}[/tex]
[tex]L = \dfrac{0.91 \times 10^{11} \pi (0.005)^2 \times 10^{-4}\times 0.0069}{0.15 \times 9.8}[/tex]
L = 3.35 m
