Question

When a 0.15 kg mass is hung from a brass wire, it stretches by 6.9 mm. If the diameter of the wire is 0.01 cm, what is its original length? Assume that the brass wire has negligible mass. The Young's modulus of a brass is 0.91×1011Pa.

Answer 1

L = 3.35 m

Step-by-step 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

`E = (stress)/(strain)`

`E = ((P)/(A))/((\Delta l)/(L))`

`E = (PL)/(A\Delta L)`

`L = (E A \Delta L)/(mg)`

`L = (0.91 * 10^(11) \pi (0.005)^2 * 0.0069)/(0.15 * 9.8)`

`L = (0.91 * 10^(11) \pi (0.005)^2 * 10^(-4)* 0.0069)/(0.15 * 9.8)`

L = 3.35 m