What are those four (4) types of defects? Briefly explain each category.2. A relatively large plate of a glass is subjected to a tensile stress of 40 MPa. If the specificsurface energy and modulus of elasticity for this glass are 0.3 J/m2 and 69 GPa,respectively, determine the maximum length of a surface flaw that is possible withoutfracture.3. What is the magnitude of the maximum stress that exists at the tip of an internalcrack having a radius of curvature of 2.5 x 10-4 mm and a crack length of 2.5 x 10-2mm when a tensile stress of 170 MPa is applied?4. A fatigue test was conducted in which the mean stress was 50 MPa and the stressamplitude was 225 MPa.(a) Compute the maximum and minimum stress levels.(b) Compute the stress ratio.(c) Compute the magnitude of the stress range.5. The fatigue data for a brass alloy are given as follows:(a) Make an S–N plot (stress amplitude versus logarithm cycles to failure) using thesedata.(b) Determine the fatigue strength at 5 x 105 cycles.(c) Determine the fatigue life for 200 MPa.

The question seems to be missing the first part where the four types of defects are mentioned. However, I can provide general information about common types of defects in materials:

1. Point defects: These are defects that occur only at or around a single lattice point. They are not extended in space in any dimension. Vacancies, interstitial atoms, or substitutional atoms are examples of point defects.

2. Line defects (Dislocations): These are one-dimensional defects around which atoms are misaligned. There are two basic types: edge dislocation and screw dislocation.

3. Surface defects: These are two-dimensional defects that occur on the surface of the material. Grain boundaries, twin boundaries, and stacking faults are examples of surface defects.

4. Volume defects: These are three-dimensional defects that occur throughout the material. They can include pores, cracks, or inclusions.

For the second question, the maximum length of a surface flaw can be calculated using the Griffith's criterion for brittle fracture:

a = [2γEσ^2]^-1

where: a = crack length γ = specific surface energy = 0.3 J/m^2 E = modulus of elasticity = 69 GPa = 69 x 10^9 Pa σ = applied stress = 40 MPa = 40 x 10^6 Pa

Substituting the given values into the equation, we can calculate the maximum length of the surface flaw.

For the third question, the maximum stress at the tip of an internal crack can be calculated using the equation:

σ_max = σ(1 + 2√(a/ρ))

where: σ = applied stress = 170 MPa a = crack length = 2.5 x 10^-2 mm ρ = radius of curvature = 2.5 x 10^-4 mm

Substituting the given values into the equation, we can calculate the maximum stress at the tip of the crack.

For the fourth question, the maximum and minimum stress levels, stress ratio, and stress range can be calculated using the following equations:

σ_max = σ_mean + σ_amplitude σ_min = σ_mean - σ_amplitude Stress ratio (R) = σ_min / σ_max Stress range = σ_max - σ_min

where: σ_mean = mean stress = 50 MPa σ_amplitude = stress amplitude = 225 MPa

Substituting the given values into the equations, we can calculate the required values.

For the fifth question, the fatigue strength at a given number of cycles and the fatigue life for a given stress can be determined from the S-N curve of the material. However, the data needed to plot the S-N curve or the equation representing the S-N curve is not provided in the question.