Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries with the help of diagrams?

Compared to High angle, relatively less presence of Crystallographic misalignment in grain boundary for small angle is reason behind its less effectiveness in interfering with slip process

Explanation:

Because of relatively much less presence of crystallographic misalignment in grain boundary for small angle, small grain boundaries are not as effective in compare to high angle in interfering with slip process

Less crystallographic misalignment in grain boundary causes not much change in slip direction and therefore interference to slip process is minimal.

this isn't the answer but they're the graphs

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.