Question

please give the explanation ans is 4th one❤❤​

Answer 1

Answer:

The correct option is B.

Given,

A force F is applied along a rod of transverse sectional area A,

Force is F

Area is A

The normal stress is given as p cos \thetapcosθ

Where, p is pressure.

And is given as p =$\dfrac{F}{A}p= A/F$

Hence, normal stress =$\dfrac{F}{A}\times cos \theta= A/F ×cosθ$

I hope this will help ❤

Answer 2

Answer:

Correct option is B F/Acos^2