In This post you will get answer of Q.
The breaking force for a wire of diameter D of a material is F. The breaking force for a wire of the same material of radius D is
Solution:
$Breaking stress=frac{breaking force}{area}$
= constant
= $frac{F}{left(frac{pi D^2}{4}right)}=frac{F’}{pi D^2}$
or $frac{4F}{pi D^2}=frac{F’}{pi D^2}$
F’ = 4F
Solution:
$Breaking stress=frac{breaking force}{area}$
= constant
= $frac{F}{left(frac{pi D^2}{4}right)}=frac{F’}{pi D^2}$
or $frac{4F}{pi D^2}=frac{F’}{pi D^2}$
F’ = 4F