Question

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

• A. F
• B. 2F
• C. $\frac{F}{4}$
• D. 4F

Answer 1

Answer: (D)

4F

Answer 2

$Breaking stress=\frac{breaking force}{area}$ = 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