Question

Wires $A$ and $B$ are connected with mass $m$ as shown in figure. Wires are of same material and have radii $r_A$ and $r_B$. End of $B$ is pulled with a force of $mg$ / 3. With reference to the above situation, which of the following statement is correct?

• A. $A$ breaks before $B$ when $r_A=r_B$
• B. $A$ breaks after $B$ when $r_A<2r_B$
• C. Neither $A$ nor $B$ will break if $r_A=2r_B$
• D. None of the above

Answer 1

Answer: (A)

Given, radii of wires $A$ and $B$ are $r_A$ and $r_B$, respectively.
$\text{ Force, }F=\frac{mg}{3}$
$\because \text{ Stress }=\frac{\text{ Force }}{\text{ Area }}$
$\therefore \text{ Stress in }B=\frac{mg}{3\pi r_B^2}\left[\because \text{ Area of wire }B=\pi r_B^2\right]$
Similarly, stress in $A=\frac{F+mg}{\pi r_A^2}=\frac{\frac{mg}{3}+mg}{\pi r_A^2}$
When $r_A=r_B$,
Stress in $B=\frac{mg}{3\pi r_B^2}$ and stress in $A=\frac{4}{3}\frac{mg}{\pi r_A^2}$
$\therefore$ Stress in $A>$ Stress in $B$
So, wire $A$ will break earlier.
If $r_A=2r_B$, then stress in $A=$ stress in $B$, it means either $A$ or $B$ may breaks.
If $r_A<2r_B$, then stress in $A$ is more than that of $B$.
$\therefore A$ may break earlier.
Thus, the statement given in option (a) is correct, rest are incorrect.