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 $m g$ / 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 $r_{A}< 2 r_{B}$
• C. Neither $A$ nor $B$ will break if $r_{A}=2 r_{B}$
• 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{m g}{3} $
$\because \text { Stress } =\frac{\text { Force }}{\text { Area }}$
$ \therefore \text { Stress in } B =\frac{m g}{3 \pi r_{B}^{2}} \left[\because \text { Area of wire } B=\pi r_{B}^{2}\right]$
Similarly, stress in $A=\frac{F+m g}{\pi r_{A}^{2}}=\frac{\frac{m g}{3}+m g}{\pi r_{A}^{2}}$
When $r_{A}=r_{B}$,
Stress in $B=\frac{m g}{3 \pi r_{B}^{2}}$ and stress in $A=\frac{4}{3} \frac{m g}{\pi r_{A}^{2}}$
$\therefore$ Stress in $A>$ Stress in $B$
So, wire $A$ will break earlier.
If $r_{A}=2 r_{B}$, then stress in $A=$ stress in $B$, it means either $A$ or $B$ may breaks.
If $r_{A}<2 r_{B}$, 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.