Question

The wires $A$ and $B$ shown in the figure are made of the same material and have radii $r_A$ and $r_B$, respectively. The block between them has a mass $m$. When the force $F$ is $mg/3$, one of the wires breaks. Then,

• A. $A$ will break before $B$ if $r_A=r_B$.
• B. $A$ will break before $B$ if $r_A<2r_B$.
• C. either $A$ or $B$ may break if$r_A=2r_B$.
• D. the lengths of $A$ and $B$ must be known to predict which wire will break.

Answer 1

Answer: (A), (B), (C)

either $A$ or $B$ may break if$r_A=2r_B$.


$T_B=\frac{mg}{3}$..........(1)
$T_A=mg+\frac{mg}{3}=\frac{4mg}{3}$
$T_A=4T_B$.......(2)
Breaking stress for wire B
$S=\frac{T_B}{\pi r_B^2}$........(3)
Breaking stress of wire A
$S^{\prime }=\frac{T_A}{\pi r_A^2}$............(4)
If $r_A=r_B$
Divide (3) by (4), we get
$S^{\prime }=4S$
Both wire A and B made of same material. If stress developed in wire B is $S$ for a particular load, then stress developed in wire B is 4 times of $S$.Therefore wire A will break first.

If $r_A<2r_B$ then area of A is less than area of A and we know if area is less, stress developed is more. Hence wire A will break before B.

If $r_A=2r_B$ and if we divide (3) by (4), we get
$S^{\prime }=S$. So we cannot predict which wire will break first.