The wires `A` and `B` shown in Fig. 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

Options

A. `A` will break before `B` if `r_(A)=r_(B)`
B. `A` will break before `B` if `r_(A)lt2r_(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

A, B, C

Explanation

Tension in `B=T_(B)=(mg)/3`
Tension in `A=T_(A)=T_(B)+mg=(4mg)/3`
`:. T_(A)=4T_(B)`
Stress `T/(pir^(2))=S`
Wire breaks When `S` is equal to breaking stress
For `r_(A)=r_(B), S_(A)=4S_(B)`
Therefore, `A` breaks before `B`
For `r_(A)=2r_(B)`
`S_(B)=(T_(B))/(pir_(A)^(2))-(T_(B))/(pir_(A)^(2))=(4T_(B))/(pi(2r_(B))^(2))=(T_(B))/(pir_(B)^(2))=S_(B)`
As the stresses area equal, either of the wires way break.