Question

If in order to break a wire of length, minimum 40 kg-wt is required, then to break a wire of same material 6 m in length and double in radius, the breaking weight required will be:
A. 40 kg-wt
B. 80 kg-wt
C. 160 kg-wt
D. 320 kg-wt

Answer 1

Hint: In the above question the stress produced will be the same in both conditions because the same material of wire is used. Therefore by equating their pressure we can easily find out the breaking weight required to break the wire.

Complete step-by-step answer:
We know that
According to question,
Let the breaking weight or force of the wire in first case, ${F_1}$= 40 kg-wt
In second case be ${F_2}$
Relation between their radii in two cases given by,
${r_2} = 2{r_1}$
Let ${A_1}$ and ${A_2}$ be the cross section area in first and second case respectively.
Let ${\sigma _1}$ and ${\sigma _2}$ be the stress produced in the first and second case respectively. But the same stress is produced in both cases because the same material is being used.
$\therefore {\sigma _1} = {\sigma _2}$
$\dfrac{{{F_1}}}{{{A_1}}} = \dfrac{{{F_2}}}{{{A_2}}} \Leftrightarrow \dfrac{{40}}{{\pi {r_{^1}}^2}} = \dfrac{{{F_2}}}{{\pi r_2^2}}$
On putting the value of ${r_2} = 2{r_1}$ we get
$\dfrac{{40}}{{\pi {r_{^1}}^2}} = \dfrac{{{F_2}}}{{\pi {{\left( {2{r_1}} \right)}^2}}}$
On further solving we get
${F_2} = 4 \times 40$kg-wt
$\therefore {F_2} = 160$kg-wt
Hence the breaking weight in the second case when the radius of the wire is doubled is 160 kg-wt.
Hence the correct option is C.

Note: If the material of the wire is changed in the second case then the stress produced in that material will be different hence this formula cannot be applied to find the breaking weight or force of a particular wire. Breaking stress is the maximum load wire can take before failure. It is also known as the tensile strength for a wire of unit cross section.