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**Hint**: This question can be solved by using the concept that the breaking stress of a material is the ratio of the breaking force to the cross-sectional area. Thus, for the same material (same breaking stress), breaking force will be directly proportional to the cross-sectional area.

Formula used:

$\text{Breaking stress = }\dfrac{\text{Breaking force}}{\text{Cross-sectional area}}$

$\text{Breaking force }\propto \text{ Cross-sectional area }\left( \text{for same material} \right)$

$\text{Cross-sectional area of wire (cylindrical) = }\dfrac{\pi {{d}^{2}}}{4}$

where d is the diameter of the wire.

Complete step by step answer:

The breaking stress of a material is the ratio of the breaking force to the cross-sectional area. Thus, for the same material (same breaking stress), breaking force will be directly proportional to the cross-sectional area. --(1)

Now, let us analyze the given information. Let the wire with diameter 2mm be wire 1 and the other one be wire 2.

Breaking force of wire 1 (F1) = $4\times {{10}^{5}}N$

Diameter of wire 1 (d1) = 2 mm

Breaking force of wire 2 (F2) =?

Diameter of wire 2 (d2) = 1.5 mm.

Now, using (1),

$\text{Breaking stress = }\dfrac{\text{Breaking force}}{\text{Cross-sectional area}}$

$\text{Breaking force }\propto \text{ Cross-sectional area }\left( \text{for same material} \right)$ ---(2)

$\text{Cross-sectional area of wire (cylindrical) = }\dfrac{\pi {{d}^{2}}}{4}$ --(3)

where d is the diameter of the wire.

Using, (2) and (3)

$\dfrac{\text{Breaking force of wire 1}}{\text{Breaking force of wire 2}}=\dfrac{\text{Cross-sectional area of wire 1}}{\text{Cross-sectional area of wire 2}}$

\[\therefore \dfrac{{{F}_{1}}}{{{F}_{2}}}=\dfrac{\dfrac{\pi {{d}_{1}}^{2}}{4}}{\dfrac{\pi {{d}_{2}}^{2}}{4}}\]

$\therefore {{F}_{2}}={{F}_{1}}\times \dfrac{{{d}_{2}}^{2}}{{{d}_{1}}^{2}}$

$=4\times {{10}^{5}}\times \dfrac{{{1.5}^{2}}}{{{2}^{2}}}=2.25\times {{10}^{5}}N$

Hence, the breaking force of wire 2 is $2.25\times {{10}^{5}}N$.

The option closest to this is the correct answer A) $2.3\times {{10}^{5}}N$.

**Note**: For wires of the same material such problems can be solved using the concept of proportionality since the breaking stress is the same for a material. However, if the materials of the wire were different, we would have to proceed by first calculating the breaking stress of the wires by use of the information in the question (or if the value is given) and then proceed to individually, find out the breaking force by using the respective formula.

Thus, it is evident that a wire with a larger diameter (and hence larger cross-sectional area) has larger breaking strength and can withstand larger loads. Thus, for high load requirements such as the cables of pulleys in elevators, the cables are made of thick iron or steel cables.

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