The cross-sectional areas of steel and brass | vedclass

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(D) Stress is defined as the ratio of force (tension) to the cross-sectional area: $	ext{Stress} = \frac{T}{A}$.Given that the stress is the same in

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$T_1/T_2 = 0.5$

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(D) Stress is defined as the ratio of force (tension) to the cross-sectional area: $	ext{Stress} = \frac{T}{A}$.Given that the stress is the same in both wires,we have:$\frac{T_1}{A_1} = \frac{T_2}{A_2}$Rearranging the terms to find the ratio of tensions:$\frac{T_1}{T_2} = \frac{A_1}{A_2}$Substituting the given values $A_1 = 0.1 \,cm^2$ and $A_2 = 0.2 \,cm^2$:$\frac{T_1}{T_2} = \frac{0.1}{0.2} = 0.5$Therefore,the ratio of the tensions is $0.5$.

The cross-sectional areas of steel and brass wires are $0.1 \,cm^2$ and $0.2 \,cm^2$ respectively. If the same stress is produced in both by a weight $W$,what is the ratio of the tensions $T_1/T_2$?

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