The area of a cross-section of a steel wire i | vedclass

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(B) Given:Area of cross-section $A = 0.1 \, cm^2 = 0.1 	imes 10^{-4} \, m^2 = 10^{-5} \, m^2$.Young's modulus $Y = 2 	imes 10^{11} \, N \, m^{-2}$.F

Vedclass · Accepted Answer

$2000$

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(B) Given:Area of cross-section $A = 0.1 \, cm^2 = 0.1 	imes 10^{-4} \, m^2 = 10^{-5} \, m^2$.Young's modulus $Y = 2 	imes 10^{11} \, N \, m^{-2}$.Fractional change in length $\frac{\Delta L}{L} = 0.1 \% = \frac{0.1}{100} = 10^{-3}$.Using the formula for Young's modulus: $Y = \frac{F \cdot L}{A \cdot \Delta L}$.Rearranging for force $F$: $F = Y \cdot A \cdot \frac{\Delta L}{L}$.Substituting the values: $F = (2 	imes 10^{11}) 	imes (10^{-5}) 	imes (10^{-3})$.$F = 2 	imes 10^{11-5-3} = 2 	imes 10^3 = 2000 \, N$.

The area of a cross-section of a steel wire is $0.1 \, cm^2$ and Young's modulus of steel is $2 \times 10^{11} \, N \, m^{-2}$. The force required to stretch it by $0.1 \%$ of its original length is ......... $N$.

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