On increasing the length by $0.5\ mm$ in a steel wire of length $2\ m$ and area of cross-section $2\ mm^{2}$, the force required is [$Y$ for steel $= 2.2 imes 10^{11}N/m^{2}$]

Question

On increasing the length by $0.5\ mm$ in a steel wire of length $2\ m$ and area of cross-section $2\ mm^{2}$, the force required is [$Y$ for steel $= 2.2 	imes 10^{11}N/m^{2}$]

Dr. Abubacker · Accepted Answer

(d) $F = \frac{YAl}{L} = \frac{2.2 	imes 10^{11} 	imes 2 	imes 10^{- 6} 	imes 5 	imes 10^{- 4}}{2} = 1.1 	imes 10^{2}N$

Q. On increasing the length by $0.5\ mm$ in a steel wire of length $2\ m$ and area of cross-section $2\ mm^{2}$, the force required is [$Y$ for steel $= 2.2 \times 10^{11}N/m^{2}$]

Explanation

Reviewed by Dr. Abubacker

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