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 $\lbrack Y\ for\ steel = \ 2.2 imes 10^{11}Nm^{- 2} brack$

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 $\lbrack Y\ for\ steel = \ 2.2 	imes 10^{11}Nm^{- 2}brack$

Dr. Abubacker · Accepted Answer

(d) $Y = \frac{Fl}{A\mathrm{\Delta}l}\ or\ \ F = \ \frac{YA\ \mathrm{\Delta}l}{l}$
Or $F = \ \frac{2.2 	imes 10^{11} 	imes 2 	imes 10^{- 6} 	imes 0.5 	imes 10^{- 3}}{2}$
= $1.1\  	imes 10^{2}N$

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 $\lbrack Y\ for\ steel = \ 2.2 \times 10^{11}Nm^{- 2}\rbrack$

Explanation

Reviewed by Dr. Abubacker

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