A steel wire of cross-sectional area $3\ imes 10^{- 6}m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young’s modulus of steel is $2 imes 10^{11}{Nm}^{- 2}$. The maximum mass the wire can hold is$\ \ (take\ g = 10\ {ms}^{- 2\ })$

Question

A steel wire of cross-sectional area $3\ 	imes 10^{- 6}m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young’s modulus of steel is $2 	imes 10^{11}{Nm}^{- 2}$. The maximum mass the wire can hold is$\ \ (take\ g = 10\ {ms}^{- 2\ })$

Dr. Abubacker · Accepted Answer

(b) Young’s modulus $Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{Strain}$
or $Y\frac{mg}{A 	imes strain}$
or $\ m = \frac{Y 	imes A 	imes strain}{g}$
$= \frac{2 	imes 10^{11} 	imes 10^{- 3} 	imes 10^{- 6}}{10} = 60kg$

Q. A steel wire of cross-sectional area $3\ \times 10^{- 6}m^{2}$ can withstand a maximum strain of $10^{- 3}$ . Young’s modulus of steel is $2 \times 10^{11}{Nm}^{- 2}$. The maximum mass the wire can hold is$\ \ (take\ g = 10\ {ms}^{- 2\ })$

Explanation

Reviewed by Dr. Abubacker

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