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The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is _____mm². (Given, y = 2.0 x 10¹¹ Nm^-2)

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answer is 40.

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Detailed Solution

Energy stoned in wire $u = \frac{1}{2} {kx}^{2}$ $= \frac{1}{2} (\frac{YA}{l}) x^{2}$
$80 = \frac{1}{2} . \frac{(2 \times 10) (A)}{20} {(2 \times 10^{- 2})}^{2}$ $A = 4 \times 10^{- 5} m^{2} = 40 {mm}^{2}$

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