The work done in increasing the length of a one metre long wire of cross-sectional area `1 mm^(2)` through 1 mm will be `(Y = 2 xx 10^(11) Nm^(-2))`

Question

The work done in increasing the length of a one metre long wire of cross-sectional area `1 mm^(2)` through 1 mm will be `(Y = 2 xx 10^(11) Nm^(-2))`

The work done in increasing the length of a one metre long wire of cross-sectional area `1 mm^(2)` through 1 mm will be `(Y = 2 xx 10^(11) Nm^(-2))`
A. 0.1 J
B. 5 J
C. 10 J
D. 250 J

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - a
Work done, `W = (1)/(2) F xx I = (1)/(2) xx` stress `xx` strain `xx` volume
or `W = (1)/(2) xx Y xx ("stress")^(2) xx` volume
`= (1)/(2) Y ((Delta I)/(I))^(2) xx AI = (1)/(2) Y (Delta^(2) A)/(I)`
`= (2 xx 10^(11) xx 10^(-6) xx 10^(-6))/(2 xx 1) = 0.1 J`

4 views Answered 2023-02-05 15:29