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A 1 m long steel wire of cross-sectional area `1 mm^(2)` is extended 1 mm. If `Y = 2 xx 10^(11) Nm^(-2)`, then the work done isA. 0.1 JB. 0.2 JC. 0.3 JD. 0.4 J |
Answer» Correct Answer - a Work done, `W = (1)/(2) Y (A)/(L) (Delta L)^(2)` Given, `A = 1 mm^(2) = 10^(-6) m`, `I = 1mm = 10^(-3) m Y = 2 xx 10^(11) Nm^(-2), L = 1m` `:. W = (1)/(2) xx 2 xx 10^(11) xx 10^(-6) xx (10^(-3))^(2) = 0.1 J` |
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