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The volume of a block of a metal changes by `0.12%` when it is heated through `20^(@)C` . The coefficient of linear expansion of the metal isA. `4 xx 10^(-5)"^(@) C^(-1)`B. `2 xx 10^(-5)"^(@) C^(-1)`C. `0.5 xx 10^(-5)"^(@) C^(-1)`D. `4 xx 10^(-4)"^(@) C^(-1)` |
Answer» Correct Answer - B Here, `(DeltaV)/(V) = 0.12%, = (0.12)/(100)` `DeltaT = 20 .^(@)C` Now, `gamma = (DeltaV)/(VDeltaT) = (0.12)/(100 xx 20) = 6 xx 10^(-5) .^(@)C^(-1)` `:. alpha = (gamma)/(3) = 2 xx 10^(-5) .^(@)C^(-1)` |
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