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A bar of iron is 10 cm at `20^(@)C`. At `19^(@)C` it will be `(alpha_(Fe)=11xx10^(-6)//.^(@)C)`A. `11xx10^(-6)`cm longerB. `11xx10^(-6)cm` cm shorterC. `11xx10^(-5)cm` shorterD. `11xx10^(-5)cm` longer |
Answer» Correct Answer - C |
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