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At `50^(@)C` , a brass rod has a length 50 cm and a diameter 2 mm . It is joined to a steel rod of the same length and diameter at the same temperature . The change in the length of the composite rod when it is heated to `250^(@)C` is (Coefficient of linear expansion of brass = `2.0 xx 10^(-5)"^(@) C^(-1)` , coefficient of linear expansion of steel = `1.2 xx 10^(-5) "^(@) C^(-1)`)A. `0.28` cmB. `0.30` cmC. `0.32` cmD. `0.34` cm |
Answer» Correct Answer - C Change in length of the brass rod is `DeltaL_(b) = alpha_(b) L_(b) DeltaT` `= 2.0 xx 10^(-5).^(@)C^(-1) xx 50 cm xx (250 .^(@)C - 50 .^(@)C)` `= 0.2 cm` Change in length of the steel rod is `DeltaL_(S) = alpha_(S)L_(S)DeltaT` `= 1.2 xx 10^(-5).^(@)C^(-1) xx 50 cm xx (250 .^(@)C - 50^(@)C)` `= 0.12 cm` `:.` Change in length of the combined rod `= DeltaL_(b) xx DeltaL_(S) = 0.2 cm + 0.12 cm = 0.32 cm` |
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