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A long mercury glass tube with a uniform capillary bore has in it a thread of mercury which is `1m` long at `0^(@)C` . What will be its length at `100^(@)C` if the real coefficient of expansion of mercury is `0.000182` and coefficient of cubical expansion of glass equal to `0.000025` ? |
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Answer» `V_(0)` , volume of mercury thread at `0^(@)C` `V_(theta)` : volume of mercury thread at `theta^(@)C` `A_(0)` : cross-sectional area of tube at `0^(@)C` `A_(theta)` : cross-sectional area of tube `theta^(@)C` The length of thread at `theta^(@)C` `l_(theta)=V_(theta)/(A_(theta))=(V_(0)(1+gamma_(m)theta))/(A_(0)(1+betatheta)` `=l_(0)(1+gamma_(m)theta)(1+betatheta)^(-1)` `=l_(0)(1+gamma_(m)theta)(1+betatheta)` `=l_(0)[1+(gamma_(m)-beta)theta]["Neglectting small terms"]` `=l_(0)[1+(gamma_(m)-(2gamma_(g))/(3))theta]` `=100[1+(0.000182-(2)/(3)xx0.000025)xx100]` `=101.654cm` |
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