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A glass bulb is filled in at 0^(@)C by 350 g of mercury. When a few steel balls are put in the bulb, it can then hold only 265 g of mercury. When the bulb is heated to 100^(@)C, with the steel ballsin mercury, 5 g of mercury flows out. Find the coefficient of linear expansion of steel. Given, the coefficient of real expansion of mercury =18 times 10^(-5@)C^(-1). Neglect expansion of glass. |
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Answer» Solution :Let `rho_(0) " and " rho_(100)` be the densities of mercury at `0^(@)C " and " 100^(@)C` respectively. `THEREFORE` Volume of mercury in the glass bulb at `0^(@)C=350/rho_(0)` Volume of mercury in the glass after the steel balls are put `=265/rho_(0)` `therefore` Volume occupied by the steel balls at `0^(@)C` ` ""=350/rho_(0)-265/rho_(0)=85/rho_(0)` Expansion of mercury at `100^(@)C` `=265/rho_(0) times 18 times 10^(-5) times 100=4.77/rho_(0)` Expansion of the steel balls at `100^(@)C` `""=85/rho_(0) times gamma_(s) times 100=(8500gamma_(s))/rho_(0)` `""[gamma_(s)=` coefficient of volume expansion of steel] Now, expansion of mercury + expansion of the steel balls =volume of mercury expelled i.e, `4.77/rho_(0)+(8500gamma_(s))/rho_(0)=5/rho_(100)` or, `4.77+8500gamma_(s)=(5rho_(0))/rho_(100)=(5rho_(100)(1+18 times 10^(-5) times 100))/rho_(100)` `""=5 times 1.018` `""=[therefore rho_(0)=rho_(100)(1+18 times 10^(-5) times 100)]` or, `8500gamma_(s)=0.32 " or, " gamma_(s)=0.32/8500=37.65 times 10^(-6@)C^(-1).` `therefore` Coefficient of LINEAR expansion `alpha_(s)=gamma_(s)/3=37.65/3 times 10^(-6)=12.55 times 10^(-6@)C^(-1).` |
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