A) \[1.5\times {{10}^{-2}}\]
B) \[2\times {{10}^{-4}}\]
C) \[3.5\times {{10}^{-3}}\]
D) \[2.5\times {{10}^{-2}}\]
Correct Answer: A
Solution :
Let \[{{V}_{0}}\] be the initial volume of glycerine, i.e. at\[{{0}^{o}}C\] (dry). If V^ be its volume at \[{{30}^{o}}C\]. then \[{{V}_{t}}={{V}_{0}}(1+v\Delta t)\] \[={{V}_{0}}(1+49\times {{10}^{-5}}\times 30)\] \[{{V}_{t}}={{V}_{0}}(1+0.01470)=1.0147070\,{{V}_{0}}\] \[\Rightarrow \] \[\frac{{{V}_{0}}}{{{V}_{t}}}=\frac{1}{1.01470}\] Let \[{{\rho }_{0}}\] and \[{{\rho }_{t}}\] be the initial and final densities of glycerine then initial density, \[{{\rho }_{0}}=\frac{m}{{{v}_{0}}}\] and final density, \[{{\rho }_{t}}=\frac{m}{{{V}_{t}}}\] where, m = mass of glycerine \[\frac{\Delta \rho }{{{\rho }_{0}}}=\] fractional change in the density \[\frac{{{\rho }_{t}}-{{\rho }_{0}}}{{{\rho }_{0}}}=\frac{m\left( \frac{1}{{{V}_{t}}}-\frac{1}{{{V}_{0}}} \right)}{\frac{m}{{{V}_{0}}}}=\left( \frac{{{V}_{0}}}{{{V}_{t}}}-1 \right)\] \[\Rightarrow \] \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=\left( \frac{1}{1.01470}-1 \right)=-0.0145\] Here, negative sign shows that density decreases with rise in temperature. \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=0.0145=1.45\times {{10}^{-2}}\] \[\Rightarrow \] \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=1.5\times {{10}^{-2}}\]
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