A) \[\text{1}.\text{5}\times \text{1}{{0}^{\text{-2}}}\]
B) \[\text{2}\times \text{1}{{0}^{\text{-4}}}\]
C) \[\text{3}.\text{5}\times \text{1}{{0}^{\text{-3}}}\]
D) \[2.5\times {{10}^{-2}}\]
Correct Answer: A
Solution :
Let \[{{V}_{0}}\] be the initial volume of glycerine, i.e. at \[0{}^\circ \text{C}\] If \[{{\text{V}}_{\text{t}}}\] be its volume at\[\text{3}0{}^\circ \text{ C}\]. 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 \[\frac{{{\rho }_{t}}-{{\rho }_{0}}}{{{\rho }_{0}}}=\frac{m\left( \frac{1}{{{V}_{t}}}-\frac{1}{{{V}_{0}}} \right)}{\frac{m}{{{V}_{0}}}}=\left( \frac{{{V}_{0}}}{{{V}_{t}}}-1 \right)\] \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=\left( \frac{1}{1.01470}-1 \right)=-0.0145\] Here, negative sign shows that density decreases with rise in temperature. \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=0.0145=1.45\times {{10}^{-2}}\]\[\Rightarrow \] \[\frac{\Delta \rho }{\Delta {{\rho }_{0}}}=1.5\times {{10}^{-2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
