A) 15 min
B) 10 min
C) 5 min
D) 12 min
Correct Answer: A
Solution :
Amount of A left in \[{{n}_{1}}\] halves \[={{\left( \frac{1}{2} \right)}^{{{n}_{1}}}}[{{A}_{0}}]\] |
Amount of B in \[{{n}_{2}}\] halves \[={{\left( \frac{1}{2} \right)}^{{{n}_{2}}}}[{{A}_{0}}]\] |
At the end halves, \[\frac{[{{A}_{0}}]}{{{2}^{ni}}}=\frac{[{{B}_{0}}]}{{{2}^{{{n}_{2}}}}}\] |
\[\Rightarrow \] \[\frac{4}{{{2}^{{{n}_{1}}}}}=\frac{1}{{{2}^{{{n}_{2}}}}}\] As \[{{A}_{0}}=4{{B}_{0}}\] |
\[\therefore \] \[\frac{{{2}^{{{n}_{1}}}}}{{{2}^{_{2}}}}=4\] |
\[\Rightarrow \] \[{{2}^{{{n}_{1}}-{{n}_{2}}}}={{(2)}^{2}}\] |
\[\therefore \] \[{{n}_{1}}-{{n}_{2}}=2\] so \[{{n}_{2}}={{n}_{1}}-2\] (i) |
Also, \[t={{n}_{1}}\times {{t}_{1/2}}(A)\] |
\[t={{n}_{1}}\times {{t}_{1/2}}(A)\] |
\[t={{n}_{2}}\times {{t}_{1/2}}(B)\] |
Let conc. of both become equal after time (t) |
\[\therefore \] \[\frac{{{n}_{1}}\times {{t}_{1/2}}(A)}{{{n}_{2}}\times {{t}_{1/2}}(B)}=1\] |
\[\Rightarrow \] \[\frac{{{n}_{1}}\times 5}{{{n}_{2}}\times 5}=1\] |
\[\frac{{{n}_{1}}}{{{n}_{2}}}=3\] (ii) |
From equation (i) and (ii), |
\[{{n}_{1}}=3\] and \[{{n}_{2}}=1\] |
\[t=3\times 5=15\,\min \] |
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