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 \[{{n}_{2}}=1\] |
| \[t=3\times 5=15\,\min \] |
You need to login to perform this action.
You will be redirected in
3 sec
