A) 25 min
B) 15 min
C) 8 min
D) 4 min
Correct Answer: C
Solution :
| [c] Let \[{{R}_{1}}\] and \[{{R}_{2}}\] be the resistances of the coils, V the supply voltage, H the heat required to boil the water. |
| For first coil, \[H=\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}\] (i) |
| For second coil, \[H=\frac{{{V}^{2}}{{t}_{2}}}{{{R}_{2}}}\] (ii) |
| Equating Eqs. (i) and (ii), we get |
| \[\frac{{{t}_{1}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{R}_{2}}}\] |
| i.e., \[\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{40}{10}=4\] |
| \[\Rightarrow \] \[{{R}_{2}}=4{{R}_{1}}\] ....(iii) |
| When the two heating coils are in parallel, |
| \[R=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{R}_{1}}+4{{R}_{1}}}{{{R}_{1}}+4{{R}_{1}}}=\frac{4{{R}_{1}}}{5}\]c |
| and \[H=\frac{{{V}^{2}}t}{R}\] (iv) |
| Comparing Eqs. (i) and (iv), we get |
| \[\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}=\frac{{{V}^{2}}t}{R}\] |
| \[\Rightarrow \] \[t=\frac{R}{{{R}_{1}}}\times {{t}_{1}}=\frac{4}{5}\times 10=8\,\min \] |
You need to login to perform this action.
You will be redirected in
3 sec
