A) 25 min
B) 15 min
C) 8 min
D) 4 min
Correct Answer: C
Solution :
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, \[R=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{R}_{1}}+4{{R}_{1}}}{{{R}_{1}}+4{{R}_{1}}}=\frac{4{{R}_{1}}}{5}\] and \[H=\frac{{{V}^{2}}t}{R}....(iv)\] Comparing Eqs. (i) and (iv), we get \[\frac{{{V}^{2}}{{t}_{1}}}{{{R}_{1}}}=\frac{{{V}^{2}}t}{R}\] \[\Rightarrow t=\frac{R}{{{R}_{1}}}\times {{t}_{1}}=\frac{4}{5}\times 10=8\,\min \]
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