A) 6
B) 12.5
C) 25
D) 7.5
Correct Answer: A
Solution :
For first filament \[H=\frac{{{V}^{2}}}{{{R}_{1}}}{{t}_{1}}\Rightarrow {{R}_{1}}=\frac{{{V}^{2}}}{H}{{t}_{1}}\] ... (i) For second filament \[H=\frac{{{V}^{2}}}{{{R}_{2}}}{{t}_{2}}\Rightarrow {{R}_{2}}=\frac{{{V}^{2}}}{H}{{t}_{2}}\] ? (ii) When placed in parallel \[H=\frac{{{V}^{2}}}{{{R}_{p}}}{{t}_{p}}\Rightarrow {{R}_{p}}=\frac{{{V}^{2}}}{H}{{t}_{p}}\] ? (iii) From Eqs. (i), (ii), (iii) we get \[\frac{1}{{{R}_{p}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\Rightarrow \] \[\frac{H}{{{V}^{2}}{{t}_{p}}}=\frac{H}{{{V}^{2}}{{t}_{1}}}+\frac{H}{{{V}^{2}}{{t}_{2}}}\] \[\Rightarrow \] \[\frac{1}{{{t}_{p}}}=\frac{1}{10}+\frac{1}{15}\] \[\Rightarrow \] \[{{t}_{p}}=6\min \]
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