A) 2R
B) R
C) RJ2
D) R/4
Correct Answer: D
Solution :
Power of bulb \[P=\frac{{{V}^{2}}}{R}\] \[\frac{{{P}_{india}}}{{{P}_{USA}}}={{\left( \frac{{{V}_{India}}}{{{V}_{USA}}} \right)}^{2}}\left( \frac{{{R}_{USA}}}{{{R}_{India}}} \right)\] \[\left( \frac{60}{60} \right)={{\left( \frac{220}{110} \right)}^{2}}\left( \frac{{{R}_{USA}}}{R} \right)\] \[1=4\left( \frac{{{R}_{USA}}}{R} \right)\] \[{{R}_{USA}}=\frac{R}{4}\]You need to login to perform this action.
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