A) 3:2
B) 3:8
C) 4:3
D) 9:4
Correct Answer: A
Solution :
Power of a bulb \[(P)=\frac{{{V}^{2}}}{R}\] \[\therefore \] \[R=\frac{{{V}^{2}}}{P}\] \[\therefore \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{60}{40}=\frac{3}{2}\]
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