Answer Posted / p,btech student , e&e engg

Resistance of bulb
1(R=(V)^2/P)
=(110)^2/100 =121
ohm
resistance of blb 2
=(230)^2/200=264.5
ohm
so voltag drop
across bulb 2 will
be more (series
connection) .So
when connected in
series ,bulb 2 will
glow brighter.

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