Answer:a
Explanation:
Given
Potential difference [tex]V=110\ V[/tex]
Power of bulb A [tex]P_A=60\ W[/tex]
Power of bulb B [tex]P_B=100\ W[/tex]
If voltage is same for both the bulbs then Power is given by
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]P_A=\dfrac{(110)^2}{R_A}[/tex]
[tex]60=\dfrac{110^2}{R_A}[/tex]
[tex]R_A=201.66\ \Omega[/tex]
similarly
[tex]R_B=\dfrac{110^2}{100}[/tex]
[tex]R_B=121\ \Omega[/tex]
[tex]R_A>R_B[/tex]
so current in bulb A is smaller than B
Thus the 60 W bulb has a greater resistance.
Thus option (a) is correct