Answer:a
Step-by-step explanation:
Given
Potential difference
![V=110\ V](https://img.qammunity.org/2021/formulas/engineering/college/nuq0rofl1w0o6d91sed8gjplbpbbfdlpjb.png)
Power of bulb A
![P_A=60\ W](https://img.qammunity.org/2021/formulas/physics/college/n9fx5w10v8yytwrcjt1pafa1cyw4o3gw56.png)
Power of bulb B
![P_B=100\ W](https://img.qammunity.org/2021/formulas/physics/college/p5kw04wgxgui3pg5zugw5ze7zae6umutfy.png)
If voltage is same for both the bulbs then Power is given by
![P=(V^2)/(R)](https://img.qammunity.org/2021/formulas/physics/college/k11k514ix0o50pzx20rhfb7s0boudkj9rp.png)
![P_A=((110)^2)/(R_A)](https://img.qammunity.org/2021/formulas/physics/college/cicvs02qz39l0rlxnr7a41paaa17f4bz8w.png)
![60=(110^2)/(R_A)](https://img.qammunity.org/2021/formulas/physics/college/g9fzyvzsctk8s9vwgt39o4rz4ufwkgm2x5.png)
![R_A=201.66\ \Omega](https://img.qammunity.org/2021/formulas/physics/college/y7iktyl6eb0lflqcm0dcu4raefzw7j4z7o.png)
similarly
![R_B=(110^2)/(100)](https://img.qammunity.org/2021/formulas/physics/college/wuxvuqxrpcm89271v8ifk4v8ujd2n9m2ke.png)
![R_B=121\ \Omega](https://img.qammunity.org/2021/formulas/physics/college/cet8kjh5s9m5v0bhbfyen83sucpblpxok1.png)
![R_A>R_B](https://img.qammunity.org/2021/formulas/physics/college/7m2zko96zonk3p039ptn8c68qj6xotnqhh.png)
so current in bulb A is smaller than B
Thus the 60 W bulb has a greater resistance.
Thus option (a) is correct