Correct Answer - B
For power to be consumed at the rate of
`1100/5=220 W`, we have `P=E_(v)I_(v)cos theta`
`220=(220xx220)/(sqrt(R^(2)+L^(2)omega^(2)))xxR/(sqrt(R^(2)+L^(2)omega^(2)))`
where `R=(V^(2))/P=(220^(2))/1100=44 Omega`
`220=((220)^(2)xx44)/(44^(2)+(Lomega)^(2)) , 44^(2)+(Lomega)^(2)=22xx44`
`(Lomega)^(2)=sqrt(220xx44-44^(2))`
`=sqrt(44(220-44))=sqrt(44xx176)=88 Omega`
`L=88/(2pixxf)=88/(2pixx50)=88/(2xx22)xx7/50=0.28 H`