Correct Answer - D
In the first case:
Power, `P = 100W`
Potential difference, `V = 220V`
And, Resistance, `R = ?` (To be calculated)
Now, `P = (V^(2))/(R)`
SO, `100 = ((220)^(2))/(R)`
And `R = (220 xx 220)/(100) = 484 Omega`
The resistance of `484 Omega` of the bulb will remain unchanged.
In the second case:
Power, `P = ?` (To be calculated)
Potential difference, `V = 110V`
And, Resistance, `R = 484 Omega` (Calculated above)
Now, `P = (V^(2))/(R)`
`P = ((110)^(2))/(484) = (110 xx 110)/(484) = 25 W`
Thus, the correct answer is : (d) 25 W.
