Correct Answer - `(5)/(2),(2)/(5)`
resistance of bulb `500 watt R_(1) = (220)^(2) // 500 Omega`
resistance of bulb `200 watt R_(2) = (220)^(2) // 200 Omega`
so ` R_(1)//R_(2) = 2//5`
When the bulb are connected in parallel , then potential difference across bulb the bulb will be the same
Ratio of heat produced `= (V^(2)//R_(1))/(V^(2)//R_(2)) = (R_(2))/(R_(1)) = (5)/(2)`
When the bulb are joined inseries , the same current will flow through both the bulb
Ratio of heat produced `= (l^(2)R_(1))/(l^(2)R_(2)) = (R_(1))/(R_(2)) = (2)/(5)`