Let total heat supply to the water by kettle for boiling is HNow,Voltage in heater = V=220V ( In household circuit, each aplliance is linked in parallel combination and so, each appliance would have same voltage and it will be equal to the voltage of supply line = 220V)Let the respective resistance of 2 coils of heater be r' and r" respectively.Now, heat supply by both coil will be the same (because it has to boil the same water) = HNow, heat supplied = electric power × time = (V^2/R) × t = (V^2 × t)/RHeat by first coil = HVoltage = 220 VResistance = r'time = 15 min. = 15×60 = 900 sso, H = (220^2 × 900)/r' = (4.356 × 10^7)/r' ...(1)Again,Heat by 2nd coil = HVoltage = 220VResistance = r"time = 10 min. = 600 sso, H = (220^2 × 600)/r" = (2.904×10^7)/r" ...(2)From (1) and (2),(4.356 × 10^7)/r' = (2.904×10^7)/r"=》r'/r" = (4.356×10^7)/(2.904×10^7) = 3/2=》r'/r" = 3/2Now, if these coils must have same resistivity and thicknessso, r' = p × l'/AAnd, r" = p × l"/Aso, r'/r" = (p × l'/A )/(p × l"/A ) = l'/l"Now, r'/r" = l'/l" = 3/2so, l'/l" = 3/2=》l" = (2l')/3so, length of second coil is two-third of the length of first coil