Water boils in an electric kettle in 15 mins, after switching on. If the length of the heating wire is decreased to `(1)/(3)` of its initial value ,then in how much time will the same amount of water boil with the same supplyvoltage ?

Explanation

Supply Voltage is same
Amount of heat produced is also same
Time taken to heat water t = 15min
`H = (V^(2))/("R") .t [H = I^(2)Rt implies (V = IR) therefore I = (V)/("R") ]`
`therefore (R_(1))/(R_(2)) = (t_(1))/(t_(2))`
`R prop l [R = rho(l)/(A)]`
`(R_(1))/(R_(2)) = (l_(1))/(l_(2))`
From (1) and (2)
`(l_(1))/(l_(2)) = (t_(1))/(t_(2)) = (15)/(t_(2))`
`l_(2) = (1)/(3)l_(1) implies (l_(1))/(l_(2)) = 3`
sub in (3) 3 =`(15)/(t_(2))`
`therefore t_(2) = (15)/(3) = 5 `min
Time taken to heat `(1)/(3)` length of the wire = 5 min