Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in

H=V2RtSince supply voltage is same and equal amount of heat will be produced, thereforeR1t1 = R2t2  or  R1R2+t1t2But  R ∝ l  ⇒ R1R2=l1l2By (i) and (ii),l1l2 = t1t2              …(iii)Now  l2 = 23 l1  ⇒  l1l2 =  32∴   By  equation iii,  32  =  15t2  ⇒  t2 =10  minutes