Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _____ to _____ times of its initial length if the water is to be boiled in 15 minutes.

Water boils in 20 min. How should the heating element length change to boil in 15 min?

Energy to boil water is fixed: $$Q = Pt$$. For the same $$Q$$, $$P_1 t_1 = P_2 t_2$$, which yields $$P_2 = P_1 \times \frac{t_1}{t_2} = P_1 \times \frac{20}{15} = \frac{4}{3}P_1$$.

Since $$P = \frac{V^2}{R}$$ for the same supply voltage, increasing power requires decreasing resistance, so $$R_2 = R_1 \times \frac{P_1}{P_2} = R_1 \times \frac{3}{4}$$.

Because $$R = \frac{\rho L}{A}$$ implies $$R \propto L$$, it follows that $$L_2 = \frac{3}{4}L_1$$.

The length must be decreased to $$\frac{3}{4}$$ of its original value.

The correct answer is Option (1): decreased, 3/4.