Question

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.

• A. increased, \(\frac{3}{4}\)
• B. increased, \(\frac{4}{3}\)
• C. decreased, \(\frac{3}{4}\)
• D. decreased, \(\frac{4}{3}\)

Answer 1

The Correct Option is C

To resolve this issue, we must establish the connection between the time required for water to boil and the heating element's resistance. The fundamental principle is the power formula: \(P = \frac{V^2}{R}\), where \(P\) denotes power, \(V\) represents voltage, and \(R\) signifies the heating element's resistance.

The energy needed to boil water is constant, meaning that \(P \times t\) (power multiplied by time) must remain invariant irrespective of the heating element's length. Consequently, we have: \(\frac{V^2}{R} \times t = \text{Constant}\).

Initially, boiling takes 20 minutes with resistance \(R_1\). When the time is reduced to 15 minutes, the resistance becomes \(R_2\). Assuming the main voltage supply is consistent, the equation transforms into: \(\frac{1}{R_1} \times 20 = \frac{1}{R_2} \times 15\).

Simplification yields: \(R_2 = \frac{3}{4}R_1\). This indicates that the resistance must decrease to \( \frac{3}{4} \) of its original value to shorten the boiling time from 20 minutes to 15 minutes.

The resistance \(R\) of a wire is directly proportional to its length \(L\), as described by the formula \(R = \rho \frac{L}{A}\), where \( \rho \) is resistivity and \(A\) is cross-sectional area. This implies: \(\frac{L_2}{L_1} = \frac{R_2}{R_1}\).

Therefore, we find: \(\frac{L_2}{L_1} = \frac{3}{4}\). This means the heating element's length must be reduced to \( \frac{3}{4} \) times its initial length to achieve boiling in 15 minutes.

The correct answer is: decreased, \( \frac{3}{4} \).