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.
- increased, \(\frac{3}{4}\)
- increased, \(\frac{4}{3}\)
- decreased, \(\frac{3}{4}\)
- decreased, \(\frac{4}{3}\)
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to understand the relationship between the time taken to boil water and the resistance of the heating element. The key concept here is the formula for power, which is given by:
\(P = \frac{V^2}{R}\)
where \(P\) is the power, \(V\) is the voltage, and \(R\) is the resistance of the heating element.
The energy required to boil the water remains constant, meaning that \(P \times t\) (power multiplied by time) must remain the same for different lengths of the heating element. Therefore, we have:
\(\frac{V^2}{R} \times t = \text{Constant}\)
Initially, the time taken is 20 minutes with a certain resistance \(R_1\). When the time changes to 15 minutes, the resistance becomes \(R_2\). Given that the main voltage supply remains the same, the equation can be rewritten as:
\(\frac{1}{R_1} \times 20 = \frac{1}{R_2} \times 15\)
Simplifying this, we get:
\(R_2 = \frac{3}{4}R_1\)
This implies that the resistance should decrease to \(\frac{3}{4}\) of its initial value to decrease the boiling time from 20 minutes to 15 minutes.
The resistance \(R\) of a wire is directly proportional to its length \(L\) (given by the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity and \(A\) is the cross-sectional area). This means that:
\(\frac{L_2}{L_1} = \frac{R_2}{R_1}\)
Therefore, we find that:
\(\frac{L_2}{L_1} = \frac{3}{4}\)
This means the length of the heating element should be decreased to \(\frac{3}{4}\) times its initial length in order to boil water in 15 minutes.
Thus, the correct answer is: decreased, \(\frac{3}{4}\).
Approach Solution -2
The resistance of the heating element is:
\[R = \rho \frac{\ell}{A}.\]
The power is inversely proportional to the length:
\[P \propto \frac{1}{\ell}.\]
From the relation \(P_1 \times t_1 = P_2 \times t_2\):
\[\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{15}{20}.\]
Substituting \(P \propto \frac{1}{\ell}\):
\[\frac{\ell_2}{\ell_1} = \frac{t_2}{t_1} = \frac{15}{20} = \frac{3}{4}.\]
Thus, the new length should be decreased to:
\[\ell_2 = \frac{3}{4} \ell_1.\]
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