# A kettle has a coil that operates at 1200 V when connected to a 240 W supply. The resistivity ρ and area of cross-section A of unit length of the coil

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A kettle has a coil that operates at 1200 V when connected to a 240 W supply. The resistivity ρ and area of cross-section A of unit length of the coil are related by the equation-
1. $A =\frac{\rho }{7000}$
2. $A =\frac{\rho }{6000}$
3. $A =\frac{\rho }{1200}$
4. $A =\frac{\rho }{240}$

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Correct Answer - Option 2 : $A =\frac{\rho }{6000}$

The correct answer is option 2) i.e. $A =\frac{ρ }{6000}$

CONCEPT:

• Resistance: The hindrance to the flow of current offered by a material is called electrical resistance.

The resistance of a conducting wire is given by the equation:

$R = ρ\frac{l}{A}$

Where R is the resistance, l is the length of the wire and A is the cross-sectional area of the material.

• Power: The rate of work done by an electric current is called power. It is denoted by P. The SI unit of power is the watt (W).

Power dissipation is given by:

$⇒ P = VI = V.\frac{V}{R} = \frac{V^2}{R}$

Where V is the potential difference across resistance, I is current flowing and R is resistance.

EXPLANATION:

Given that:

Voltage, V = 1200 V

Power, P = 240 W

Using, $P = \frac{V^2}{R}$

$\Rightarrow R = \frac{V^2}{P} = \frac{1200^2}{240} =6000\: \Omega$

We know that, $R = ρ\frac{l}{A}$

$\Rightarrow 6000 = ρ\frac{l}{A}$

For unit length, $\Rightarrow 6000 = ρ\frac{1 }{A}$       (∵ unit length = 1 m)

$\Rightarrow A = \frac{ρ}{6000}$