Question

The heater element of a 120 V toaster is a 6.4 m length of nichrome wire, whose diameter is 0.68 mm. The resistivity of nichrome at the operating temperature of the toaster is 1.3 × $10^{-6} \Omega.m$. The toaster is operated at a voltage of 120 V. The power drawn by the toaster is closest to ____ × $10^{2}$ watt.

Answer 1

6.3

Answer 2

The resistance $R$ of the nichrome wire is given by the formula:

$R = \rho \frac{L}{A}$

where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area of the wire.

The wire has a circular cross-section. The diameter is given as $d = 0.68$ mm. The radius is $r = \frac{d}{2} = \frac{0.68}{2} = 0.34$ mm. Convert the radius to meters: $r = 0.34 \times 10^{-3}$ m. The cross-sectional area $A$ is given by:

$A = \pi r^2 = \pi (0.34 \times 10^{-3})^2 = \pi (0.1156 \times 10^{-6})$ m$^2$.

The length of the wire is $L = 6.4$ m. The resistivity of nichrome is $\rho = 1.3 \times 10^{-6} \Omega.m$.

Now, calculate the resistance $R$:

$R = (1.3 \times 10^{-6} \Omega.m) \times \frac{6.4 \text{ m}}{\pi (0.1156 \times 10^{-6}) \text{ m}^2}$

$R = \frac{1.3 \times 6.4}{\pi \times 0.1156} \Omega$

$R = \frac{8.32}{0.1156\pi} \Omega$

Using the value of $\pi \approx 3.14159$:

$R \approx \frac{8.32}{0.1156 \times 3.14159} \approx \frac{8.32}{0.363168} \approx 22.908 \Omega$.

The toaster is operated at a voltage $V = 120$ V. The power $P$ drawn by the toaster is given by the formula:

$P = \frac{V^2}{R}$

$P = \frac{(120 \text{ V})^2}{22.908 \Omega} = \frac{14400}{22.908} \text{ W}$

$P \approx 628.61$ W.

The question asks for the power in the format ____ $\times 10^2$ watt.

$P \approx 628.61$ W $= 6.2861 \times 100$ W $= 6.2861 \times 10^2$ W.

We need to find the value closest to 6.2861 to fill in the blank. Considering the significant figures of the input values (2 significant figures for L, d, $\rho$, and likely V), the result should be rounded to 2 significant figures. $628.61$ rounded to 2 significant figures is $630$. In the format $\times 10^2$, this is $6.3 \times 10^2$. The value to fill in the blank is 6.3.

If we consider rounding the coefficient $6.2861$ to one decimal place, we get 6.3.

The power drawn by the toaster is approximately $6.2861 \times 10^2$ W. The value closest to this, assuming the blank expects a value with a certain precision, is likely obtained by rounding. Based on typical question formats, rounding the coefficient to one decimal place is common when the input data precision suggests a similar level of precision in the result.