Question

(a) The resistance of a wire of 0.01 cm radius and 1.0 cm length is \( 7 \, \Omega \).
Calculate its resistivity.
OR
(b) An electric heater is rated 220 V; 11 A. Calculate the power consumed if the heater is operated at 200 V.

Hint:

(a) Resistivity formula: \( \rho = \frac{R \times A}{l} \). Remember to convert units to meters. (b) For constant resistance, power \( \propto V^2 \). \( P_2 = P_1 \times \left( \frac{V_2}{V_1} \right)^2 \).

Answer 1

Part (a): Calculation of resistivity
Step 1: Write down the given data.
Resistance of wire, \( R = 7 \, \Omega \)
Radius of wire, \( r = 0.01 \, \text{cm} = 0.01 \times 10^{-2} \, \text{m} = 10^{-4} \, \text{m} \)
Length of wire, \( l = 1.0 \, \text{cm} = 1.0 \times 10^{-2} \, \text{m} = 10^{-2} \, \text{m} \)
Step 2: Recall the formula for resistance.
The resistance of a wire is given by: \[ R = \rho \frac{l}{A} \] where \( \rho \) is the resistivity, \( l \) is the length, and \( A \) is the cross-sectional area.
Step 3: Calculate the cross-sectional area.
The wire has circular cross-section, so: \[ A = \pi r^2 = \pi (10^{-4})^2 = \pi \times 10^{-8} \, \text{m}^2 \]
Step 4: Rearrange the formula to find resistivity.
\[ \rho = R \times \frac{A}{l} \]
Step 5: Substitute the values.
\[ \rho = 7 \times \frac{\pi \times 10^{-8}}{10^{-2}} \] \[ \rho = 7 \times \pi \times 10^{-8} \times 10^{2} \] \[ \rho = 7 \times \pi \times 10^{-6} \, \Omega \text{-m} \]
Step 6: Calculate numerical value.
Using \( \pi \approx 3.14 \): \[ \rho = 7 \times 3.14 \times 10^{-6} = 21.98 \times 10^{-6} \, \Omega \text{-m} \] \[ \rho \approx 2.20 \times 10^{-5} \, \Omega \text{-m} \]
Step 7: Final answer.
\[ \boxed{\rho = 7\pi \times 10^{-6} \, \Omega \text{-m} \approx 2.20 \times 10^{-5} \, \Omega \text{-m}} \] Part (b): Power consumed at different voltage
Step 1: Write down the given data.
Rated voltage, \( V_1 = 220 \, \text{V} \)
Rated current, \( I_1 = 11 \, \text{A} \)
Operating voltage, \( V_2 = 200 \, \text{V} \)
Step 2: Calculate the resistance of the heater.
Using Ohm's law, the resistance of the heater is constant and can be found from rated values: \[ R = \frac{V_1}{I_1} = \frac{220}{11} = 20 \, \Omega \]
Step 3: Calculate power at rated voltage (for verification).
Rated power: \[ P_1 = V_1 \times I_1 = 220 \times 11 = 2420 \, \text{W} \]
Step 4: Calculate power at operating voltage.
Since resistance is constant, power at voltage \( V_2 \) is: \[ P_2 = \frac{V_2^2}{R} = \frac{(200)^2}{20} \] \[ P_2 = \frac{40000}{20} = 2000 \, \text{W} \]
Step 5: Alternative method using ratio.
\[ \frac{P_2}{P_1} = \left( \frac{V_2}{V_1} \right)^2 = \left( \frac{200}{220} \right)^2 = \left( \frac{10}{11} \right)^2 = \frac{100}{121} \] \[ P_2 = P_1 \times \frac{100}{121} = 2420 \times \frac{100}{121} = 2420 \times 0.8264 \approx 2000 \, \text{W} \]
Step 6: Final answer.
\[ \boxed{P = 2000 \, \text{W}} \]