The question asks for a method used for the absolute measurement of resistances. Absolute measurement implies determining a quantity directly from fundamental units or constants, without relying on prior calibration against a standard of the same quantity. We need to evaluate the given options in this context.
B) Wheatstone bridge method. The Wheatstone bridge is a highly accurate method for measuring resistance. While it's often used for comparative measurements against a known standard, it can also be configured for absolute measurements, especially when high precision is required and the reference resistors are precisely known. It allows for the determination of an unknown resistance based on the ratios of other resistances, which can be derived from fundamental principles or precisely known standards, making it suitable for absolute measurements.
To calculate the resistance of the wire, we will use the formula relating resistance, specific resistance (resistivity), length, and cross-sectional area. First, we need to convert all given units to a consistent system (e.g., SI units or CGS units). Then, calculate the cross-sectional area of the wire using its diameter, and finally, apply the resistance formula.
Identify the given parameters and convert units:
Calculate the radius of the wire:
The radius \(r\) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.014 \text{ cm}}{2} = 0.007 \text{ cm} \]Calculate the cross-sectional area of the wire:
The wire is cylindrical, so its cross-sectional area \(A\) is given by the formula for the area of a circle:
\[ A = \pi r^2 \]\[ A = \pi (0.007 \text{ cm})^2 \]\[ A = \pi (49 \times 10^{-6}) \text{ cm}^2 \]\[ A \approx 3.14159 \times 49 \times 10^{-6} \text{ cm}^2 \]\[ A \approx 153.938 \times 10^{-6} \text{ cm}^2 \]Calculate the resistance of the wire:
The resistance \(R\) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]Substitute the calculated values into the resistance formula:
\[ R = (9.6 \times 10^{-6} \text{ ohm-cm}) \times \frac{440 \text{ cm}}{153.938 \times 10^{-6} \text{ cm}^2} \]\[ R = \frac{9.6 \times 440}{153.938} \text{ ohm} \]\[ R = \frac{4224}{153.938} \text{ ohm} \]\[ R \approx 27.439 \text{ ohm} \]Round to an appropriate number of significant figures:
The resistance is approximately \(27.4 \text{ ohm}\).
C) 27.4 ohm
The question asks to identify the type of material or circuit where Ohm's law is not applicable. Ohm's law states a linear relationship between voltage and current. We need to evaluate each option based on this principle.
A) Semiconductors. Ohm's law describes a linear relationship between voltage and current (\(V = IR\)), where \(R\) is constant. Semiconductors exhibit non-linear current-voltage characteristics, meaning their resistance is not constant but varies with the applied voltage or current. Therefore, Ohm's law is not applicable to semiconductors.
The question asks about the appropriate method for accurately measuring low resistance. Different types of bridges are used for measuring resistance, each suited for a specific range of resistance values.
Therefore, the Kelvin bridge is the correct choice for accurate low resistance measurement.
A) Kelvin bridge — The Kelvin bridge is a modification of the Wheatstone bridge specifically designed to measure low resistances (typically below \(1 \Omega\)) accurately by eliminating the effects of lead and contact resistances.
The question asks for Ohm's law in point form, which is a fundamental concept in electromagnetism and field theory. We need to identify the equation that relates current density, electric field, and conductivity at a specific point in a material.
\[El = (JA) \left(\frac{1}{\sigma} \frac{l}{A}\right)\]
\(E = \frac{J}{\sigma}\)
Rearranging this equation to solve for \(J\), we get:
\(J = \sigma E\)
Correct Option: C) J = σE
To determine which bulb has the least resistance, we need to use the relationship between power, voltage, and resistance. The power \(P\) of an electrical appliance is given by the formula \(P = \frac{V^2}{R}\), where \(V\) is the voltage and \(R\) is the resistance. From this, we can derive the formula for resistance as \(R = \frac{V^2}{P}\). We will calculate the resistance for each bulb given its voltage and power ratings.
Given \(V = 220 \text{ V}\) and \(P = 60 \text{ W}\).
\[ R_A = \frac{(220 \text{ V})^2}{60 \text{ W}} = \frac{48400}{60} \approx 806.67 \text{ \Omega} \]Given \(V = 220 \text{ V}\) and \(P = 100 \text{ W}\).
\[ R_B = \frac{(220 \text{ V})^2}{100 \text{ W}} = \frac{48400}{100} = 484 \text{ \Omega} \]Given \(V = 115 \text{ V}\) and \(P = 100 \text{ W}\).
\[ R_C = \frac{(115 \text{ V})^2}{100 \text{ W}} = \frac{13225}{100} = 132.25 \text{ \Omega} \]Comparing the calculated resistances:
The least resistance is \(132.25 \text{ \Omega}\), which corresponds to option C.
Correct Option: C) 115V, 100W