The question asks about the characteristic property of good conductors. We need to understand the definitions of conductance, resistance, reluctance, and thermal conductivity in the context of electrical conduction.
Correct Option: A) Conductance
Good conductors are characterized by their ability to allow electric current to flow with ease. This property is directly measured by conductance. A high conductance means the material readily conducts electricity.
This question tests the understanding of fundamental SI units and their derived quantities, specifically focusing on work/energy and power. We need to recall the definitions and units of work, energy, and power to determine what a Newton meter represents.
B) One joule — As derived, one Newton meter is the definition of one Joule, which is the SI unit of work or energy.
Ohm's Law describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the physical conditions and temperature remain constant. This implies that the resistance of the conductor must remain constant for the law to be valid.
A) Resistance must be uniform
This is the correct condition. Ohm's Law is valid only for ohmic materials where the resistance remains constant irrespective of the voltage applied or current flowing through it. If the resistance changes (e.g., due to temperature variations), the relationship between voltage and current will no longer be linearly proportional, and Ohm's law will not hold true.
The question asks to identify an insulating material from the given options. An insulator is a material that does not allow electricity to flow through it easily, due to the absence of free electrons. Conductors, on the other hand, have free electrons and allow electricity to pass through them readily.
D) Paper — Paper is an insulating material. It does not conduct electricity because its electrons are tightly bound and there are no free charge carriers to facilitate the flow of current.
The question asks to identify which of the given options is NOT equivalent to a Watt. A Watt is the unit of power. We will use the fundamental definitions and formulas relating power, voltage, current, resistance, energy, and time to determine the equivalence of each option to a Watt.
The Watt (\(\text{W}\)) is the SI unit of power. Power is defined as the rate at which work is done or energy is transferred.
\[ \text{Power} = \frac{\text{Energy}}{\text{Time}} \]The SI unit of energy is Joule (\(\text{J}\)) and the SI unit of time is second (\(\text{s}\)).
\[ 1 \text{ Watt} = 1 \frac{\text{Joule}}{\text{second}} \]Electrical power (\(P\)) is also given by the product of voltage (\(V\)) and current (\(I\)).
\[ P = V \times I \]\[ 1 \text{ Watt} = 1 \text{ Volt} \times 1 \text{ Ampere} \]Ohm's Law states \(V = I \times R\), where \(R\) is resistance in Ohms (\(\Omega\)).
Substitute \(V = I \times R\) into the power formula \(P = V \times I\):
\[ P = (I \times R) \times I = I^2 \times R \]\[ 1 \text{ Watt} = 1 \text{ Ampere}^2 \times 1 \text{ Ohm} \]D) Amperes/volt
Let's analyze this option: \(\frac{\text{Amperes}}{\text{Volt}}\). From Ohm's Law, \(V = I \times R\), so \(R = \frac{V}{I}\). Therefore, \[\frac{1}{R} = \frac{I}{V}\]. The unit of \(\frac{I}{V}\) is \(\frac{\text{Amperes}}{\text{Volt}}\), which is the unit of conductance (Siemens, \(\text{S}\)), not power (Watt). Thus, Amperes/volt is not equivalent to Watt.
This question tests the understanding of how a galvanometer is converted into an ammeter or a voltmeter. The key lies in understanding the function of each device and how resistance is connected to achieve that function.
A galvanometer is converted into an ammeter by connecting a low resistance (shunt) in parallel with it. However, the statement says "a galvanometer with low resistance in series is an ammeter". This is incorrect for a standard ammeter conversion. Let's re-evaluate the options carefully. The question asks which statement is TRUE. An ammeter is designed to measure current and must have very low resistance. When a low resistance is connected in series with a galvanometer, the total resistance of the combination would be the sum of the galvanometer's resistance and the low series resistance. This would still be relatively low, making it suitable for measuring current, but the standard conversion involves a parallel shunt. However, let's consider the other options. If we consider the overall resistance of the ammeter, it should be low. Connecting a low resistance in series with a galvanometer would result in a low overall resistance for the device, which is a characteristic of an ammeter. Let's re-examine the standard definitions.
An ammeter is connected in series and should have a very low resistance. To convert a galvanometer (which has some internal resistance \(R_g\)) into an ammeter, a small resistance \(R_s\) (shunt) is connected in parallel with it. The equivalent resistance of this combination is \(\frac{R_g R_s}{R_g + R_s}\), which is very low.
Let's re-read option C: "A galvanometer with low resistance in series is an ammeter\