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In a resistance thermometer the resistances at \(0^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\) are \(6.74\) and \(7.74 \mathrm{ohm}\) respectively. The temperature corresponding to \(6.53\) ohm resistance is (a) \(53^{\circ} \mathrm{C}\) (b) \(21^{\circ} \mathrm{C}\) (c) \(-53^{\circ} \mathrm{C}\) (d) \(-21^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The temperature is \(-21^{\circ} \mathrm{C}\), so the correct option is (d).

Step by step solution

01

Understand the Linear Resistance-Temperature Relationship

In a resistance thermometer, resistance changes linearly with temperature. This can be represented as: \( R_t = R_0 + \alpha t \) where \( R_0 \) is resistance at \(0^{\circ} \text{C}\), \( t \) is temperature in degree Celsius, and \( \alpha \) is the temperature coefficient of resistance.
02

Calculate the Temperature Coefficient of Resistance (α)

We know: \( R_0 = 6.74 \) ohm, \( R_{100} = 7.74 \) ohm. Now calculate \( \alpha \) using the formula: \( \alpha = \frac{R_{100} - R_0}{100} \). This gives \( \alpha = \frac{7.74 - 6.74}{100} = 0.01 \).
03

Calculate Temperature for Given Resistance

Given resistance \( R_t = 6.53 \) ohm, use the rearranged formula \( t = \frac{R_t - R_0}{\alpha} \). By substituting the values: \( t = \frac{6.53 - 6.74}{0.01} = -21 \).
04

Select the Correct Option

The calculated temperature for the given resistance is \(-21^{\circ} \mathrm{C}\). Therefore, the correct answer is option (d) \(-21^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Coefficient of Resistance
The temperature coefficient of resistance, often denoted as \( \alpha \), is a crucial parameter when dealing with resistance thermometers. It represents the rate at which resistance changes with temperature. This linear relationship is particularly useful because it allows us to predict resistance at any given temperature using a simple formula.

The formula used is:
  • \( R_t = R_0 + \alpha t \)
Here, \( R_t \) is the resistance at temperature \( t \), \( R_0 \) is the resistance at \(0^{\circ} \mathrm{C}\), and \( \alpha \) is the temperature coefficient. In the example exercise, this coefficient is calculated as \( \alpha = 0.01 \). This value indicates how much the resistance increases for each degree Celsius rise in temperature.

Understanding \( \alpha \) is essential for determining how a material's resistance will change with temperature.
Linear Resistance-Temperature Relationship
The linear resistance-temperature relationship simplifies the analysis of how resistance varies with temperature, allowing for linear equations to predict resistance changes. This linearity is crucial for designing accurate resistance thermometers.

In practice, the relationship is given by:
  • \( R_t = R_0 + \alpha t \)
This tells us that resistance increases by the same amount for each degree of temperature change. It's a straight line if you plot resistance against temperature. This concept makes calculations straightforward and predictable, crucial for many practical applications, such as in labs or industrial environments where temperature measurement needs to be both accurate and reliable.

In our example, knowing the resistance at 0 and 100 degrees, and using the \( \alpha \) value, we can easily determine the temperature for any given resistance.
Resistance Measurement
Resistance measurement in the context of resistance thermometers involves calculating the unknown temperature using the known resistances at specific temperatures. It's all about using the linear relationship and the temperature coefficient to find the desired value.

Here’s the process:
  • First, you calculate \( \alpha \) using known resistances at two temperatures (like 0 and 100 degrees).
  • Next, with any new resistance value, rearrange the formula to solve for temperature:
  • \( t = \frac{R_t - R_0}{\alpha} \)
For the exercise given, this approach led to determining that a resistance of \(6.53\) ohms corresponds to a temperature of \(-21^{\circ} \mathrm{C}\). This method allows for quick and efficient temperature determination, linking electrical measurements with thermal properties effectively.

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Most popular questions from this chapter

In the condensation of a gas the mean \(\mathrm{KE}(K)\) and potential energy \((U)\) of molecules changes; thus (a) \(K\) decreases, \(U\) decreases (b) \(K\) increases, \(U\) keeps constant (c) \(K\) keeps constant, \(U\) decreases (d) \(K\) decreases, \(U\) increases

Two identical containers \(A\) and \(B\) with frictionless pistons contain the same ideal gas at the same temperature and the same volume \(V\). The mass of gas in \(A\) is \(m_{A}\) and that in \(B\) is \(m_{B}\). The gas in each cylinder is now allowed to expand isothermally to the same final volume \(2 \mathrm{~V}\). The changes in pressure in \(A\) and \(B\) are found to be \(\Delta P\) and \(1.5 \Delta P\) respectively. Then (a) \(4 m_{A}=9 m_{B}\) (b) \(2 m_{A}=3 m_{B}\) (c) \(3 m_{A}=2 m_{B}\) (d) \(9 m_{A}=4 m_{B}\)

If for hydrogen \(C_{p}-C_{v}=m\) and for nitrogen \(C_{p}-C_{v}=\) \(n\), where \(C_{e}\) and \(C_{v}\) refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between \(m\) and \(n\) is (Molecular weight of hydrogen \(=2\) and molecular weight of nitrogen \(=14)\) (a) \(n=14 m\) (b) \(n=7 m\) (c) \(m=7 n\) (d) \(m=14 n\)

Mercury boils at \(367^{\circ} \mathrm{C}\). However, mercury thermometers are made such that they can measure temperature upto \(500^{\circ} \mathrm{C}\). this is done by (a) maintaining vacuum above mercury column in the stem of the thermometer (b) filling nitrogen gas at high pressure above the mercury column (c) filling nitrogen gas at low pressure above the mercury column (d) filling oxygen gas at high pressure above the mercury column

The gas law \(\frac{P V}{T}=\) constant is true for (a) isothermal changes only (b) adiabatic changes only (c) both isothermal and adiabatic changes (d) neither for isothermal nor for adiabatic changes

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