To find the temperature coefficient of resistance

$\alpha$

of the conductor, we use the formula for the change in resistance with temperature:

$$R_T = R_0 \left( 1 + \alpha T \right)$$

Where:

• 
  $R_T$ 

  is the resistance at temperature $T$ 

  ,

• 
  $R_0$ 

  is the resistance at the reference temperature (0°C),

• 
  $\alpha$ 

  is the temperature coefficient of resistance,

• 
  $T$ 

  is the temperature change in °C.

Step 1: Given

• The ratio of resistances at 15°C and 37.5°C is given as
  $\frac{R_{15}}{R_{37.5}} = \frac{4}{5}$ 

  .

• The resistance at temperature 15°C,
  $R_{15} = R_0 (1 + \alpha \times 15)$ 

  .

• The resistance at temperature 37.5°C,
  $R_{37.5} = R_0 (1 + \alpha \times 37.5)$ 

  .

Step 2: Set up the equation based on the given ratio

$$\frac{R_{15}}{R_{37.5}} = \frac{4}{5}$$

Substituting the expressions for

$R_{15}$

and

$R_{37.5}$

:

$$\frac{R_0 (1 + \alpha \times 15)}{R_0 (1 + \alpha \times 37.5)} = \frac{4}{5}$$

Canceling

$R_0$

from both the numerator and denominator:

$$\frac{1 + 15\alpha}{1 + 37.5\alpha} = \frac{4}{5}$$

Step 3: Solve for $\alpha$

Cross-multiply to solve for

$\alpha$

:

$$5(1 + 15\alpha) = 4(1 + 37.5\alpha)$$

Expanding both sides:

$$5 + 75\alpha = 4 + 150\alpha$$

Simplify:

$$5 - 4 = 150\alpha - 75\alpha$$

$$1 = 75\alpha$$

$$\alpha = \frac{1}{75}$$

Final Answer:

The temperature coefficient of resistance

$\alpha$

is

$\boxed{\frac{1}{75}} \, \text{per °C}$

.

When two resistors with resistances

$R_1$

and

$R_2$

and temperature coefficients of resistance

$\alpha_1$

and

$\alpha_2$

are connected in series, the effective temperature coefficient of resistance

$\alpha_{\text{eff}}$

is given by the formula:

$$\alpha_{\text{eff}} = \frac{\alpha_1 R_1 + \alpha_2 R_2}{R_1 + R_2}$$

This formula takes into account the individual resistances and temperature coefficients of the two wires, considering that their total resistance is the sum of the individual resistances.

Slope of V-I graph represents Resistance. So R1>R2

Concept: Ohm's law states \(V = IR\) with constant \(R\). Resistance of metals depends on temperature.
Formula: \(R_T = R_0(1 + \alpha T)\).
Solution: For high currents, metallic wires heat up significantly, increasing their resistance. This violates the constant resistance assumption of Ohm's law. Therefore, Ohm's law holds for small currents where heating is negligible. A and R are true, and R explains A.

Concept: Ohm's law, temperature dependence of resistance.
Formula: \(I = V/R\), \(R\) increases with \(T\) for most resistors.
Solution: For most metallic resistors, resistance \(R\) increases with increasing temperature \(T\). When connected to a battery (constant voltage \(V\)), the current \(I = V/R\) decreases as \(R\) increases. Thus, A and R are true and R explains A.

For most resistors, resistance \(R\) increases with temperature. When connected to a battery, the voltage \(V\) is constant. According to Ohm's law, \(I = V/R\). As \(R\) increases due to temperature, the current \(I\) must decrease. Thus, both assertion and reason are true, and the reason correctly explains the assertion.

When temperature increases, thermal vibrations of atoms in the conductor increase, leading to more frequent collisions for electrons. This reduces the relaxation time \(\tau\), which in turn decreases the drift velocity \(v_d = \frac{eE\tau}{m}\). Decreased drift velocity leads to decreased conductivity \(\sigma = \frac{ne^2\tau}{m}\). Thus, both (A) and (R) are true, and (R) is the correct explanation for (A).

For (A): \(\Delta T = 227 - 27 = 200^\circ C\). Given \(R=150 \Omega\) and \(R_0=100 \Omega\). Using \(R = R_0(1 + \alpha\Delta T)\) \(\Rightarrow 150 = 100(1 + \alpha \times 200)\) \(\Rightarrow 1.5 = 1 + 200\alpha\) \(\Rightarrow 0.5 = 200\alpha\) \(\Rightarrow \alpha = 2.5 \times 10^{-3} /^{\circ} C\). So, (A) is true. For (R): The formula \(R = R_0(1 + \alpha\Delta T)\) is an empirical approximation for temperature dependence of resistance. It is often used for significant temperature changes and is not strictly limited to very small \(\Delta T\) or \(\Delta R \ll R_0\). Thus, (R) is false.

Assertion (A) is true. For electrolytes, increased temperature leads to more ion dissociation and reduced viscosity. Reason (R) is true as increased temperature reduces viscosity. (R) correctly explains (A) as lower viscosity allows ions to move more freely, increasing conductivity.

Assertion (A) is true. Increased temperature in semiconductors generates more charge carriers, decreasing resistivity. Reason (R) is also true, as thermal vibrations increase electron-ion collisions in conductors, increasing resistivity. However, (R) explains conductors, not semiconductors, so it's not the correct explanation for (A).