Let’s solve the problem step by step:

1. Key Concepts

• Alpha particles (
  $\alpha$ 

  ) have a charge of $q_\alpha = 2e$ 

  , where $e = 1.6 \times 10^{-19} \, \text{C}$ 

  is the elementary charge.

• Electric current
  $I$ 

  is the rate of flow of charge: $$I = \frac{\text{Total charge passing}}{\text{Time}}$$ 

2. Total Charge Passing per Minute

We are given:

• 
  $N = 10,000 \, \alpha$ 

  -particles per minute,

• Each
  $\alpha$ 

  -particle carries a charge $q_\alpha = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{C}$ 

  .

The total charge passing in 1 minute is:

$$Q = N \cdot q_\alpha = 10,000 \cdot 3.2 \times 10^{-19} = 3.2 \times 10^{-15} \, \text{C}.$$

3. Current

Convert the time from minutes to seconds:

$$t = 60 \, \text{seconds}.$$

The current is:

$$I = \frac{Q}{t} = \frac{3.2 \times 10^{-15}}{60} = 0.533 \times 10^{-16} \, \text{A}.$$

Rounding off, the current is approximately:

$$\boxed{0.5 \times 10^{-16} \, \text{A}}.$$

Product of Resistivity and Conductivity is 1.

\[ \rho\times \sigma= 1\]

So graph will be rectangular hyperbola.

Silver is a metal which has very high electrical conductivity and hence possesses the smallest resistivity compared to semiconductors (Silicon, Germanium) and insulators (Glass).

Resistance is inversely proportional to the square of the diameter: \(R = \frac{1}{d^2}\). Since the diameter is tripled, the resistance becomes \(\frac{R}{9} = \frac{81}{9} = 9 \Omega\).

Using the relation \(I = n e A v_d ⇒ v_d = \frac{I}{n e \pi r^2}\). Substituting the given values: \(v_d = \frac{10}{10^{22} \times 1.6 \times 10^{-19} \times \pi \times (10^{-3})^2} = \frac{6.25}{\pi} \times 10^3\text{ m s}^{-1}\).

Resistance of a stretched wire of constant volume is proportional to the square of its length: \( R \propto l^2 \). Thus, \( \frac{\Delta R}{R}\text{ (%)} = 2 \frac{\Delta l}{l}\text{ (%)} = 2 \times 0.1\% = 0.2\% \).

Copper is a conductor (lowest resistivity), Nichrome is an alloy (intermediate), Germanium and Silicon are semiconductors with Silicon having a larger bandgap and thus higher resistivity.

By definition, the reciprocal of electrical resistance is electrical conductance, \(G = \frac{1}{R}\).

Using standard formulas of current electricity: drift velocity \(v_d = \frac{eE}{m}\tau\) gives (A)-(R); resistivity \(\rho = \frac{E}{J}\) gives (B)-(S); relaxation time \(\tau = \frac{m}{ne^2\rho}\) gives (C)-(P); current density \(J = ne v_d\) gives (D)-(Q).

Formula: \(v_d = \frac{i}{n A e}\). If both \(i\) and \(A\) are doubled, the drift velocity \(v_d\) remains unchanged as the ratio \(\frac{i}{A}\) is constant.