Diagram illustrating two configurations of identical metal slabs joined in series and parallel to compare their resistances.

Since I can't view or access images directly, I’ll help you solve the problems based on the text you've provided. Let's go through each question one by one and provide a clear explanation.

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1. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.

Solution:

In a series circuit, the current (I) is the same in both wires.

Current is given by:
\[
I = nAev_d
\]
where:
- \( n \) = number density of electrons,
- \( A \) = cross-sectional area,
- \( e \) = charge of electron,
- \( v_d \) = drift velocity.

Given:
- Diameter is same ⇒ Area \( A \) is same for both wires.
- \( n_X = 2n_Y \)
- Current \( I \) is same.

So,
\[
I = n_X A e v_{dX} = n_Y A e v_{dY}
\]

Cancel common terms (\( A \), \( e \)):
\[
n_X v_{dX} = n_Y v_{dY}
\Rightarrow 2n_Y v_{dX} = n_Y v_{dY}
\Rightarrow 2v_{dX} = v_{dY}
\Rightarrow \frac{v_{dX}}{v_{dY}} = \frac{1}{2}
\]

Answer: The ratio of drift velocity of electrons in X to Y is \( \boxed{1:2} \).

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2. Two wires of equal length, one of copper and the other of manganin have the same resistance. Which wire is thicker?

Solution:

Resistance is given by:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) = resistivity,
- \( L \) = length,
- \( A \) = cross-sectional area.

Given:
- \( R \) is same,
- \( L \) is same,
- So \( \rho / A \) must be constant ⇒ \( A \propto 1/\rho \)

Now:
- Copper has lower resistivity (\( \rho_{Cu} \approx 1.7 \times 10^{-8} \, \Omega\cdot m \)),
- Manganin has higher resistivity (\( \rho_{Mn} \approx 4.8 \times 10^{-7} \, \Omega\cdot m \)).

So, for same resistance, the manganin wire must have larger area (i.e., be thicker) to compensate for higher resistivity.

Answer: The manganin wire is thicker.

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3. Nichrome and copper wires of same length and same radius are connected in series. Current \( I \) is passed through them. Does the current in both wires remain the same? Justify your answer.

Solution:

Yes, the current remains the same in both wires.

When wires are connected in series, the same current flows through all components because there is only one path for the current.

Even though nichrome and copper have different resistivities, and thus different resistances, the current is determined by the total resistance and voltage. But in series, the current is identical at every point.

Answer: Yes, the current is the same in both wires because they are in series.

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4. When electric drift in a metal from lower to higher potential, does it mean that all the free electrons of the metal are moving in the same direction?

Solution:

No.

- Electric drift refers to the net movement of electrons due to an electric field.
- Electrons move from higher potential to lower potential (since they are negatively charged).
- But this does not mean all electrons are moving in the same direction.
- Electrons undergo random thermal motion, and their drift velocity is very small compared to their random speed.
- The net drift is in one direction (opposite to the electric field), but individual electrons zigzag randomly.

Answer: No, electrons have random thermal motion; only their net drift is in one direction.

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5. The emf of a cell is always greater than its terminal voltage. Why?

Solution:

EMF (\( \mathcal{E} \)) is the maximum potential difference a cell can provide when no current flows.

Terminal voltage (\( V \)) is the actual voltage measured when current flows.

When current \( I \) flows, there is a voltage drop inside the cell due to its internal resistance \( r \):
\[
V = \mathcal{E} - Ir
\]

Thus:
\[
\mathcal{E} > V \quad \text{(when } I > 0\text{)}
\]

Answer: Because of the internal resistance of the cell, some voltage is lost inside the cell as \( Ir \), so terminal voltage is less than EMF.

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6. Two identical slabs of a given metal are joined together in two different ways, as shown in figures (a) and (b). What is the ratio of their resistances?

(Assuming figure (a): side-by-side, parallel connection; figure (b): end-to-end, series connection)

Let’s suppose each slab has:
- Length \( L \),
- Cross-sectional area \( A \),
- Resistivity \( \rho \).

Then resistance of one slab:
\[
R_0 = \rho \frac{L}{A}
\]

#### Case (a): Parallel connection (side by side)
- Total area = \( 2A \),
- Length = \( L \),
- Resistance:
\[
R_a = \rho \frac{L}{2A} = \frac{R_0}{2}
\]

#### Case (b): Series connection (end to end)
- Total length = \( 2L \),
- Area = \( A \),
- Resistance:
\[
R_b = \rho \frac{2L}{A} = 2R_0
\]

Now, ratio of resistances:
\[
\frac{R_a}{R_b} = \frac{R_0/2}{2R_0} = \frac{1}{4}
\]

But the question asks: "What is the ratio of their resistances?" — likely meaning \( R_{(a)} : R_{(b)} \)

So:
\[
R_a : R_b = \frac{1}{2}R_0 : 2R_0 = 1 : 4
\]

Answer: The ratio of resistances is \( \boxed{1:4} \) (resistance in (a) to (b)).

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7. A steady current flows in a metallic conductor of non-uniform cross-section. Of these quantities, which is constant along the conductor: current, current density, electric field, drift speed?

Solution:

- Current (I): Yes, constant. In a steady state, charge conservation implies current is same everywhere.
- Current density (J): No. \( J = I/A \). Since \( A \) varies, \( J \) changes.
- Electric field (E): No. For metals, \( E = \rho J \), and since \( J \) varies, \( E \) varies.
- Drift speed (v_d): No. \( v_d = J/(ne) \), so it varies with \( J \).

Only current is constant.

Answer: Current is constant along the conductor.

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8. A resistance \( R \) is connected across a cell of emf \( \mathcal{E} \) and internal resistance \( r \). A potentiometer now measures the potential difference between the terminals of the cell as \( V \). Write the expression for \( V \) in terms of \( \mathcal{E}, V \), and \( R \).

Wait — the question says: “Write the expression for \( V \)” — but already uses \( V \) as the measured PD.

Actually, it says: “Write the expression for \( V \) in terms of \( \mathcal{E}, V \), and \( R \)” — that seems like a typo.

Likely meant: “Write the expression for \( V \) in terms of \( \mathcal{E}, r \), and \( R \).”

Let’s correct and solve:

Circuit: Cell with emf \( \mathcal{E} \), internal resistance \( r \), connected to external resistance \( R \).

Total resistance = \( R + r \)

Current:
\[
I = \frac{\mathcal{E}}{R + r}
\]

Terminal voltage:
\[
V = \mathcal{E} - Ir = \mathcal{E} - \left(\frac{\mathcal{E}}{R + r}\right)r = \mathcal{E} \left(1 - \frac{r}{R + r}\right) = \mathcal{E} \left(\frac{R}{R + r}\right)
\]

Alternatively:
\[
V = IR = \frac{\mathcal{E}}{R + r} \cdot R = \mathcal{E} \cdot \frac{R}{R + r}
\]

Answer:
\[
\boxed{V = \mathcal{E} \cdot \frac{R}{R + r}}
\]

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✔ Final Answers Summary:

1. Drift velocity ratio: \( \boxed{1:2} \)
2. Manganin wire is thicker
3. Yes, current is same in series
4. No, only net drift is in one direction
5. Due to internal resistance, \( V = \mathcal{E} - Ir \)
6. Resistance ratio: \( \boxed{1:4} \)
7. Only current is constant
8. \( \boxed{V = \mathcal{E} \cdot \frac{R}{R + r}} \)

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