A wire of diameter 1mm carries a current of 20A. What is the current density in the wire?

The change in potential energy ($\Delta U$) is related to the charge ($q$) and the change in potential ($\Delta V$) by the formula $\Delta U = q \Delta V$.
Given:
$\Delta U = -1.28 \times 10^{-4}\text{ J}$
The charge of an electron is $q = -1.602 \times 10^{-19}\text{ C}$. (The problem states $q = 1.602 \times 10^{-19}\text{ C}$, which is the magnitude of the elementary charge. For an electron, the charge is negative.)

There is a significant discrepancy between this calculated value and the provided options. If we assume a typo in the question and that the potential energy was $-1.28 \times 10^{-14}\text{ J}$ instead of $-1.28 \times 10^{-4}\text{ J}$, then:

This value is approximately $8 \times 10^4\text{ V}$, which matches option (d). Assuming this typo:
The correct option is (d).

7. Two charges $3.21 \times 10^{-9}\text{ C}$ and $5.4 \times 10^{-9}\text{ C}$ are situated at $3\text{m}$ and $6\text{m}$ respectively on the positive x-axis of a co-ordinate system. Find the total electric potential at a point $9\text{m}$ away from the origin due to these charges.

The electric potential ($V$) due to a point charge ($Q$) at a distance ($r$) is given by $V = \frac{kQ}{r}$, where $k = 9 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2$.
Given:
$Q_1 = 3.21 \times 10^{-9}\text{ C}$ at $x_1 = 3\text{m}$
$Q_2 = 5.4 \times 10^{-9}\text{ C}$ at $x_2 = 6\text{m}$
Point of interest is at $x_P = 9\text{m}$.

Distance from $Q_1$ to $P$: $r_1 = x_P - x_1 = 9\text{m} - 3\text{m} = 6\text{m}$
Distance from $Q_2$ to $P$: $r_2 = x_P - x_2 = 9\text{m} - 6\text{m} = 3\text{m}$

Total potential $V_{\text{total}} = V_1 + V_2 = 4.815\text{ V} + 16.2\text{ V} = 21.015\text{ V}$

The calculated value $21.015\text{ V}$ is approximately $21\text{ V}$.
The correct option is (b).

8. Copper contains $8.4 \times 10^{28}$ free electrons/m$^3$. A copper wire of cross-sectional area $7.4 \times 10^{-7}\text{m}^2$ carries a current of $1\text{ A}$. The electron drift speed is approximately:

The current ($I$) in a conductor is related to the number density of charge carriers ($n$), the cross-sectional area ($A$), the drift speed ($v_d$), and the charge of each carrier ($q$) by the formula $I = n A v_d q$.
Given:
$n = 8.4 \times 10^{28}\text{ m}^{-3}$
$A = 7.4 \times 10^{-7}\text{m}^2$
$I = 1\text{ A}$
Charge of an electron $q = 1.602 \times 10^{-19}\text{ C}$

The correct option is (d).

9. A given wire of resistance $10\Omega$ has a length of $5\text{m}$ and a cross-sectional area of $4.0 \times 10^{-3}\text{m}^2$. Calculate the conductivity of the wire.

Resistance ($R$) is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. Conductivity ($\sigma$) is the reciprocal of resistivity, $\sigma = \frac{1}{\rho}$.
Given:
$R = 10\Omega$
$L = 5\text{m}$
$A = 4.0 \times 10^{-3}\text{m}^2$

Rearrange the resistance formula: $\rho = \frac{R \times A}{L}$

Since $1\text{m} = 100\text{cm}$, then $1\text{m}^{-1} = 10^{-2}\text{cm}^{-1}$.

There is a significant discrepancy between this calculated value and the provided options, which are in the range of $10^6$ to $10^8$. If we assume a typo in the cross-sectional area, and it was $4.0 \times 10^{-10}\text{m}^2$ instead of $4.0 \times 10^{-3}\text{m}^2$, then:

Converting to $\Omega^{-1}\text{cm}^{-1}$:

This matches option (b). Assuming this typo:
The correct option is (b).

10. A wire of diameter $1\text{mm}$ carries a current of $20\text{A}$. What is the current density in the wire?

Current density ($J$) is defined as current ($I$) per unit cross-sectional area ($A$), $J = \frac{I}{A}$.
Given:
Diameter $d = 1\text{mm}$
Current $I = 20\text{A}$

The calculated value $2.546 \times 10^7\text{ A/m}^2$ is approximately $2.55 \times 10^7\text{ A/m}^2$.
The correct option is (a).

11. An electric bulb is rated $60\text{W}, 220\text{V}$. Calculate the resistance of its Filament when it is operating normally.

The power ($P$) dissipated by a resistor is related to the voltage ($V$) across it and its resistance ($R$) by the formula $P = \frac{V^2}{R}$.
Given:
Power $P = 60\text{W}$
Voltage $V = 220\text{V}$

The options are not visible for this question, but the calculation is straightforward.