The current in $$2\Omega $$  resistor will be zero because it is not a part of any closed loop.

KEY CONCEPT : Resistance of Galvanometer,
$$G = \frac{{{\text{Current sensitivity}}}}{{{\text{Voltage sensitivity}}}} \Rightarrow G = \frac{{10}}{2} = 5\Omega $$
Here $${i_g} =$$  Full scale deflection current $$= \frac{{150}}{{10}} = 15mA$$
$$V =$$  voltage to be measured $$= 150\,volts$$
(such that each division reads $$1\,volt$$ )
$$ \Rightarrow R = \frac{{150}}{{15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega $$

For ammeter,

$$\eqalign{ & 0.002I \times G = 0.998I \times {r_s} \cr & {r_s} = \frac{{0.002}}{{0.998}}G \cr & \Rightarrow {r_s} = 0.002004G = \frac{1}{{499}} \times G \cr} $$
Equivalent resistance of ammeter,
$$\eqalign{ & \frac{1}{R} = \frac{1}{G} + \frac{1}{{{r_s}}} \cr & \therefore \frac{1}{R} = \frac{1}{G} + \frac{1}{{\frac{G}{{499}}}} \cr & \Rightarrow R = \frac{G}{{500}} \cr} $$

From Ohms law, resistance $$R = \frac{{\Delta V}}{{\Delta I}}$$
Resistance is said to be negative if on increasing voltage or temperature, current decreases.
In $$V-I$$  graph, for portion $$CD$$ of graph, the current decreases with increasing voltage.
Thus, portion $$CD$$ corresponds to negative resistance.
Alternative
We know that the slope of graph $$V-I$$  will give the resistance, since slope of $$CD$$ curve is negative.
So, it has negative resistance.

$$\eqalign{ & {R_1} + {R_2} = 1000 \cr & \Rightarrow {R_2} = 1000 - {R_1} \cr} $$
On balancing condition
$${R_1}\left( {100 - l} \right) = \left( {1000 - {R_1}} \right)l\,......\left( {\text{i}} \right)$$

On Interchanging resistance balance point shifts left by $$10\,cm$$

On balancing condition
$$\eqalign{ & \left( {1000 - {R_1}} \right)\left( {110 - l} \right) = {R_1}\left( {l - 10} \right) \cr & {\text{or,}}\,{R_1}\left( {l - 10} \right) = \left( {1000 - {R_1}} \right)\left( {110 - l} \right)\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing eqn (i) by (ii)
$$\eqalign{ & \frac{{100 - l}}{{l - 10}} = \frac{l}{{110 - l}} \cr & \Rightarrow \left( {100 - l} \right)\left( {110 - l} \right) = l\left( {l - 10} \right) \cr & \Rightarrow 11000 - 100l - 110l + {l^2} = {l^2} - 10l \cr & \Rightarrow 11000 = 200l \cr & {\text{or,}}\,l = 55 \cr} $$
Putting the value of $$'l'$$ in eqn (i)
$$\eqalign{ & {R_1}\left( {100 - 55} \right) = \left( {1000 - {R_1}} \right)55 \cr & \Rightarrow {R_1}\left( {45} \right) = \left( {1000 - {R_1}} \right)55 \cr & \Rightarrow {R_1}\left( 9 \right) = \left( {1000 - {R_1}} \right)11 \cr & \Rightarrow 20{R_1} = 11000 \cr & \therefore {R_1} = 550K\Omega \cr} $$

Potential gradient $$ = \frac{{{\text{Potential}}\,{\text{applied}}}}{{{\text{Length of wire}}}}$$
$$\therefore k = \frac{V}{l}$$
Given, $$V = 2\,V,\,l = 4\,m$$
$$\therefore k = \frac{2}{4} = 0.5\,V/m$$

Total external resistance will be the total resistance of whole length of box. It should be connected between $$A$$ and $$D.$$

To carry a current of $$4\,A$$  we need four path, each carrying a current of $$1\,A.$$  Let $$r$$ be the resistance of each path. These resistances are connected in parallel. So, their equivalent resistance.
$$\eqalign{ & \frac{1}{{{r_p}}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} \cr & {\text{or}}\,\,{r_p} = \frac{r}{4}. \cr & {\text{But,}}\,\,{r_p} = \frac{r}{4} = 5\,\,\left( {{\text{given}}} \right) \cr & \therefore r = 20\,\Omega \cr} $$
For this purpose two resistances should be connected is series. There are four such combinations in parallel. Hence, the total number of resistances $$ = 4 \times 2 = 8.$$

$$\eqalign{ & J = \frac{I}{A} = \frac{Q}{{tA}} = \frac{{ne}}{{tA}} \cr & = 9.6 \times {10^3}A/{m^2} \cr} $$

Neutral temperature is the temperature of a hot junction at which $$E$$ is maximum.
$$ \Rightarrow \frac{{dE}}{{d\theta }} = 0\,{\text{or}}\,a + 2b\theta = 0 \Rightarrow \theta = \frac{{ - a}}{{2b}} = - 350$$
Neutral temperature can never be negative hence no $$\theta $$ is possible.