Given, length of wire, l = 1m = 100 cm
Resistance, R = 10\(\Omega \)
The emf of a battery, E₁ = 6 V
R₁ = 5 \(\Omega \) and x = 40 cm
Current, I = \(\frac { { E }_{ 1 } }{ R+{ R }_{ 1 } } =\frac { 6 }{ 10+5 } =\frac { 6 }{ 15 } =\frac { 2 }{ 5 } A\) 
\(\\ { V }_{ AB }=IR=\frac { 2 }{ 5 } X10=4V\)
The emf of the primary cell = \(\frac { { V }_{ AB } }{ l } \times x=\frac { 4 }{ 100 } \times 40\)
= 1.6 V

The high resistance voltmeter means that current will flow through it. Hence, there is no potential difference across it. So, the reading shown by the high resistance voltmeter can be taken as the emf of the cell.
The internal resistance of a cell depends on
(i) the concentration of electrolyte and
(ii) distance between the two electrodes.
The emf of cell, E = 2.2 V
The terminal voltage across cell, when 5\(\Omega \) resistance R is connected across it, V = 1.8 V
Let internal resistance = r
\(\because \)  Internal resistance,\(r=R\left( \frac { E }{ V } -1 \right) \) 
\(=5\left( \frac { 2.2 }{ 1.8 } -1 \right) =5\times \frac { 0.4 }{ 1.8 } =\frac { 2 }{ 1.8 } =\frac { 10 }{ 9 } \Omega \)

In the closed loop ABDA,
100 I₁ + 15 I_g - 60 I₂ = 0
20 I₁ + 3 I_g - 12 I₂ = 0       ......(i)
In the closed loop BCDB,
10 (I₁ - I_g) - 5 (I₂ + I_g) - 15 I_g = 0
or  10 I₁ - 30 I_g - 5 I₂ = 0
or  2 I₁ - 6I_g - I₂ = 0           .......(ii)
In the closed loop ADCEA,
60 I₂ + 5 (I₂ + I_g) = 10
or  65 I₂ + 5 I_g = 10
or  13 I₂ + Ig = 2              ......(iii)
On solving Eqs. (i) , (ii) and (iii), we get
I_g = 4.87 mA

Given, diameter = 0.24 cm,
radius, \(r=\frac { 0.24\times { 10 }^{ -2 } }{ 2 } \) = 0.12 x 10^-2 m
and Current, I = 10A
\(\therefore\) Current density, \(J=\frac { I }{ A } =\frac { I }{ \pi { { r }^{ 2 } } } \)
\(=\frac{10}{3.14 \times\left(0.12 \times 10^{-2}\right)^2}=2.2 \times 10^6 \mathrm{Am}^{-2}\)

According to Wheatstone bridge circuit,
\(\frac { P }{ Q } =\frac { R }{ S } ,S=\frac { RQ }{ P } =\frac { 8X12 }{ 10 } =\frac { 48 }{ 5 } ,S=9.6\Omega \)