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Given, Emf of cell, E₁ = 1.25 V
Balance point is obtained at, l₁ = 35 cm
When, the cell is replaced by another cell 
New balance point is, l₂ = 63 cm
Emf of the second cell can be found out using the relation,
                       $\boxed{\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1} = \frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$

Now, substituting values we get,

$\frac{{\mathrm{E}}_2}{1.25} = \frac{63}{35}$ 
 $$\begin{gathered} \Rightarrow {\mathrm{E}}_2 = 1.25 \times \frac{63}{35}V \\ = 2.25 \mathrm{volt} \end{gathered}$$
                         

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Given,
Number density of electrons, n = 8.5 x 10²⁸ m^–3  Current carried by the wire, I = 3.0 A
Area of cross-section of the wire, A = 2.0 x 10^–6 m²
Length of the wire, l = 3.0 m
Charge on electron, e = 1.6 x 10^–19 C 

Using the formula for Drift velocity  $\boxed{{\mathrm{\nu }}_{\mathrm{d}} = \frac{\mathrm{I}}{\mathrm{neA}}}$$$\begin{gathered} = \frac{3}{8.5 \times {10}^{28} \times 1.6 \times {10}^{-19} \times 2.0 \times {10}^{-6}}\mathrm{m} {\mathrm{s}}^{-1} \\ = 1.103 \times {10}^{-4} \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$ 

Therefore,
Time taken by an electron to drift from one end to another end is,
$$\begin{gathered} \mathrm{t} =\frac{\mathrm{l}}{{\mathrm{\nu }}_{\mathrm{d}}} \\ = \frac{3.0}{1.103 \times {10}^{-4}} \mathrm{s} \\ = 2.72 \times {10}^4\mathrm{s} \\ (\approx 7.5 \mathrm{h}). \end{gathered}$$

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Given,
Surface charge density of earth's surface = 10^–9 C/m² 
Current flowing across the surface of earth, I = 1800 A
The radius of earth,r= 6370 km = 6.37 x 10⁶ m

Charge on entire surface of the earth, q= $\mathrm{\sigma }.\mathrm{A}$

Area of earth's surface, A = 4${\mathbf{\pi r}}^{\mathbf{2}}$

This implies, 
q=4$\mathrm{\pi }$ (6.37 x 10⁶)² x 10^–9 C 

Using the formula , I=$\frac{\mathrm{q}}{\mathrm{t}}$ we get,
                           
                           t=$\frac{\mathrm{q}}{\mathrm{I}}$ 
Therefore,
Time required for the flow of entire charge is,

$$\begin{gathered} = \frac{4\times 3.14\times {\left(6.37 \times {10}^6\right)}^2 \times {10}^{-9}}{1800} \\ = 283 \mathrm{seconds}. \end{gathered}$$

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Applying Kirchhoff’s second law to the mesh ABDA
– 10I₁ – 5I_g + (I – I₁) 5 = 0
or 3I₁ – I + I_g = 0 ...(i)

Again, applying Kirchhoff’s second law to the mesh BDCB,
– 5I_g – 10(1 – I_{l +} I_g) + 5(I₁ – I_g) = 0
or 3I₁ – 21 – 4I_g = 0 ...(ii)
Applying Kirchhoff’s second law to the mesh ABCEA,
– 10I₁ – 5(I₁ – I_g) –10I + 10 = 0
or 3I₁ + 2I – I_g = 2 ...(iii)
Adding (i) and (iii), we get
6I₁ + I = 2 ...(iv)
Multiplying (i) by 4 and adding in (ii), we get
15I₁ – 6I0 = 0 ...(v)
Solving equations (iv) and (v), we get
I₁ = 
So, current in branch AB is 0.235 A.
Putting the value of I₁ in equation (v) and simplifying, we get
Total current,                    
Putting the values of I and I₁ in equation (iii) and simplifying, we get

The negative sign indicates that the direction of current is opposite to that shown in Fig. above.
So, current in branch BD is “– 0.118 A”.
Current in branch BC is (I_{1 -} I_g) i.e.,  
i.e.,  
Current in branch AD is (I – I₁)
i.e.,   i.e.,    or 0.353 A
Current in branch DC is (I₁ – I₁ + I_g)
i.e.,  or    or  0.235 A

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Given,
Emf of secondary cells, $\mathrm{E} = 2.0 \mathrm{V}$
Internal resistance of cell, r= 0.015 $\mathrm{\Omega }$ 
Number of secondary cells, n=6
External resistance, R = 8.5 $\mathrm{\Omega }$
               
Current drawn from the supply, I = $\boxed{\mathrm{I} = \frac{\mathrm{nE}}{\mathrm{R}+\mathrm{nr}}}$                                               
                                                $$\begin{gathered} = \frac{6 \times 2.0}{8.5 +6 \times 0.015} \\ = 1.4 \mathrm{A} \end{gathered}$$

And, terminal voltage of the supply is, V=IR   
                                                        $$\begin{gathered} = 1.4 \times 8.5 \\ = 11.9 \mathrm{V} \end{gathered}$$