The electric fuse is a device which is used to limit the current in an electric circuit. Thus, the use of fuse safeguards the circuit and the appliances connected in the circuit from being damaged. It is always connected with the live (or phase) wire. The fuse wire is a short piece of wire made of a material of high resistance and low melting point so that it may easily melt due to overheating when excessive current passes through it.

According to question, emf of the cell is directly proportional to the balancing length i.e.
$$E \propto l\,......\left( {\text{i}} \right)$$
Now, in the first case, cells are connected in series to support one another i.e.
Net $$emf = {E_1} + {E_2}$$
From Eq. (i), $${E_1} + {E_2} = 50\,cm\left( {{\text{given}}} \right)\,......\left( {{\text{ii}}} \right)$$
Again cells are connected in series in opposite direction i.e.
$${\text{Net}}\,emf = {E_1} - {E_2}$$
From Eq. (i), $${E_1} - {E_2} = 10\,......\left( {{\text{iii}}} \right)$$
From Eqs. (ii) and (iii)
$$\frac{{{E_1} + {E_2}}}{{{E_1} - {E_2}}} = \frac{{50}}{{10}} \Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{5 + 1}}{{5 - 1}} = \frac{6}{4} = \frac{3}{2}$$

Resistance of wire $$\left( R \right) = \rho \frac{l}{A}$$
If wire is bent in the middle then
$$l' = \frac{l}{A},A' = 2A$$
∴ New, $$R' = \rho \frac{{l'}}{{A'}} = \frac{{\rho \frac{l}{A}}}{{2A}} = \frac{{\rho l}}{{4A}} = \frac{R}{4}.$$

Current from $$D$$ to $$C = 1\,A$$
$$\eqalign{ & \therefore {V_D} - {V_C} = 2 \times 1 = 2V \cr & {V_A} = 0 \cr & \therefore {V_C} = 1V, \cr & \therefore {V_D} - {V_C} = 2 \cr & \Rightarrow {V_D} - 1 = 2 \cr & \therefore {V_D} = 3V \cr & \therefore {V_D} - {V_B} = 2 \cr & \therefore 3 - {V_B} = 2 \cr & \therefore {V_B} = 1V \cr} $$

Power, $$P = \frac{{\Delta U}}{{\Delta t}} = V\frac{{\Delta q}}{{\Delta t}} = Vi$$
$${\text{or}}\,\,P = Vi = \frac{{{V^2}}}{R}\,\,\left( {\because V = iR} \right)$$
When resistors are in series, then
$$\eqalign{ & {R_1} = R + R + R \cr & = 3R \cr} $$
$$\therefore $$ Power dissipated,
$${P_1} = \frac{{{V^2}}}{{{R_1}}} = \frac{{{V^2}}}{{3R}}\,......\left( {\text{i}} \right)$$
When resistors are in parallel, then
$$\eqalign{ & \frac{1}{{{R_2}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} \cr & = \frac{3}{R} \cr & \Rightarrow {R_2} = \frac{R}{3} \cr & \therefore {P_2} = \frac{{{V^2}}}{{{R_2}}} = \frac{{{V^2}}}{{\frac{R}{3}}} = \frac{{3{V^2}}}{R}\,.......\left( {{\text{ii}}} \right) \cr} $$
By taking ratio of Eqs. (i) and (ii)
$$\eqalign{ & \frac{{{P_2}}}{{{P_1}}} = \frac{{3{V^2}}}{R} \times \frac{{3R}}{{{V^2}}} \cr & = 9 \cr & {P_2} = 9{P_1} \cr & = 9 \times 10 \cr & = 90\,W \cr} $$

The mass liberated $$m,$$ electrochemical equivalent of a metal $$Z,$$ are related as
$$\eqalign{ & m = Zit \cr & \Rightarrow m = 3.3 \times {10^{ - 7}} \times 3 \times 2 = 19.8 \times {10^{ - 7}}kg \cr} $$

Using colour code we have
$$\eqalign{ & R = 27 \times {10^3}\Omega \pm 10\% \cr & = 27\,k\Omega \pm 10\% \cr} $$

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

For various combinations equivalent resistance is maximum between $$P$$ and $$Q.$$