Potential across $$PM,$$
$${V_{{\text{PM}}}} = 4 \times 1 = 4\,V$$
Now, equivalent resistance across $$PN$$
$$\eqalign{ & \frac{1}{{{R_{{\text{eq}}}}}} = \frac{1}{{0.5}} + \frac{1}{{0.5}} \cr & {R_{{\text{eq}}}} = \frac{{0.5}}{2} = 0.25\,\Omega \cr} $$
Effective resistance between $$P$$ and $$M$$
$$\eqalign{ & {R_{{\text{PM}}}} = 0.25 + 1 \cr & = 1.25\,\Omega \cr} $$
For balancing condition,
$$\eqalign{ & \frac{{{V_{{\text{MN}}}}}}{{{V_{{\text{PM}}}}}} = \frac{{{R_{{\text{MN}}}}}}{{{R_{{\text{PM}}}}}} \cr & {V_{{\text{MN}}}} = \frac{{{V_{{\text{PM}}}} \times {R_{{\text{MN}}}}}}{{{R_{{\text{PM}}}}}} = \frac{1}{{1.25}} \times 4 \cr & = 3.2\,V \cr} $$

It is found that temperature of inversion $$\left( {{T_i}} \right)$$  is as much above the neutral temperature $$\left( {{T_n}} \right)$$  as neutral temperature is above the temperature of the cold junction $${T_0},$$  i.e.
$$\eqalign{ & {T_i} - {T_n} = {T_n} - {T_0} \cr & {\text{or}}\,\,{T_i} = 2{T_n} - {T_0} \cr} $$
But, here the cold junction is kept at $${0^ \circ }C.$$
Thus, $${T_i} = 2{T_n}\,\,{\text{or}}\,\,{T_n} = \frac{{{T_i}}}{2}$$

Applying Kirchhoff's voltage law in the given loop and going in direction of current $$PSTQ$$   total voltage is equal to zero

So, $$ - 2i + 8 - 4 - 1 \times i - 9i = 0$$
$$\therefore i = \frac{1}{3}A$$
Potential difference across $$PQ = \frac{1}{3} \times 9$$
$$ = 3\,V$$

$${R_{{\text{eq}}}} = \frac{{\left( {R + \frac{R}{2}} \right) \times R}}{{\left( {R + \frac{R}{2} + R} \right)}} = \frac{3}{5}R$$

Temperature coefficient of resistance is defined as the increase in resistance per unit original resistance per degree rise of temperature and is given by
$$\alpha = \frac{{{R_t} - {R_o}}}{{{R_o} \times t}}\,\,{\text{or}}\,\,{R_t} = {R_o}\left( {1 + \alpha t} \right)$$
$${R_t} =$$  Resistance at final temperature
$${R_o} =$$  Resistance at initial temperature
$$t =$$  Change in temperature
Case I
$$5 = {R_o}\left[ {1 + \alpha \left( {50} \right)} \right]\,.......\left( {\text{i}} \right)$$
Case II
$$7 = {R_o}\left[ {1 + \alpha \left( {100} \right)} \right]\,.......\left( {{\text{ii}}} \right)$$
Dividing Eq. (i) by Eq. (ii)
$$\eqalign{ & \frac{5}{7} = \frac{{1 + 50\alpha }}{{1 + 100\alpha }} \cr & \therefore 5 + 500\alpha = 7 + 350\alpha \cr & \therefore \alpha = \frac{2}{{150}} = 0.01{/^ \circ }C \cr} $$

Let the resistor to be connected across $$CD$$  be $$x.$$ Then the equivalent resistance across $$EF$$  should be $$x$$ and also across $$AB$$  should be $$x.$$ So we get
$$\frac{{\left( {2R + x} \right)R}}{{3R + x}} = x$$

solve to get
$$x = \left( {\sqrt 3 - 1} \right)R$$

Copper is a metal whereas Germanium is Semi-conductor.
NOTE : Resistance of metal decreases and semiconductor increases with decrease in temperature.

In the first case $$\frac{X}{Y} = \frac{{20}}{{80}} = \frac{1}{4}$$
In the second case $$\frac{{4X}}{Y} = \frac{\ell }{{100 - \ell }} \Rightarrow \ell = 50$$

The given circuit can be redrawn as,

as $$4\,\Omega $$  and $$x\,\Omega $$  are parallel
$$\eqalign{ & x' = \frac{1}{4} + \frac{1}{x} = \frac{{\left( {4 + x} \right)}}{{4x}} \cr & x' = \frac{{4x}}{{4 + x}} \cr} $$
$$\& 1\,\Omega $$  and $$1\,\Omega $$  are also parallel $$x'' = 2\,\Omega $$
Now equivalent resistance of circuit
$$\eqalign{ & x = \frac{{4x}}{{4 + x}} + 2 = \frac{{8 + 6x}}{{4 + x}} \cr & 4x + {x^2} = 8 + 6x \cr & \Rightarrow {x^2} - 2x - 8 = 0 \cr & x = \frac{{2 \pm \sqrt {4 - 4\left( 1 \right)\left( { - 8} \right)} }}{2} \cr & = \frac{{2 \pm \sqrt {36} }}{2} = 4\Omega \cr} $$
Reading of Ammeter $${A_1} = \frac{V}{{\left( {R + r} \right)}}$$
$${A_1} = \frac{9}{{4 + 0.5}} = 2\,{\text{Ampere}}$$

$$\eqalign{ & Vit = ms\Delta t \cr & \Rightarrow t = \frac{{ms\Delta t}}{{Vi}} = \frac{{1 \times 4200 \times 80}}{{220 \times 4}} = 381.82\,s \cr & = 6.3\,\min \cr} $$