Current in $$AB = I = \frac{e}{{R + r}}$$
potential difference across $$AB = IR = \frac{{eR}}{{R + r}}$$
To obtain balanced point,
$$\frac{{eR}}{{R + r}} > \frac{e}{3}\,\,{\text{or}}\,\,R > \frac{r}{2}$$

Resistance, $$R = \frac{{\rho \ell }}{A}$$
$$R = \rho \frac{\ell }{A} \times \frac{\ell }{\ell } = \frac{{\rho {\ell ^2}}}{V}$$
$$\left[ {\because {\text{Volume}}\left( V \right) = A\ell } \right]$$
Since resistivity and volume remains constant therefore $$\% $$ change in resistance
$$\frac{{\Delta R}}{R} = \frac{{2\Delta \ell }}{\ell } = 2 \times \left( {0.5} \right) = 1\% $$

After adding two conductors, the circuit will acquire the form shown in figure. From symmetry considerations, we calculate that the central conductor will not participate in electric charge transfer.

Therefore, if the initial resistance $${{R_1}}$$ of the circuit was $$5r,$$  where $$r$$ is the resistance of a conductor, the new resistance of the circuit will become
$${R_2} = 2r + \frac{{2r}}{2} = 3r$$
Therefore, $$\frac{{{R_2}}}{{{R_1}}} = \frac{3}{5}$$

Current through $$50\,V$$  and $$30\,V$$  batteries are respectively $$2.5\,A$$  and $$3\,A.$$

$$\eqalign{ & H = {I^2}Rt.{\text{Here}}\,{R_1} = \rho \frac{1}{{\pi {r^2}}}\,\,{\text{and }}{R_2} = \rho \frac{1}{{\pi {{\left( {2r} \right)}^2}}} \cr & {\text{That}}\,{\text{is,}}\,{R_1} = 4{R_2}. \cr} $$
Hence, $$\frac{{{H_1}}}{{{H_2}}} = 4$$

For first case, balanced condition of meter bridge will be
$$\frac{5}{{{l_1}}} = \frac{R}{{\left( {100 - {l_1}} \right)}}\,......\left( {\text{i}} \right)$$
Now, by shunting resistance $$R$$ by an equal resistance $$R,$$ new resistance in that arm become $$\frac{R}{2}.$$
So, new balanced condition will be
$$\frac{5}{{1.6{l_1}}} = \frac{{\frac{R}{2}}}{{\left( {100 - 1.6{l_1}} \right)}}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{ & \frac{{1.6}}{1} = \frac{{\left( {100 - 1.6{l_1}} \right)}}{{100 - {l_1}}} \times 2 \cr & \Rightarrow 160 - 1.6{l_1} = 200 - 3.2{l_1} \cr & 1.6{l_1} = 40 \cr & {l_1} = \frac{{40}}{{1.6}} = 25\,m \cr & \Rightarrow {\text{From}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right),\,\,\frac{5}{{25}} = \frac{R}{{75}} \Rightarrow R = 15\,\Omega \cr} $$

Circumference of circle $$ = 2\pi r = 2 \times \pi \frac{{10}}{{100}}$$
$$ = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$$
Resistance of wire $$ = 12 \times \frac{\pi }{5} = \frac{{12\pi }}{5}$$
Resistance of each section $$ = \frac{{12\pi }}{{10}}\Omega $$
$$\therefore $$ Equivalent resistance $$ = \frac{{\frac{{12\pi }}{{10}} \times \frac{{12\pi }}{{10}}}}{{\frac{{12\pi }}{{10}} + \frac{{12\pi }}{{10}}}}$$
$$ = \frac{{6\pi }}{{10}} = 0.6\,\pi \,\Omega $$

$$\eqalign{ & J = \frac{I}{{2\pi r\ell }} = \frac{{\frac{{dV}}{{dR}}}}{{2\pi r\ell }}\,......\left( {\text{i}} \right) \cr & dR = \rho \frac{{dr}}{{2\pi r\ell }} = \frac{1}{\sigma } \times \frac{{dr}}{{2\pi r\ell }}\,......\left( {{\text{ii}}} \right) \cr & {\text{Now}}\,E = - \frac{{dV}}{{dr}} \cr & \therefore dV = - Edr = - \frac{\lambda }{{2\pi \in r}}dr\,......\left( {{\text{iii}}} \right) \cr} $$

From (i), (ii), and (iii)
$$\eqalign{ & J = \frac{1}{{2\pi r\ell }}\left[ {\frac{{\lambda dr}}{{2\lambda \in r}} \times \frac{{\sigma 2\pi r\ell }}{{dr}}} \right] = \frac{{\lambda \sigma }}{{2\pi \in r}}\,......\left( {{\text{iv}}} \right) \cr & {\text{Also}}\,I = \frac{{dV}}{{dR}} = \frac{{ - \lambda }}{{2\pi \in r}}dr \times \frac{{\sigma \times 2\pi r\ell }}{{dr}} = \frac{{ - \lambda \sigma \ell }}{ \in }\,......\left( {\text{v}} \right) \cr} $$
Here negative sign signifies that the current is decreasing
$${\text{But}}\,I = \frac{{d\left( q \right)}}{{dt}} = \frac{{d\left( {\lambda \ell } \right)}}{{dt}} = \ell \frac{{d\lambda }}{{dt}}\,......\left( {{\text{vi}}} \right)$$
From (v) and (vi)
$$\ell \frac{{d\lambda }}{{dt}} = - \frac{{\lambda \sigma \ell }}{ \in } \Rightarrow \frac{{d\lambda }}{\lambda } = \frac{{ - \sigma }}{{ \in \ell }}dt$$
on integrating
$$\eqalign{ & \int\limits_{{\lambda _0}}^\lambda {\frac{{d\lambda }}{\lambda }} = - \frac{\sigma }{ \in }\int\limits_0^t {dt} \cr & \therefore {\log _e}\frac{\lambda }{{{\lambda _0}}} = - \frac{{\sigma t}}{ \in }\,\,\,\,\,\therefore \lambda = {\lambda _0}{e^{ - \frac{\sigma }{ \in }t}} \cr} $$
Substituting this value in (iv) we get
$$J = \frac{{\sigma {\lambda _0}}}{{2\pi \in r}}{e^{ - \frac{\sigma }{\smallint }t}}$$

Radiated power of a black body,
$$P = \sigma A{T^4}$$
where, $$A =$$  surface area of the body
$$T =$$  temperature of the body
and $$\sigma =$$  Stefan's constant
When radius of the sphere is halved, new area,
$$A' = \frac{A}{4}$$
∴ Power radiated,
$$\eqalign{ & P' = \sigma \left( {\frac{A}{4}} \right){\left( {2T} \right)^4} = \frac{{16}}{4} \cdot \left( {\sigma A{T^4}} \right) \cr & = 4P = 4 \times 450 = 1800\,watts \cr} $$

$$\eqalign{ & {R_t} = {R_0}\left( {1 + \alpha t} \right){\text{at}}\,{t^ \circ }C\,{R_t} = 3{R_0} \cr & \alpha = 4 \times {10^{ - 3}}{/^ \circ }C \cr & 3{R_0} = {R_0}\left( {1 + 4 \times {{10}^{ - 3}} \times t} \right) \cr & \therefore 3 - 1 = 4 \times {10^{ - 3}}t \cr & \therefore t = \frac{2}{{4 \times {{10}^{ - 3}}}} = {500^ \circ }C \cr} $$