At Null point

$$\frac{X}{{{\ell _1}}} = \frac{{10}}{{{\ell _2}}}$$
$$\eqalign{ & {\ell _1} = 52 + {\text{End}}\,{\text{correction}} = 52 + 1 = 53cm \cr & {\ell _2} = 48 + {\text{End}}\,{\text{correction}} = 48 + 2 = 50cm \cr} $$
$$\therefore \frac{X}{{53}} = \frac{{10}}{{50}} \Rightarrow X = \frac{{53}}{5} = 10.6\,\Omega $$

If potential difference across $$R\,\Omega $$  resistor is equal to potential difference of cell $$B,$$ galvanometer shows no deflection.
Applying Kirchhoff's law, $$500I + 100I = 12$$
So, $$I = \frac{{12 \times {{10}^{ - 2}}}}{6} = 2 \times {10^{ - 2}}A$$
Hence, $${V_B} = 100\left( {2 \times {{10}^{ - 2}}} \right) = 2\,V$$

$$\eqalign{ & {v_d} = \frac{I}{{neA}}\,{\text{Here,}}\,I = 5.4A,n = 8.4 \times {10^{28}},\,{\text{per}}\,{m^3} \cr & A = {10^{ - 6}}{m^2},e = 1.6 \times {10^{ - 19}}C \cr & \therefore {v_d} = \frac{{5.4}}{{8.4 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times {{10}^{ - 6}}}} = 0.4\,mm/s \cr} $$

Between any two terminals, two resistors of two arms are in series i.e. between $$B$$ and $$C,$$ equivalent resistance is
$$\eqalign{ & \frac{1}{{{R_{BC}}}} = \frac{1}{4} + \frac{1}{8} \cr & \frac{1}{{{R_{BC}}}} = \frac{{2 + 1}}{8} \cr & \therefore {R_{BC}} = \frac{8}{3}\Omega \cr} $$

Since, the voltage is same for the two combinations, therefore $$H \propto \frac{1}{R}.$$  Hence, the combination of $$39$$ bulbs will glow more.

Since the opening or closing the switch does not affect the current through $$G,$$ it means that in both the cases there is no current passing through $$S.$$ Thus potential at $$A$$ is equal to potential at $$B$$ and it is the case of balanced wheatstone bridge..
$${I_P} = {I_Q}$$  and $${I_R} = {I_G}$$

Resistance is directly proportional to length
$$\eqalign{ & \frac{1}{{{R_{AB}}}} = \frac{1}{3} + \frac{1}{{4 + 5}} = \frac{{\left( {4 + 5} \right) + 3}}{{\left( 3 \right)\left( {4 + 5} \right)}} \cr & {R_{AB}} = \frac{{3 \times \left( {4 + 5} \right)}}{{3 + \left( {4 + 5} \right)}} = \frac{{27}}{{12}} \cr} $$
Similarly,
$$\eqalign{ & {R_{BC}} = \frac{{4 \times \left( {3 + 5} \right)}}{{4 + \left( {3 + 5} \right)}} = \frac{{32}}{{12}} \cr & {R_{AC}} = \frac{{5 \times \left( {3 + 4} \right)}}{{5 + \left( {3 + 4} \right)}} = \frac{{32}}{{12}} \cr & \therefore {R_{AB}}:{R_{BC}}:{R_{AC}} = 27:32:35 \cr} $$

The effective circuit is shown in figure.
$${R_{AB}} = \frac{{3R}}{8}$$

Motion of conduction electrons due to random collisions has no preferred direction and average to zero. Drift velocity is caused due to motion of conduction electrons due to applied electric field $$\vec E.$$

We know that $$R = \rho \frac{l}{a}$$  
Where $$l$$ is the length of the conductor through which the current flows and a is the area of cross section.
$$\eqalign{ & {\text{Here}}\,l = L\,{\text{and}}\,a = L \times t \cr & \therefore R = \frac{{\rho L}}{{L \times t}} = \frac{\rho }{t} \cr} $$
$$\therefore $$ $$R$$ is independent of $$L$$