Resistance of a conductor of length I and radius r is given by
\(\mathrm{R}=\rho \frac{1}{\pi r^2}\)

\({ N }_{ 1 }.2\pi R={ N }_{ 2 }.2\pi (R/2)\)
  \(\therefore \ { N }_{ 2 }=2{ N }_{ 1 }\)
Magnetic moment of a coil, M =NAI
For the coil of radius 'R'
  \({ M }_{ 1 }={ N }_{ 1 }{ IA }_{ 1 }={ N }_{ 1 }I\pi { R }^{ 2 }\)
For the coil of radius R/2
\({ M }_{ 2 }={ N }_{ 2 }{ IA }_{ 2 }=2{ N }_{ 1 }I\pi { R }^{ 2 }/4={ N }_{ 1 }.\pi { R }^{ 2 }/2\)
 \({ M }_{ 1 }:{ M }_{ 2 }=1:2\) 

From the first balance point, we get
\(\frac{R}{S}=\frac{33.7}{66.3}\)
After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S_eq, where
\(S_{e q}=\frac{12 S}{S+12}\)
and hence the new balance condition now gives
\(\frac{51.9}{48.1}=\frac{R}{S_{e q}}=\frac{R(S+12)}{12 S}\)
Substituting the value of R/S from we get
\(\frac{51.9}{48.1}=\frac{S+12}{12} \cdot \frac{33.7}{66.3}\)
which gives S = 13.5Ω. Using the value of R/S above, we get R = 6.86 Ω.

As, \({ \mu }_{ r }=1+{ \chi }_{ m }=0,\) for superconductors,
\(\therefore { \ \chi }_{ m }-1\)

The surface integral of electric field intensity over a closed surface in free space is \(1\epsilon_o\) times the total charge q enclosed by the surface \(\phi =\oint { \overrightarrow { E } .\overrightarrow { ds } = } q/\epsilon _{ o }\)

\(1.88\times {{10}^{5}}\)Nm²C^-1

Here, \(C_1=25\mu F, C_2=100\mu , C_s=20\mu F\)
\(q=C_sV_s=2400\mu C\) on each capacitor
\(C_p=125\mu F, V_p={2400+2400\over 125}=38.4V\)
Loss of energy = \({1\over 2}C_sV_s^2-{1\over 2}C_pV_p^2\) 

Let \(\sigma\) be the surface charge density of capacitor plates.
for mica \(E_1={\sigma\over K_1\epsilon_0}\)
and for fibre \(E_2={\sigma\over K_2\epsilon_o}\)  or \({E_1\over E_2}={K_2\over K_1}\)
As \(E_s=6.4\times 10^6V/m\)
\(E_1={K_2\over K_1}\times E_2={2.5\over 8}\times 6.4\times 10^6=2\times 10^6V/m\)
Maximum voltage or capacitor
\(V=E_1d_1+E_2d_2=2\times 10^3+3.2\times 10^3=5200V\)