Given: \(q_{1}=4 \mu \mathrm{C}=4 \times 10^{-6} \mathrm{C}\)
\(q_{2}=-6 \mu \mathrm{C}=-6 \times 10^{-6} \mathrm{C}, r=20 \mathrm{~cm}\)
Let the electric potential be zero at a point P, a distance x (in cm) from q₁ Then
\(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{x}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{(r-x)}=0\)
\( \therefore \frac{4}{x} =-\frac{(-6)}{(20-x)} \)
\(\Rightarrow 4(20-x=6 x \)
\(x =8 \mathrm{~cm}\)
i.e. 8 cm from 4 μC charge.

(a) Team work, concern, respect to teacher and responsibility.
(b) Electrical energy in the dielectric of the capacitor

Initially, if we consider a charged capacitor, then its charge would be

and energy stored is
U₁ = \(\frac { 1 }{ 2 } { CV }^{ 2 }\)  ...........(i)
Then, this charged capacitor is connected to unchanged capacitor.
Let the common potential be V₁. The charge flows from first capacitor to the other capacitor unless both the capacitors attain common potential

Q₁ = CV₁ and Q = CV₂
Applying conservation of charge,
Q = Q₁ + Q₂
\(\Rightarrow\) CV = CV₁ + CV₂
\(\Rightarrow\) V = V₁+ V₂
\(\Rightarrow \ V_{1}=\frac{V}{2}\)
Total energy stored, \(U_{2}=\frac{1}{2} C V_{1}^{2}+\frac{1}{2} C V_{2}^{2}\)
\(=\frac{1}{2} C\left(\frac{V}{2}\right)^{2}+\frac{1}{2} C\left(\frac{V}{2}\right)^{2} \Rightarrow U_{2}=\frac{1}{4} C V^{2}\)    ...........(ii)
From Eqs. (i) and (ii), we get
U₂ < U₁
Hence, energy stored in the combination is less than that stored initially in single capacitor.

As \(\tau =nIBA \ sin \ \theta ,\ therefore,\ (i) \ \tau =0, \ when \ sin\theta =0 \ or \ \theta =0°, \ i.e.\)when the plane of coil is perpendicular to the direction of magnetic field. (ii) \(\tau =\) maximum, when \(sin \ \theta \)=maximum=1 or \(\theta =90°\)
\({ \tau }_{ max }=nIBA\times 1=nIBA\)
It will be so when the plane of coil is parallel to the direction of magnetic field.
(ii) Here, n = 200; r = 0.05m; I = 2.0 A; B = 0.20 T; \(\theta =90°-60°\)=\(30°\)
\( \tau =nIBA \ sin \ \theta ,\ therefore,\ (i) \ \tau =0, \ when \ sin\theta =0 \ or \ \theta =30°-60°, \ i.e.,\)
\( \\ { \tau }_{ max }=nIBA\times 1=nIBA\)
\( \tau =nIBA \ sin \ \theta =nIB({ \pi r }^{ 2 })sin\theta =200\times 2.0\times 0.20\left[ (22/7)\times { \left( 0.05 \right) }^{ 2 } \right] \times sin30°\)
\(=0.314\quad N-m=0.31\ Nm\)
(iii) From the above study, we find that when potential energy of the coil is minimum, both force and torque acting on the coil are zero. The same is true in real life. a person who is humble and boasts of nothing, would be a happy person, with no pulls and pressure of life.

\(Here, \ L=80mH=80\times { 10 }^{ -3 }H, \ C=50\times { 10 }^{ -6 }F, \ R=60\Omega \ { E }_{ v }=220V, \ v=?\)
At resource, \( \ { X }_{ L }={ X }_{ C }\)
\( i.e., \ \ \omega L=\frac { 1 }{ \omega C } , \ \omega =\frac { 1 }{ \sqrt { LC } }\)
\(\omega =\frac { 1 }{ \sqrt { LC } } =\frac { 1 }{ \sqrt { 80\times { 10 }^{ -3 }\times 50\times { 10 }^{ -6 } } } = \ 500 \ rad/s\)
\(v=\frac { \omega }{ 2\pi } =\frac { 500 }{ 2\times 3.14 } =79.6Hz\)
\(\\ Q=\frac { { \omega }_{ r }L }{ R } =\frac { 500\times 80\times { 10 }^{ -3 } }{ 60 } =0.67\)

Number of Uranium atoms fissioned
\(n=\frac { total \ energy \ released }{ energy \ released/fission } \)
\( =\frac { 7.6\times { 10 }^{ 13 } }{ 200\times 1.6\times { 10 }^{ -13 } } =2.375\times { 10 }^{ 24 }\)
\( =\frac { 7.6\times { 10 }^{ 13 } }{ 200\times 1.6\times { 10 }^{ -13 } } =2.375\times { 10 }^{ 24 }\)
\( Mass \ of \ Uranium=\frac { Mass \ number }{ Avogadro's \ number } \times n\)
\(=\frac { 235\times 2.375\times { 10 }^{ 24 } }{ 6.023\times { 10 }^{ 23 } } =926.66g\)

Here, h = 125 m ;
R = 6400 km = 6.4 x 10⁶ m.
Area covered = \(\pi { d }^{ 2 }=\pi \times 2hR\ \left[ \because \ \sqrt { 2hR } \right] \)
= 3.14 x 2 x 125 x 6.4 x 10⁶
= 5024 x 10⁶ m² = 5024 km²

Here, \(A=2°,\delta =1°,\mu =?\)
As prism is thin, therefore, \(\delta =(\mu -1)A\)
\(1=(\mu -1)2,\quad \mu =1+\frac { 1 }{ 2 } =\frac { 3 }{ 2 } =1.5\)