Total target power = 200000 = \(2\times { 10 }^{ 5 }MW\) 
Total nuclear power = 10% of total target power
\(=\frac { 10 }{ 100 } \times 2\times { 10 }^{ 5 }=2\times { 10 }^{ 4 }MW\) 
Energy produced/fission = 200 MeV
Efficiency of power plant = 25%
Energy converted into electrical energy per fission
\(=\frac { 25 }{ 100 } \times 200=50MeV\)   
\( =50\times 1.6\times { 10 }^{ -13 }J\)
Total electrical energy to be produced per year
\(=2\times { 10 }^{ 4 }MW=2\times { 10 }^{ 4 }\times { 10 }^{ 6 }W=2\times { 10 }^{ 10 }W\) 
\( =2\times { 10 }^{ 10 }J/s=2\times { 10 }^{ 10 }\times 60\times 60\times 24\times 365J/yr\)
Number of fission in one year,
\(n=\frac { 2\times { 10 }^{ 10 }\times 60\times 60\times 24\times 365 }{ 50\times { 1.6\times 10 }^{ -13 } }\) 
\( n=\frac { 2\times 36\times 24\times 365 }{ 8 } \times { 10 }^{ 24 }\)
Mass of \(6.023\times { 10 }^{ 23 }\) atoms of \(^{ 235 }{ U }=235g\) 
\(=235\times { 10 }^{ -3 }kg\) 
Mass of \(_{ 92 }^{ 235 }{ U }\) required to produce
\(=\frac { 2\times 36\times 24\times 365 }{ 8 } \times { 10 }^{ 24 }\) atoms
\(=\frac { 235\times { 10 }^{ -3 }\times 2\times 36\times 24\times 365\times { 10 }^{ 24 } }{ 6.023\times { 10 }^{ 23 }\times 8 } \) 
\(=3.08\times { 10 }^{ 4 }kg\) 
Thus, the mass of uranium needed per year is \(=3.08\times { 10 }^{ 4 }kg\)

Given, the equation of carrier wave,
 c(t) = 1 sin (8\(\pi \) t)        ..........(i)
(i) According to the figure,
Amplitude of modulating signal,
A_m = 1 V
Amplitude of carrier wave,
A_C =2
T_m = 1 s
From Eq.(i), we get
 \({ \omega }_{ m }=\frac { 2\pi }{ { T }_{ m } } =\frac { 2\pi }{ 1 } =2\pi \ rad/s\)    ....(ii)
 c(t) = 2 sin (8 \(\pi \) t)
So,  \({ \omega }_{ c }=4{ \omega }m_{ }\)
From Eq. (ii) , we get
So,   \({ \omega }_{ c }=4{ \omega }_{ m }\)
Amplitude of modulated wave,
A = A_m + A_c
 = 1 + 2= 3 V
The sketch of the amplitude modulated waveform is shown below:

For carrier signal, \({ \omega }\) = 8 \(\pi \) 
\(T=\frac { 2\pi }{ \omega } =\frac { 2\pi }{ 8\pi } =\frac { 1 }{ 4 } =0.25s\)
(ii) Modulation index, \(m=\frac{A_{m}}{A_{c}}=\frac{1}{2}=0.5\) 

The magnifying power of a telescope is equal to the ratio of the visual angle substended at the eye by final image formed at least distance of distinct vision to the visual angle subtended at naked eye by the object at infinity.

(i) Average power delivered by an AC circuit is
\({ P }_{ av }={ V }_{ rms }{ I }_{ rms }cos\phi \)
where is minimum, the power delivered is minimum and hence, power dissipated will be maximum for the circuit.

A metre bridge with resistors X and Y is represented in the given figure.

a) Balance point from end A, l₁ = 39.5 cm
Resistance of the resistor Y = 12.5 Ω
Condition for the balance is given as,
\(\frac{X}{Y}=\frac{100-l_{1}}{l_{1}}\)
\(x=\frac{100-39.5}{39.5} \times 12.5=8.2 \Omega\)
Therefore, the resistance of resistor X is 8.2 Ω.
The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to minimize the resistance, which is not taken into consideration in the bridge formula.
(b) If X and Y are interchanged, then l₁ and 100 - l₁ get interchanged.
The balance point of the bridge will be 100 - l₁ from A.
100 - l₁ = 100 - 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer will show no deflection. Hence, no current would flow through the galvanometer.

(i) Electrical energy is stored in the capacitor.It is stored in the dielectric.
(ii) No effect. When the metal sheet is placed in the middle the new arrangement is equivalent to a old combination of two capacitors each of plate separation \(\frac{d}{2}\) and hence capacitance 2C.
\({ C }_{ S }=\frac { 2C\times 2C }{ 2C+2C } =C\)
(iii) Team work, concern, respect to teacher and responsibility.

(i) Niyaz shows the values of courage to tell truth and determination.
(ii) It is based on the principle when a current carrying coil placed in external magnetic field, it develops torque.
(iii) Sensitivity of galvanometer can be increased by 
(a) increasing number of turns in the coil and
(b) by increasing current in the coil.

(i) Priyanshu's father is responsible, makes his child understood the concepts of solenoid and toroid.
(ii) A solenoid has magnetic field straight within the turns and in toroid, it is in form of concentric circles.
(iii) The magnitude of magnetic field in solenoid is given by \(B={ { \mu } }_{ \circ }nI\)
(where, n = number of turns/length, I = current in coil)

Current sensitivity, \({ I }_{ s }=\frac { NAB }{ k } \) and
Voltage sensitivity, \(V_{ s }=\frac { NAB }{ kR } \)
Since, the resistance of the coil may vary, it implies an increase in current sensitivity may not necessarily increase voltage sensitivity.
Thus, the trajectory of both the particles will be same.

(i)When As is implanted in Si-crystal, n-type wafer is created. The no of majority carrier electrons due to doping of As is
\(n_{ e }={ N }_{ D }=\frac { 1 }{ 10^{ 6 } } \times5\times10^{ 28 }=5\times10^{ 22 }/m^{ -3 }\)
No of minority carriers (holes) in \(n-\)type wafer is 
\({ n }_{ h }=\frac { { n }_{ 1 }^{ 2 } }{ n_{ e } } =\frac { (1.5\times10^{ 16 })^{ 2 } }{ 5\times10^{ 22 } } =0.45\times10^{ 10 }/m^{ 3 }\)
when B is implanted in Si-crystal p-type wafer is created with no of holes
\({ n }_{ h }=N_{ A }=\frac { 200 }{ 10^{ 6 } } \times(5\times10^{ 28 })=1\times10^{ 25 }/m^{ 3 }\)
Minority carriers (electrons) created in the p-type wafer is
\({ n }_{ e }=\frac { { n }_{ 1 }^{ 2 } }{ n_{ h } } =\frac { (1.5\times10^{ 16 })^{ 2 } }{ 1\times10^{ 25 } } =2.25\times10^{ 10 }/m^{ 3 }\)
When \(p-n\)a junction is reverse biased, the minority carrier holes of \(n-\)region wafer \((n_{ h }=0.45\times10^{ 10 }/m^{ 3 })\) would contributte  more  to the reverse saturation current than minority carrier than minority carrier electrons \((n_{ e }=2.25\times10^{ 7 }/m^{ 3 })\)