(i) Current is induced in the anticlockwise direction in the upper coil. Then only as per Lenz's law, the induced current will oppose the motion of the magnet.
(ii) In the lower coil, current is induced in the clockwise direction. As per Lenz's law the direction of induced emf will in opposite direction of motion of the magnet.

Magnetic field B = 1 T
Mass of the proton, m_p = 1.67 x 10⁻²⁷kg
Charge of the proton, q = 1.60 x 10⁻¹⁹C
\(f=\frac { qB }{ { { 2\pi m }_{ p } } } =\frac { \left( 1.60\times { 10 }^{ -19 } \right) \left( 1 \right) }{ 2\left( 3.14 \right) \left( 1.67\times { 10 }^{ -27 } \right) } \)
= 15.3 x 10⁶ Hz = 15.3 MHz

In the given problem, since the potential depends only on x, we can use \(\vec { E } =\frac { dV }{ dx } \hat { i } \) (the other two terms \(\frac { \eth V }{ \eth y } \) and \(\frac { \eth V }{ \eth z } \) are zero)
From 0 to 1 cm, the slope is constant and so \(\frac { dV }{ dx } \) = 25V cm^-1, So \(\vec { E } \)= -25V cm^-1\(\hat { i } \)
From 1 to 4 cm, the potential is constant
V = 25 V. It implies that \(\frac { dV }{ dx } \) = 0, So \(\vec { E } \) = 0
From 4 to 5 cm, the slope \(\frac { dV }{ dx } \) = -25 V cm^-1
So \(\vec { E } \) = + 25Vcm^-1\(\hat { i } \)
The plot of electric field for the various points along the x-axis is given below

f_o = 125 cm; f_e = 2 cm; m = ? L = ?; θ_i = ?
(a) Equation for magnification of telescope,
\(m=\cfrac { { f }_{ o } }{ { f }_{ e } } \) 
Substituting, \(m=\cfrac { 125 }{ 2 } =62.5\) 
(b) Equation for approximate length of telescope, L = f_o+ f_e
Substituting, L = 125 + 2 = 127 cm = 1.27 m
(c) Equation for angular magnification,\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \) 
Rewriting, \({ \theta }_{ 1 }=m\times { \theta }_{ 0 }\) 
Substituting,
\({ \theta }_{ i }=62.5\times 1'=62.5'=\cfrac { 62.5 }{ 60 } =1.04^{ o }\) = 1^o2'30''

\(\beta =50 \)
\(V_{\beta E} =600 \mathrm{mV} \)
\(=0.6 \mathrm{~V} \)
\(\mathrm{V}_{\mathrm{B}} =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{EB}} \)
\(\mathrm{V}_{\mathrm{B}} =3-0.6 \)
\(=2.4 \mathrm{~V} \)
\(\mathrm{I}_{\mathrm{B}} =\frac{\mathrm{V}_{\mathrm{B}}}{R_B}=\frac{2.4}{60 \times 10^3}=40 \mu \mathrm{A} \)
\(\mathrm{I}_{\mathrm{C}} =\beta \mathrm{I}_{\mathrm{B}}=50 \times 40 \mu \mathrm{A} =2 \mathrm{~mA} \)
\(V_C=R_FI_C=500 \times 2 \times10^{-3}=1 V\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=3-1=2V\)

Power of the pulse P = 60 x 10^-3 W
Time internal t = 10^-6 S
Energy U = Power x time
U = p x t
= 60 x 10^-3 x 10^-6
U = 60 x 10^-9 J
Lineral momentum, \(\mathrm{P}=\frac{\text { Energy }}{\text { speed }}=\frac{U}{C} \ \)
\(\mathrm{P}=\frac{60 \times 10^{-9}}{3 \times 10^{8}} \)
P = 20 x 10^-17 kg ms^-1

L = 40 x 10^-3 H; i = 0.3 sin (200t – 40^o)
X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90^o Therefore,
v = V_msin (ωt + 90^o)
v = 2.4sin (200t - 40^o + 90^o)
v = 2.4sin (200t + 50^o)volt

(a) The capacitance of the capacitor in the presence of dielectric is
C = \(\frac { { \varepsilon }_{ r }{ \varepsilon }_{ 0 }A }{ d } =\frac { 5\times 8.85\times 10^{ -12 }\times 6 \times 10^{-4}}{ 6\times 10^{ -3 } } \)
= 44.25 x 10^-13F = 4.425 pF
The stored charge is
Q = CV = 44.25 x 10^-13 x 10
= 442.5 x 10^-13C = 44.25pC
The stored energy is
\(U=\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 44.25 x 10^-13 x 100
= 2.21 x 10^-10 J
(b) After the removal of the dielectric, since the battery is already disconnected the total charge will not change. But the potential difference between the plates increases. As a result, the capacitance is decreased.
New capacitance is
C₀=\(\frac { C }{ { \varepsilon }_{ r } } =\frac { 44.25\times 10^{ -12 } }{ 5 } \)
= 0.885 x 10^-12 F = 0.885 pF
The stored charge remains same and 44.25 pC. Hence newly stored energy is
U₀ =\(\frac { { Q }^{ 2 } }{ 2{ C }_{ 0 } } =\frac { { Q }^{ 2 }{ \varepsilon }_{ r } }{ 2C } =\varepsilon _{ r }U\)
= 5 x 2.21 x 10^-10J = 11.05 x 10^-10J
The increased energy is ΔU = (11.05 - 2.21) x 10^-10 J  = 8.84 x 10^-10 J
When the dielectric is removed, it experiences an inward pulling force due to the plates. To remove the dielectric, an external agency has to do work on the dielectric which is stored as additional energy. This is the source for the extra energy 8.84 x 10^–10 J.

\(\frac { P }{ Q } =\frac { R }{ S } \)
\(S=\frac { Q }{ P } \times R\)
\(S=\frac { 1000 }{ 100 } \times 40S=400\Omega \)

Thus applying Kirchoff’s second law to the closed loop EACE
I₁R₁ + I₂R₂ + I₃R₃ = ξ
and for the closed loop ABCA
I₄R₄ + I₅R₅ - I₂R₂ = 0

For the fish, the distance of the bird from the free surface is \(\cfrac { 4h }{ 3 } \) or 1.33 h, Distance of fish from the free surface = 0.5 h. So, the required distance is 1.83 h.

Again, effective distance of bird from the mirror is NH + h or 2.33 h. So, the distance of the bird's image from the mirror is 2.33 h. Distance from fish is 2.33 h + 0.5 h i.e. 2.83 h.
(b) For the bird, the distance of fish from the free surface is 
\(\cfrac { h }{ 2n } \) or \(\cfrac { h }{ 2\times \cfrac { 4 }{ 3 } } \) or \(\cfrac { 3h }{ 8 } \) or 0.375 h
Distance of fish from bird is 0.375 h + h i.e. 1.375 h.
Again, distance of image of fish from mirror is \(\cfrac { h }{ 2 } \) 
Effective distance (for the bird) of the image of fish is the mirror is
\(\cfrac { 1 }{ n } \left[ \cfrac { 3h }{ 2 } \right] +h.i.e,\cfrac { 3 }{ 4 } \times \cfrac { 3h }{ 2 } +h\) or \(\cfrac { 9h }{ 8 } +h\) or  \(\cfrac { 17h }{ 8 } \)  or 2.125 h.

\(\frac { 1 }{ 2 } { mv }^{ 2 }=hv\quad or\quad { E }_{ k }=\frac { hc }{ \lambda } \)
\(or\quad \frac { { p }^{ 2 } }{ 2m } =\frac { hc }{ \lambda } or\quad p=\sqrt { \frac { 2mhc }{ \lambda } } \)
de Broglie wavelength\(=\frac { h }{ p } =\frac { h }{ \sqrt { \frac { 2mhc }{ \lambda } } } \)
\(\\ =\sqrt { \frac { h\lambda }{ 2mc } } \)

Advantages of robotics:
(i) The robots are much cheaper than humans.
(ii) Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
(iii) Robots are more precise and error free in performing the task.
(iv) Stronger and faster than humans.
(v) Robots can work in extreme environmental conditions: extreme hot or cold, space or underwater. In dangerous situations like bomb detection and bomb deactivation.
(vi) In warfare, robots can save human lives.
(vii) Robots are significantly used in handling materials in chemical industries especially in nuclear plants which can lead to health hazards in humans.
Disadvantages of Robotics:
(i) Robots have no sense of emotions or conscience.
(ii) They lack empathy and hence create an emotionless workplace.
(iii) If ultimately robots would do all the work, and the humans will just sit and monitor them, health hazards will increase rapidly.
(iv) Unemployment problem will increase.
(v) Robots can perform defined tasks and cannot handle unexpected situations.
(vi) The robots are well programmed to do a job and if a small thing goes wrong it ends up in a big loss to the company.
(vii) If a robot malfunctions, it takes time to identify the problem, rectify it, and even reprogram if necessary. This process requires signi cant time.
(viii) Humans cannot be replaced by robots in decision making.
(ix) Till the robot reaches the level of human intelligence, the humans in work place will exit.

Total rest mass (initial) = 236.0526 u
Total rest mass (final) = 235.8293 u
Decrease in rest mass due to fission = 0.2233 u
Energy released
= 0.2233 u x 931 \(\frac{MeV}{u}\) = 207.9 MeV.

Electric field between the capacitor plates
\(E=\frac { \sigma }{ \varepsilon _{ 0 } } =\frac { q }{ \varepsilon _{ 0 }A } \)
Where q is the charge accumulated on the positive plate.
The electric flux through this plate
\({ \phi }_{ E }=EA=\frac { q }{ \varepsilon _{ 0 }A } .A=\frac { q }{ \varepsilon _{ 0 } } \)
∴ Displacement current
\(\\ { I }_{ d }=\varepsilon _{ 0 }.\frac { d\phi }{ dt } =\varepsilon _{ 0 }\frac { d }{ dt } \left[ \frac { q }{ \varepsilon _{ 0 } } \right] =\frac { dq }{ dt } \)
\(\frac { dq }{ dt } \) is the rate at which charge flows to a positive plate through the conducting wire.
I_d = I_c
i.e. displacement current between the capacitor plates = conduction current through the wire.

Given:
We know that e = -\(\frac { d\Phi }{ dt } \)
e = \(\frac{d}{dt}\) (5t³ + 4t² + 2t)
e = 15t² + 8t + 2
for t = 2s, e = 15 x (2)² + 8 (2) + 2
e = 78V

The first two colour bands
for yellow = 4
for Violet = 7
for Brown = 10¹
The value of carbon resistor = 47 x 10 = 470\(\Omega \)
The gold ring showing tolerance of ± 5%
R = (470 ± 5%)\(\Omega \)

Let the currents flowing through wires XY and PQ be I₁ and I₂
Magnetic field along PQ is \(\mathrm{B}_{1}=\frac{\mu_{o} I_{2}}{2 \pi r}\)
Force per unit length on PQ is \(\frac{F_{2}}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)
If the wire XY is slightly displaced and released, it executes simple harmonic motion with the condition that acceleration is directly proportional to the displacement y
\(\therefore a=-\omega^{2} y\)  .....(1) 
The distance between two wires = d
Time period
\(T=\frac{2 \pi}{\omega} \)
\(a=\frac{g}{d} y \) .....(2)
By comparing the equations (1) and (2) we get
\(\omega^{2} =-\frac{g}{d} \ \therefore \omega=\sqrt{\frac{g}{d}} \)
\(\text { Time period } =\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{d}{g}} \)
\(\therefore T =2 \pi \sqrt{\frac{d}{g}} \)

In a right angle triangle OPN let the angle \(\angle\)OPN = \(\theta \)₁ which implies, \({ \varphi }_{ 1 }=\frac { \pi }{ 2 } -{ \theta }_{ 1 }\) and also in a right angle triangle OPM, 
\(\angle\)OPN = \(\theta \)₂ which implies, \({ \varphi }_{ 2 }=\frac { \pi }{ 2 } +{ \theta }_{ 2 }\)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a } \left( cos\left( \frac { \pi }{ 2 } -{ \theta }_{ 1 } \right) -cos\left( \frac { \pi }{ 2 } +{ \theta }_{ 2 } \right) \right) \hat { n } \)
\(=\frac { { \mu }_{ ° }I }{ 4\pi a } (si{ n }_{ 1 }+{ sin }_{ 2 })\hat { n } \)

Charge Q = Current x Time interval
= I x t
At t = 0 s,  
dq = dI x t
= 5 x 0
dq = 0 C
At t = 2 s,    
dg = dI x t
=5 x 2
dq = 10 C
At t = 5 s,
dg = dl x t
=0 x 5
dq = 0 C
At t= 0 s, dg = 0 C; At t = 2 s, dg = 10 C; At t= 5 s, dg = 0 C.