Magnetic field induction at a point on the axis at distance x from the centre of the circular coil carrying current is
\({ B }_{ 1 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { { 2\pi nIa }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } \)
Magnetic field induction at the centre of the circular coil carrying current is
\({ B }_{ 2 }=\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { 2\pi nI }{ a } \)
But as per question,\({ B }_{ 1 }=\cfrac { { B }_{ 2 } }{ 8 } \)
\(\Rightarrow \ \cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { { 2\pi nIa }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } =\cfrac { { \mu }_{ 0 } }{ 4\pi } .\cfrac { 2\pi nl }{ a } X\frac { 1 }{ 8 } \)
\(\Rightarrow \ \cfrac { { a }^{ 2 } }{ ({ a }^{ 2 }+{ x }^{ 2 })^{ 3/2 } } =\cfrac { 1 }{ 8a } \quad \Rightarrow { 8 }a^{ 3 }=({ a }^{ 2 }+{ x }^{ 3 })^{ 3/2 }\)
\(\Rightarrow \)  2a = (a² + x²)^1/2
\(\Rightarrow \)  4a² = a²+x²  \(\Rightarrow x=\sqrt { 3a }\)

Here r = 0.05 m d = 1 mm = 10^-3m
Capacitance \(C={\epsilon_oA\over d}={\epsilon_o\pi r^2\over d}\)
= \(8.85\times 10^{-12}\times{22\over 7}\times {(0.05)^2\over 10^{-3}}=0.69\times 10^{-10}\) 

Induced emf  = (1/2)\(\omega\)BR²
= (1/2) \(\times\)4\(\pi\)\(\times\)0.4\(\times\)10^-4 \(\times\)(0.5)²
= 6.28 \(\times\)10^-5 V
The number of spokes is immaterial because the emf’s across the spokes are in parallel.

Clearly , the fish can see the outside view of the cone with semi vertical angle
But \(\mu \) = 1.sin i_c
or 1/3 = 1/ sin i_c
or sin i_c = 3/4 = 0.75 
\(\theta\)/2  =i_c = sin^-1 (0.75 ) = 48.6⁰

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

Consider a plane perpendicular to the direction of propagation of the wave. An electric charge, on the plane, will be set in motion by the electric and magnetic fields of em wave, incident on this plane. This illustrates that em waves carry energy and momentum.

A superconducting material and nitrogen are diamagnetic in nature. when a ball of superconducting material is dipped in liquid nitrogen, it behaves as a diamagnetic material. When it is placed near a bar magnet, it will be feebly magnetized opposite to the direction of the magnetising field. Due to it,
(i) it will be repelled, i.e., moving away from the magnet
(ii) its direction of magnetic field of the magnet. 

Range of frequency used for TV transmission is 54 MHz to 890 MHz (VHF and UHF).These waves and light waves are electromagnetic waves.The ionosphere is unable to reflect back these waves to earth.

Welcome bridge is said to be balanced, when no current flows through the galvanometer arm of Wheatstone bridge. In balanced bridge,
\(\frac{P}{Q}=\frac{R}{S}\)

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 \(\phi =\oint { \overrightarrow { E } .\overrightarrow { ds } = } q/\epsilon _{ o }\)

Energy is not lost. It gets transferred from regions of destructive interference to the regions of constructive interference.

Because in n-type semiconductor, electrons (having negative charge) are majority carriers which are responsible for the conduction

The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R₁ = 220 mm
Hence, focal length of the objective mirror,  \({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =110\)
Radius of curvature of the secondary mirror, R₁ = 140 mm
Hence, focal length of the secondary mirror, \({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } \) = 70 mm
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, u = f₁ - d
= 110 - 20
= 90 mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \)
\(\frac { 1 }{ 70 } -\frac { 1 }{ 90 } =\frac { 9-7 }{ 630 } =\frac { 2 }{ 630 } \)
∴ v = \(\frac{630}{2}\) = 315 mm
Hence, the final image will be formed 315 mm away from the secondary mirror.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

(a) \(\overrightarrow{F}=I(\overrightarrow{l} \times \overrightarrow{B})\), where \(\overrightarrow{l}\) is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. According to Fleming's left hand rule,\(\overrightarrow{B}\) must act horizontally in a direction perpendicular to the wire carrying current.
(b) Adaptation to different situations and flexible and adjustable attitude.
(c) Avoiding unnecessary arguments in conflicting situations in everyday life.

(i) Here,     \(\Delta m=1\quad a.m.u.=1.66\times { 10 }^{ -27 }kg\)
\(E=\left( \Delta m \right) { c }^{ 2 }=1.66\times { 10 }^{ -27 }{ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }=1.49\times { 10 }^{ -10 }J\)
(ii) Yes, in the phenomenon of pair production. Under suitable conditions, a photon materialises into an electron and a position :   \(\gamma ={ e }^{ -1 }+{ e }^{ +1 }\)
(iii) Einstein's relation, \(E=\left( \Delta m \right) { c }^{ 2 }\) emphasis that when certain mass disappears, an equivalent amount of energy appears. The reverse is also true. It Implies that to gain something, you have to lose another in equivalent amount. No one can have all gains together or all losses together. It also implies that nothing come for free. You have to pay the price in one form and acquire something in the desired form.

Since the electron has been localised within the nucleus (of size about \({ 10 }^{ -14 }\) m)of an atom, so \({ \lambda =10 }^{ -14 }\)
Momentum of electron,
  \(p=\frac { h }{ \lambda } =\frac { 6.63\times { 10 }^{ -34 } }{ { 10 }^{ -14 } } =6.63\times { 10 }^{ -20 }kg{ ms }^{ -1 }\)
The relativistic relation for the energy of an electron is
\(E=\sqrt { { p }^{ 2 }{ c }^{ 2 }+{ { m }_{ 0 } }^{ 2 }{ c }^{ 4 } } \)
Neglecting the rest-mass energy term(i.e., second term), we have
 \(E=pc=(6.63\times { 10 }^{ -20 })\times (3\times { 10 }^{ 8 })J\)
\( =\frac { 6.63\times 3\times { 10 }^{ -12 } }{ 1.6\times { 10 }^{ -13 } } MeV=124.3MeV\)
This energy E(=124.3MeV) is very large as compared to the binding energy which is provided by Coulomb's force within the nucleus. Since the energy of electron inside the nucleus, hence the electron does not reside the nucleus.