Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T₂ = 100°C
Resistance of the silver wire at T₂, R₂ = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as 
\(\alpha=\frac{R_{2}-R_{1}}{R_{2}\left(T_{2}-T_{1}\right)}\)
\(=\frac{2.7-2.1}{2.1(100-27.5)}=0.0039^{\circ} \mathrm{C}^{-1}\)
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.

Not necessarily, The basic requirement is that the total electric flux through the area enclosed by the loop should vary with time. The flux change may arise from any portion of the area.
Elsewhere E or dE/dt may be zero. In particular. there need be no electric field at points which make the loop.

Here, M = 2.5 Am², m = 6.6 g = 6.6 x 10^-3 kg
\(\rho =7.9\times { 10 }^{ 3 }kg/{ m }^{ 3 },\) 
\(V=\frac { m }{ \rho } =\frac { 6.6\times { 10 }^{ -3 } }{ 7.9\times { 10 }^{ 3 } } =0.835\times { 10 }^{ -6 }{ m }^{ 3 }\) 
\(I=\frac { M }{ V } =\frac { 2.5 }{ 0.835\times { 10 }^{ -6 } } =3.0\times { 10 }^{ 6 }{ Am }^{ -1 }\)

(i) Magnetic flux versus current

(ii) 

Alternatively 

When Iis increasing at constant rate
(iii) Magnetic energy stored

Since magnetic field is brought to zero in time \(\triangle t\), the magnetic flux also reduces from maximum to zero and hence induced emf is produced in the ring.
Induced emf  \(=E\left( 2\pi b \right) \)   ........(i)       \(\left[ \therefore \ V=Ed \right] \)
E is electric field generated around the ring.
Also induced emf = rate of change of magnetic field \(\times\) area
\(=\frac { B\pi { r }^{ 2 } }{ \triangle t } \)               .......(ii)
From eqn. (i) and (ii), we have
\(2\pi bE=\frac { B\pi { a }^{ 2 } }{ 2\triangle t } \)
or  \(bE=\frac { B{ a }^{ 2 } }{ 2\triangle t } \)    .......(iii)
Torque acting on the ring
\(\tau =b\times force=bQE\)
Using eqn. (iii), we get
\(\tau =\frac { QB{ a }^{ 2 } }{ 2\left( \Delta t \right) } \)
If \(\Delta L\) is the change in angular momentum, then
\(\Delta L=\tau \times \Delta t=\frac { QB{ a }^{ 2 } }{ 2 } \)
Initial angular momentum = 0
Final angular momentum = \(m{ b }^{ 2 }\omega =\frac { QB{ a }^{ 2 } }{ 2 } \)
or    \(\omega =\frac { QB{ a }^{ 2 } }{ 2m{ b }^{ 2 } } \)

\(C=2.0\mu F=2\times { 10 }^{ -6 }F,\)
\(\frac { dV }{ dt } =6V{ s }^{ -1 }\)
Displacement current,
\({ I }_{ D }={ \epsilon }_{ 0 }A\frac { dE }{ dt } ={ \epsilon }_{ 0 }A\frac { d }{ dt } \left( \frac { V }{ d } \right) \)
\(=\frac { { \epsilon }_{ 0 }A }{ d } \frac { dV }{ dt } =C\frac { dV }{ dt } \)
\(=\left( 2\times { 10 }^{ -6 } \right) \times 6=12\times { 10 }^{ -6 }A\)
\(=12\mu A\)

Here, I = 3.0 x 10¹⁴ M m^-2 , E₀ = ?, B₀ = ?
Intensity of the plane electromagnetic wave is
\(I={ u }_{ av }c=\frac { 1 }{ 2 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }c\)
\(\therefore { E }_{ 0 }=\sqrt { \frac { 2I }{ { \epsilon }_{ 0 }c } = } \sqrt { \frac { 2\times 3\times { 10 }^{ 14 } }{ \left( 8.85\times { 10 }^{ -12 } \right) \times \left( 3\times { 10 }^{ 8 } \right) } } \)
= 4.75 x 10⁸ V m^-1
\({ B }_{ 0 }=\frac { { E }_{ 0 } }{ c } =\frac { 4.75\times { 10 }^{ 8 } }{ 3\times { 10 }^{ 8 } } =1.58T\)

The coil draws lesser current in 2nd case because of inductive reactance of the coil.
In 1st case,\( \ R=\frac { V }{ I } =\frac { 10 }{ 2 } =5\Omega \)
In second case,\( \ Z=\frac { V }{ I } =\frac { 10 }{ 1 } =10\Omega \)
Inductive reactance,
\({ X }_{ L }=\sqrt { { Z }^{ 2 }-{ R }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 5 }^{ 2 } } =5\sqrt { 3 } \)
\(\\ 2\pi vL=5\sqrt { 3 } \ L=\frac { 5\sqrt { 3 } }{ 2\pi v } =\frac { 5\sqrt { 3 } }{ 2\times 3.14\times 50 } =0.0288 \ H\)

\(Here, \ R=200 \ ohm, \ C=15.0\mu F=15\times { 10 }^{ -6 }F\)
\({ E }_{ v }=220V, \ v=50Hz, \ { I }_{ v }=? \ { V }_{ R }=?, \ { V }_{ C }=?\)
\( Now \ { X }_{ C }=\frac { 1 }{ \omega C } =\frac { 1 }{ 2\pi vC } =\frac { 1 }{ 2\times 3.14\times 50\times 15\times { 10 }^{ -6 } } =212.2\Omega \)
\((a) \ Impedance \ of \ the \ circuit,\)
\(Z=\sqrt { { R }^{ 2 }+{ X }_{ C }^{ 2 } } =\sqrt { { 200 }^{ 2 }+\left( 212.3 \right) ^{ 2 } } =291.7 \ ohm\)
\( \therefore \ \ Current \ in \ the \ circuit, \ { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 220 }{ 291.7 } =0.75A\)
\( (b) \ { V }_{ R }={ I }_{ v }\times R=0.75\times 200=150.8 \ V\)
\({ V }_{ C }={ I }_{ v }{ X }_{ C }=0.75\times 212.3=159.2 \ V\)
\( { V }_{ R }+{ V }_{ C }=150.8+159.2=310V, \ which \ is \ more \ than \ the \ source \ voltage \ of \ 220 \ V.\)
This paradox is resolved by the fact that the two voltage are not in same phase. Therefore, they cannot be added like ordinary numbers. As V_R and V_C are out of phase by \({ 90 }^{ \circ }\), therefore
\({ V }_{ RC }=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ C }^{ 2 } } =\sqrt { \left( 150.8 \right) ^{ 2 }+\left( 159.2 \right) ^{ 2 } }\)
\(=220V,\ the\ source\ voltage\)

The point (0, 0, -a) lies on z-axis. Therefore the magnetic field induction at the given point due to current along z-axis is zero.
The magnetic field induction due to current along x-axis at the given point is \(\overset { \rightarrow }{ { B }_{ x } } ={ \frac { { \mu }_{ 0 } }{ 4\pi } }\frac { I }{ a } \overset { \Lambda }{ j } \)  
The magnetic field induction due to current along y-axis at the given point is 
\(\overset { \rightarrow }{ { B }_{ y } } ={ \frac { { \mu }_{ 0 } }{ 4\pi } }\frac { I }{ a } \overset { \Lambda }{ (-i) } \) 
Total magnetic field induction,
\(\overset { \rightarrow }{ B } =\overset { \rightarrow }{ { B }_{ x } } +\overset { \rightarrow }{ { B }_{ y } } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { I }{ a } (\overset { \Lambda }{ j } -\overset { \Lambda }{ i } )\) 

(i) Respecting law and obeying the rules enacted by the nation; self discipline and avoiding arguments
(ii) The change in the magnetic field in a metal detector due to the metals carried by the contents of checking which, induces the change in current in the coil of the metal detector. This detection of change sets up an alarm.

(a) The values shown by Shalini are :
(i) high degree of general awareness,
(ii) ability to take quick decisions,
(iii) concern for her aunt,
(iv) helping and caring nature.
(b) In microwave oven, microwaves of required wavelength get strongly absorbed by water due to its energy which heated the water. Since every food items has lot of water, so they can be heated and cooked quickly in a microwave over. With this, the microwaves heat the food item completely, not simply from outside as in case of normal oven.

The ultraviolet rays were discovered by Ritter. The frequency range of ultraviolet rays is \(8\times { 10 }^{ 14 } \ Hz \ to \ 3\times { 10 }^{ 17 } \ Hz\) ultraviolet rays are used
(i) to preserve the food stuff
(ii) in burglar alarm

Let r be the radius of each small drop and R be the radius of bigger drop.
As volume of bigger drop = volume of n small drops
\({4\over 3}\pi R^3=n\times{4\over 3}r^3\)
\(R=n^{1/3}r\)
Also, potential of bigger drop = \(total \ charge\over capacity\)
\(V={nq\over 4\pi\epsilon_o R}={nq\over 4\pi\epsilon_on^{1/3}r}=n^{2/3}{q\over 4\pi\epsilon_o r}\)
\(=n^{2/3}\) times the potential of each small drop.

Work done by electrical forces in the process
= Final stored energy-Initial stored energy
\(=\frac { { Q }^{ 2 } }{ { 2C }_{ 2 } } -\frac { { Q }^{ 2 } }{ { 2C }_{ 1 } }\\ =\frac { { Q }^{ 2 } }{ 2\left( 4\pi { \varepsilon }_{ 0 }a \right) } -\frac { 1 }{ 2 } .\frac { { Q }^{ 2 } }{ 2\left( 4\pi { \varepsilon }_{ 0 }b \right) }\\ =\frac { { Q }^{ 2 } }{ \left( 8\pi { \varepsilon }_{ 0 } \right) } \left( \frac { 1 }{ a } -\frac { 1 }{ b } \right) \)

Graph of resistivity of copper as a function of temperature is given below (resistivity of metals increases with increase in temperature).

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