(i)  E₁ + E₂ = 300k
k is potential gradient in volt/cm
E₁ - E₂ = 120k
⇒ \(\frac { { E }_{ 1 } }{ { E }_{ 2 } } =7/3\)
(ii) E₁ + E₂ = 300k
∴ E₁ + 3/7 = 300k
E₁ = 210k
Therefore, balancing length for cell is E₁ = 210 cm (Award this mark even if the student writes E₁ = 240k)
(Award full marks for any other correct method)
(iii) By decreasing potential gradient.
[Or through increasing length, reducing potential drop across wire, increasing resistance put in series with the main cell etc.]

According to the colour codes, resistances of two wires are as given below:
(i) Code of brown = 1, Code of green = 5
Code of blue = 6, R₁ = \(15\times 10^{ 6 }\Omega \pm 20%\)%
(ii) Code of orange = 3, Code of black = 0
Code of green = 5, R₂ = \(30\times 10^{ 5 }\Omega \pm 20%\)%
\(\therefore Ratio \ of \ resistance,\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { 15\times { 10 }^{ 6 } }{ 30\times { 10 }^{ 5 } } =\ 5\)
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =5\)

H = \(\frac{V^2t_1}{R_1}=\frac{V^2t_2}{R_2}\)
or  \(\frac{R_1}{R_2}=\frac{t_1}{t_2}\)    .......(i)
But  \(R \propto l\),
so \(\frac{R_1}{R_2}=\frac{l_1}{l_2}\)   ...........(ii)
Now, l₂ = (2/3)l₁
From (i) and (ii);
\(\frac{l_1}{l_2}=\frac{t_1}{t_2}\)
or   \(\frac{3}{2}=\frac{15}{t_2}\)
or  t₂ = 10 minutes

Resistance of resistor = \(0.34 \ \Omega\) = 34 x 10^-2 \(\Omega\) 
The colour of the third band of a coded resistor is due to multiplier 10^-2 which is for silver.

t = \(\frac{l}{v_d}=\frac{l}{I/n Ae}=\frac{lnAe}{I}\)
= \(\frac{1 \times (8 \times 10^{28})\times(5\times10^{-7})\times(1.6\times10^{-19})}{1}\)
= 6.4 x 10³ s

The arrangement of resistance is shown in fig and its equivalent circuit. Since the wire has a resistance 1 ohm per cm, the resistance of the cross-piece DE is 10 ohm. Since apex angle is \(60°\) and the cross-piece must also be the mid-points of the legs AB and AC so ADE becomes an equilateral triangle.

As is clear from fig DA and AE are in series and, therefore resistance of the combination of DA and AE = 10 + 10 = 20 ohm. This resistance of 20 ohm is in parallel to DE.
The effective resistance r of the portion DAED is given by
\(\frac { 1 }{ r } =\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\( =\frac { 2+1 }{ 20 } =\frac { 3 }{ 20 }\)
\(r=20/3 \ ohm\)
Now BD and EC are in series with the resistance of portion DAED
Resistance between B and C
\(=10+\frac { 20 }{ 3 } +10=26.67 \ ohm\)

Relaxation time is inversely proportional to the velocities of electrons and ions. The applied electric field produces the insignificant change in velocities of electrons of the order of 1 mm/s, whereas the change in temperature T affects velocities at the order of 10² m/s
This decreases the relaxation time cinsiderably in metals and consequently resistivity of metal or conductor increases as, \(\rho=\frac{1}{\sigma}=\frac{m}{n e^2 \tau}\)

Number density of free electrons in a copper conductor, n = 8.5 x 10²⁸ m^-3 Length of the copper wire, l = 3.0 m
Area of cross-section of the wire, A = 2.0 x 10^-6 m²
Current carried by the wire, I = 3.0 A, which is given by the relation,
I = nAeV_d
Where,
e = Electric charge = 1.6 x 10⁻¹⁹ C
V_d = Drift velocity = \(\frac{\text { Length of the wire (l) }}{\text { Time taken to cover l(t) }}\)
\(I=n A e \frac{l}{t}\)
\(t=n A e \frac{l}{I}\)
\(=\frac{3 \times 8.5 \times 10^{28} \times 2 \times 10^{-6} \times 1.6 \times 10^{-19}}{3.0}\)
\(=2.7 \times 10^{4} s\)
Therefore, the time taken by an electron to drift from one end of the wire to the other is 2.7 x 10⁴ s.

When a potentiometer is connected between A and B,
\(E_{1}=k \times 300\) ...(i)
where k is potential gradient.
When the potentiometer is connected between A and C,
\(E_{1}-E_{2}=k \times 100 \ \left(\because E_{1}>E_{2}\right)\) ...(ii)
Dividing equations (i) and (ii), we get
\(\frac{E_{1}}{E_{1}-E_{2}}=\frac{300}{100} \Rightarrow \frac{E_{1}}{E_{2}}=\frac{3}{2}\)

The discrepancy is due to (i) non-idealness of voltmeter, and (ii) an internal resistance of cell.
Now, \(\frac{V}{E}=\frac{1.4}{1.5} \Rightarrow \frac{I R}{I(R+r)}=\frac{1.4}{1.5}\)
\(\Rightarrow \ 1+\frac{r}{R}=\frac{15}{14} \Rightarrow \frac{r}{R}=\frac{1}{14} \Rightarrow \frac{R}{r}=14\)
Here, R is the resistance of the voltmeter and r is the internal resistance of the cell.

(i) When constant current flows through a conductor of uniform area of cross-section ,the potential difference, across a length l of the wire, is directly proportional to that length of the wire.[\(V\propto l\)] (provided current and area are constant] 
(ii) Current flowing in the potentiometer wire
 \(i=\frac { E }{ { R }_{ total } } =\frac { 2.0 }{ 15+10 } =\frac { 2 }{ 25 } A\)
∴Potential difference across the two ends of the wire
\({ V }_{ AB }=\frac { 2 }{ 25 } \times 10V=\frac { 20 }{ 25 } =0.8volt\)
Hence potential gradient K = \(\frac { { V }_{ AB } }{ { l }_{ AB } } =\frac { 0.8 }{ 1.0 } =0.8\frac { V }{ m } \)
Current flowing in the circuit containing experimental cell, = \(\frac { 1.5 }{ 1.2+0.3 } =1A\)
Hence, potential difference across length AO of the wire \(=0.3\times 1V=0.3V\)
\(\Rightarrow \)  \(0.3=K\times { l }_{ AO }\)
\(=0.8\times { l }_{ AO }\)
\(\Rightarrow \)\({ l }_{ AO }=\frac { 0.3 }{ 0.8 } m=0.375m\)
\(=\) 37.5cm

(a) To obey elders.
(b) Using, I = V/R
(i) 147mA;
(ii) 2.2mA
(iii) Wet skin with 147mA,
(c) When the current flows, the result is fatal

(i) Lat carbon resistor S is given to the bridge. Then,
\(\frac { 2R }{ R } =\frac { 2R }{ S } \Rightarrow \frac { R }{ S } =1\)
R = S = 22 x 10³ \(\Omega =22k\Omega \)
(ii) After interchanging the resistance, the balanced bridge would be
\(\frac { 2R }{ X } =\frac { 22\times { 10 }^{ 3 } }{ 2\times 22\times { 10 }^{ 3 } } =\frac { 1 }{ 2 }\)
X = 4R = 4 x 22 x 10³
= 88 x 10³ \(\Omega \)
The colour sequence of X is grey, grey and orange. Thus, equivalent resistance of Wheatstone bridge,
\(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ 3R } +\frac { 1 }{ 6R } =\frac { 3 }{ 6R } \)     
\( { R }_{ eq }=2R\)
\(\therefore \ Current \ through \ the \ circuit,I=\frac { 1 }{ 3 } \times \frac { V }{ 2R } =\frac { V }{ 6R } A\)

(i) Acknowledging the help from others with gratitude. Awareness of the technology, helping tendency, practical knowledge of the subject.
(ii) Vasu checked the wiring because he has practical knowledge and he can rectify the problem.
(iii) An electric fuse is a wire used as a safety device, which melts when current exceeds the limit. Fuse wire has low melting point, high resistivity 

Here,
 R = \(5.0 \ \Omega\);
r₁ = 5 x 10^-2 m;
D = 2.0 x 10^-3 m;
\(\rho= 2.0 \times 10^{-7} \ \Omega m\).
R = \(\frac{\rho l}{\pi D^2/4}\)
or l = \(\frac{R \pi D^2}{4 \rho}\)
Let n be the number of turns in the coil. Then total length of the wire used, l = \(2 \pi r_1 n\)
or n = \(\frac{l}{2 \pi r_1}=\frac{R \pi D^2}{4 \rho \times 2 \pi r_1}=\frac{RD^2}{8 \rho r_1}\)
 = \(\frac{5.0 \times (2.0 \times 10^{-3})^2}{8 \times (2.0 \times 10^{-7})\times (5 \times 10^{-2})^2}\)
 = 250

From Kirchhoff's junction rule, we have
\( { I }_{ 1 }={ I }+{ I }_{ 2 }\)   ...........(i)
Applying Kirchhoff's loop rule to outer loop containing 10V cell, we get
\(10=IR+{ 10I }_{ 1 }\) ............(ii)
Applying Kirchhoff's loop rule to outer loop containing 2V cell, we get
\(2={ 5I }_{ 2 }-RI\)
\(2=5\left( { I }_{ 1 }-I \right) -RI\)
\(4={ 10I }_{ 1 }-10I-2RI\)
Subtracting eq (ii) from eq (i), we get
\(6=3RI+10I\)
\(2=I\left( R+\frac { 10 }{ 3 } \right) \)
From Ohm's law, we have
\(V=I\left( R+{ R }_{ off } \right) \)
Comparing eq (iii) and (iv), we get
\({ R }_{ ef }=\frac { 10 }{ 3 } \Omega \)
If \({ E }_{ eff }\) and \({ R }_{ eff }\) are the effective voltage and effective internal resistance of the combination, then the equivalent circuit is shown.

\(I=\frac { \varepsilon }{ R+\frac { { \rho }_{ 1 }l }{ { A }_{ 1 } } } \)for set A
\(I=\frac { \varepsilon }{ R+\frac { { \rho }_{ 1 }l }{ 2{ A }_{ 1 } } +\frac { { \rho }_{ 2 }l }{ { 2A }_{ 2 } } } \)for set B
Equating the above two expressions and simplifying
\(\frac { { \rho }_{ 1 } }{ A_{ 1 } } =\frac { { \rho }_{ 2 } }{ { A }_{ 2 } } \)
(ii) Potential gradient of the potentiometer wire for Set A K = \(I\frac { { \rho }_{ 1 } }{ A_{ 1 } } \)
Potential drop across the potentiometer wire in Set B
\(V=I\left( \frac { { \rho }_{ 1 } }{ 2A_{ 1 } } +\frac { { \rho }_{ 2 }l }{ 2{ A }_{ 2 } } \right) \)
\(V=\frac { I }{ 2 } \left( \frac { { \rho }_{ 1 } }{ 2A_{ 1 } } +\frac { { \rho }_{ 2 } }{ { A }_{ 2 } } \right) l\)
\(K'=\frac { I }{ 2 } \left( \frac { { \rho }_{ 1 } }{ A_{ 1 } } +\frac { { \rho }_{ 2 } }{ { A }_{ 2 } } \right) \),using the condition obtained in part (i)
\(K'=I\frac { { \rho }_{ 1 } }{ { A }_{ 1 } } \),which is equal to K.Therefore, balancing length obtained in the two sets is same.

Potentiometer does not draw any (net) current front-the cell. Voltmeter draws some current from cell, when connected across at, hence measures terminal voltage.

We apply Kirchoff's current law in the shown circuit.
At junction B,
i₁​=i_g​+i₃
At junction D,
i_2​+i_g​=i₄​
If current through the galvanometer is zero,
i_g = 0
thus i₁ = i₃
and i₂ = i₄
Applying Kirchoff's voltage law for loop ABDA,
i₁P + i_gG = i₂R
Applying Kirchoff's voltage law for loop BCDB,
i₃Q + i₄S + i_gG
When i_g = 0,
i₁P = i₂R
and i₃Q = i₄S
But i₁ = i₃ and i₂ = i₄,
Therefore \(\frac{P}{Q}=\frac{R}{S}\)

The resistivity of a metallic conductor is given by
ρ = ρ₀[1 + ∝(T-T₀)]
Where, ρ₀ = Resistivity at reference temperature
T₀ = Reference temperature
∝ = Coefficient of resistivity
From the above relation, we can say that the graph between resistivity of a conductor with temperature is straight line. But, at temperatures much lower than 273 K ( i.e. 0°C), the graph deviates considerably from a straight line as shown in the figure.

Let us redraw circuit as shown.

Using Kirchoff's voltage law in closed loop DABCD
\(\left( { I }_{ 1 }+{ I }_{ 2 } \right) { R }_{ 3 }+{ I }_{ 1 }{ R }_{ 1 }-{ E }_{ 1 }=0\)      ..........(1)
In a closed loop ABFEA
\(\left( { I }_{ 1 }+{ I }_{ 2 } \right) { R }_{ 3 }+{ I }_{ 2 }{ R }_{ 2 }-{ E }_{ 2 }=0\)      .........(2)
Multiplying (1) by (R₂ + R₃) and (2) by R₃ and subtracting, we get
\({ I }_{ 1 }=\frac { { E }_{ 1 }{ R }_{ 2 }+{ E }_{ 1 }{ R }_{ 1 }-{ E }_{ 2 }{ R }_{ 3 } }{ { R }_{ 1 }{ R }_{ 2 }+{ R }_{ 2 }{ R }_{ 3 }+{ R }_{ 1 }{ R }_{ 3 } } \)
\(=\frac { 5\times 60+5\times 20-4\times 20 }{ 200\times 60+60\times 20+200\times 20 } \)
\(=\frac { 300+100-80 }{ 12000+1200+4000 } \)
\(=\frac { 320 }{ 17200 } A\)
Similarly,I₂ =\(\frac { { E }_{ 2 }{ R }_{ 1 }+{ E }_{ 2 }{ R }_{ 3 }-{ E }_{ 1 }{ R }_{ 3 } }{ { R }_{ 1 }{ R }_{ 2 }+{ R }_{ 2 }{ R }_{ 3 }+{ R }_{ 1 }{ R }_{ 3 } } \)
= \(\frac { 4\times 200\times 4\times 20-5\times 20 }{ 200\times 60+60\times 20+200\times 20 } \)
= \(\frac { 800+80-100 }{ 17200 } =\frac { 780 }{ 17200 } A\)
\(\therefore\) Total current in mA
= \({ I }_{ 1 }+{ I }_{ 2 }\)
= \(\frac { 320 }{ 17200 } +\frac { 780 }{ 17200 } =\frac { 1100 }{ 17200 } \)
= \(\frac { 11 }{ 172 } A=0.06395A\)
= 63.95mA
= 64 mA (approx)

\(P_1=\frac{V^2}{R^1}\)
\(R_2=\frac{V^2}{R_2}\)
In series
\(P=\frac{V^2}{R_1+R_2}\)

By Kirchhoff's first law at D,
 \({ I }_{ DC }=1A\)          \([\because { I }_{ Dc }+1=2]\)
    
Along ACDBA,                                                             
\({ V }_{ A }+1+1\times 2-2={ V }_{ B }\)  (V_A = 0)
But,  \({ V }_{ B }=1+2-2=1V\)
\({ V }_{ B }=1V\)                           

Since, slope of 1 > slope of 2
\(\therefore\) R₁ < R₂
Also, we know that resistance is directly proportional to the temperature.
Therefore, T₂ > T_1.

For a given supply voltage, heat produced in time t is 
\(H =\frac{V^2}{R}t=\frac{V^2}{\rho l/A}=\frac{V^2At}{\rho l} \ \ i.e., \ \ H \propto \frac{t}{l}\)
\(\therefore \frac{H_1}{H_2}=\frac{t_1}{t_2}\times\frac{l_2}{l_1}\)
As H₁ = H₂,
so \(1 = \frac{t_1}{t_2}\times\frac{l_2}{l_1}\)
or \(\frac{l_2}{l_1}=\frac{t_2}{t_1}=\frac{8}{10}<1\)
or  l₂ < l₁
It means the length of the heating wire should be decreased.

Unit of potential gradient is V cm^-1 or V m ^-1. Since fall of potential, V = K l where K is the potential gradient, therefore, for the given value of V, if K is decreased, l will increase.

The material must have high resistivity and low temperature coefficient of resistance.

For student A, the numbers attached to brown, green and orange are 1, 5 and 3. So the value of resistor is R_A = 15 x 10³ \(\Omega\) =\(15 \ k \Omega\).
For student B, the numbers attached to black, green and red are 0, 5 and 2 respectively. So the value of resistor is, R_B = 05 x 10² \(\Omega\) =\(500 \ k \Omega\).
Thus, student A picked up the correct resistor of \(15 \ k \Omega\).

Mobility,  \(\mu = \frac{drift \ \ velocity}{electric \ \ field}=\frac{v_d}{E}\)
(a) The charge carriers in an electrolyte are positive and negative ions.
(b) The charge carriers in an ionised gas are electrons and positively charged ions.

Equivalent resistance of two cells in parallel
\(r_{e q}=\frac{r_1 r_2}{r_1+r_2}\)
Equivalent emf of two cells in parallel
\(r_{e q}=\frac{\epsilon_1 r_2+\epsilon_2 r_1}{r_1+r_2}\)
Current in circuit,
\(I=\frac{\epsilon_{e q}}{R+r_{e q}}=\frac{\frac{\epsilon_1 r_2+\epsilon_2 r_1}{r_1+r_2}}{R+\frac{r_1 r_2}{\left(r_1+r_2\right)}}=\frac{\epsilon_1 r_2+\epsilon_2 r_1}{R\left(r_1+r_2\right)+r_1 r_2}\)

The internal resistance of a cell depends upon:
(i) Nature, concentration and temperature of electrolyte,
(ii) the nature of electrodes,
(iii) the distance between the electrodes and 
(iv) area of the electrodes immersed in the electrolyte.

Mobility of charge carriers inside conductor is defined as the magnitude of drift velocity of charge per unit electric field applied.
SI unit of mobility is m²s^-1V^-1 or ms^-1N^-1C.

The electrical conductivity of copper is next only to silver which is costly. Therefore, it conducts the current without offering much resistance. The copper being diamagnetic material does not get magnetised due to the current through it and hence does not disturb the current in the circuit.

Drift velocity,
\(v_d=\frac{1}{n A \varepsilon}=\frac{I}{n\left(\pi D^2 / 4\right) e}, \text { i.e., } v_d \propto \frac{1}{n}\)
Since the wires are connected in series, current I through both is same. Therefore,
\(\frac{\left(v_d\right)_x}{\left(v_d\right)_y}=\frac{n_y}{n_x}=\frac{1}{2}\)(vd)x(vd)y=nynx=12

With the increase in temperature, the free electrons collide more frequently with the ions/atoms of conductor, resulting decrease in relaxation time.