Internal resistance of the cell = r
Balance point of the cell in open circuit, l₁​ = 76.3cm
An external resistance (R) is connected to the circuit with R = 9.5Ω
New balance point of the circuit, l₂ ​= 64.8cm
Current flowing through the circuit = l
Using the relation connecting resistance and emf is,
\(\mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{2}}\right) \mathrm{R}\)
\(=\frac{76.3-64.8}{64.8} \times 9.5=1.68 \Omega\)
Therefore, the internal resistance of the cell is 1.68Ω.

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have much lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10²².

Potential gradient \(=\frac{2.5}{8}=0.3125 \mathrm{~V} / \mathrm{m}\). As potential drop across the balancing length is equal to the emf of the
cell, i.e. emf = 0.3125 x 1 = 0.3125 V.

(a) (i) Presence of mind.
(ii) High degree of general awareness.
(iii) Helping and caring nature. 2
(b) For maximum current, external resistance,
R = 0
\(I=\frac{E}{R+r}=\frac{12}{0.4}=\frac{12}{0.4}=30A\)

Principle : When a steady current flows through a wire of uniform cross-section, the  potential drop across any segment is directly proportional to the length of the segment of the wire i.e. V a l
Potential gradient is the potential drop across the wire per unit length of the wire i.e
\(K=\frac{V}{l}\)
Potential gradient \(K=\frac{V}{l}=\frac{IR}{l}\)
\(K=\frac{\frac{I\times pl}{A}}{l}\)
\(K=\frac{IP}{A}\)
(b) (i)  \(\frac{\varepsilon _1-\varepsilon _2}{\varepsilon _1-\varepsilon _2}=\frac{120}{300}=\frac{2}{5}\)
\(\frac{\varepsilon _1}{\varepsilon _2}=\frac{7}{3}\)
(iii)  \(\frac{e_1+e_2}{\varepsilon _2}=\frac{300}{x}\Rightarrow \ \frac{300}{x}=\frac{10}{7}\)
\(\Rightarrow\ \ x=210\ cm\)

(i) Values displayed by Vishwajeet are as follows
(a) General awareness
(b) Presence of mind
(c) Use of scientific knowledge
(ii) Potentiometer is based on null point method. Voltmeter draws current at the time of measurement of potential difference across the cell. So, it does not give the correct value of emf. For more precise measurements, it should be low.

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.

Superconductors are the materials that lose all its resistance at very low temperature = 0 K
Applications of Super conductor are used
a) In making very strong electromagnets
b) In producing veru high speed computers

a) in parallel, power dissipation \(\alpha\) 1/R 
Therefore 3\(\Omega \) wire will dissipate more power
b) In series, power dissipation \(\alpha\) R
Therefore 9\(\Omega \) wire will dissipate more power

Now, 1/R = 1R₁ + 1R₂
Because 1 = 48/240 = 0.2 m

\(\mathrm{R}=\rho(1 / \mathrm{A})\)
\( =\rho\left(1^2 / \mathrm{A} 1\right) \\ =\rho 12 / \mathrm{V} \)
Since \(\rho\) and V are constans therefore
\(\rightarrow\left(R_2 / R_1\right)=\left(12 / l_1\right)^2=9\)
Because R₂ = 9R₁
\(=9 \times 10=90 \Omega\)

The end error in the meter bridge is due to the following reasons:
(i) The zero mark of the scale provided along the bridge wire may not start from the position where the bridge wire leaves the copper strip and 100 cm mark of the scale may not be at position, where the bridge wire just touches the other copper strip.
(ii) The resistances of the connecting wires and copper strips of meter bridge have not been taken into account.
The end error can be removed by repeating the experiment by interchanging the known and unknown resistances and taking the mean of the resistances determined.

As the temperature of certain metals and alloys decreases, their resistance also decreases. When the temperature reaches a certain critical value called critical temperature, the resistance of material completely disappears, i.e., it becomes zero. Then the material behaves as a superconductor.
Thus superconductors are those material conductors whose resistances disappear at critical temperature. The critical temperature is different materials.
Superconductors are used
(i) in power transmission
(ii) to produce very high speed computers.

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.

Value of carbon resistance = \(01×10oΩ±5% = 1Ω±5%\)

A constant current can be kept inside a conductor by maintaining a constant potential difference across the two ends of conductor.

Yes. A steady current in a cylindrical conductor if electric force is acting on free electrons which make the electrons to move in a particular direction. It is possible due to electric field within the conductor.