Let the original dimensions of length be l, diameter D etc. Since the volume remains constant
\(\frac { \pi { D }^{ 2 } }{ 4 } l=\frac { \pi { \left( D/2 \right) }^{ 2 }l' }{ 4 } \)
\( \frac { \pi { D }^{ 2 } }{ 4 } l=\frac { \pi { D }^{ 2 }l' }{ 4\times 4 } \ or \ l'=4l\)
When diameter is reduced to one half, the length is increased to 4 times.
In first case
\(R=\frac { \rho l }{ A } \)
\(=\frac { \rho l }{ \pi { D }^{ 2 }/4 } =\frac { 4\rho l }{ \pi { D }^{ 2 } } \)
\(In \ second \ case,\)
\(R'=\frac { \rho l }{ A' } \)
\( =\frac { \pi (4l) }{ \frac { \pi { \left( D/2 \right) }^{ 2 } }{ 4 } } =\frac { \rho 4l4 }{ \pi { D }^{ 2 }4 }\)
\( R'=\left( \frac { 4\rho l }{ \pi { D }^{ 2 } } \right) 16=16 \ R\)
\(R'=16R\)
Resistance becomes 16 times the original.

Since in this experiment, the deflection is one sided, it means \({ E }_{ 1 }>E\) in the galvanometer.
(i) Since deflection in the galvanometer decreases while moving from one end A of the wire to the end B, it means positive of \({ E }_{ 1 }\) is connected to A and negative of \({ E }_{ 1 }\) is connected to B.
(ii) Since deflection in the galvanometer increases while moving from one end A of the wire to end B, it means negative of \({ E }_{ 1 }\)is connected to A and positive of \({ E }_{ 1 }\) is connected to B.

The Cu wires or Al wires are used for wiring in the home.The main considerations involved in this process are cost of metal and good conductivity of metal.

The resistance of thick metallic strips is negligible and hence do not affect the resistance of the potentiometer wire. So we can increase the length of potentiometer wire hence sensitivity of potentiometer can be increase d which leads to accurate measurement.

Resistivity of a conductor \(\rho =\frac { m }{ { ne }^{ 2 }\tau } \) 
So as temperature changes n and \(\tau \) changes.

(b) current density \(j\left( \frac { I }{ A } \right) \)is also directed along E and the relation is:
\( j=\sigma E\)

Resistance of the standard resistor, R = 10.0Ω
Balance point for this resistance, l₁​ = 58.3cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂​ = 68.5cm
Hence, potential drop across X, E₂​ = iX
The relation connecting emf and balance point is,
\(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}\)
\(\frac{\mathrm{iR}}{\mathrm{iX}}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}}\)
\(\mathrm{X}=\frac{\mathrm{l}_{1}}{\mathrm{l}_{2}} \times \mathrm{R}\)
\(=\frac{68.5}{58.3} \times 10=11.749 \Omega\)
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.

(a) Since current is given to be steady, it is constant. The current density, electric field and drift speed are inversely proportional to area of cross section.
(b) No, examples of non-ohmic elements are vacuum diode,semiconductor diode, etc.
(c) By Ohms Law, I = V / R. Now, if current required is high, the voltage should be high and the resistance should be low. Hence, a low voltage supply from which one needs high currents must have very low internal resistance.
(d) Any high tension supply must have a large internal resistance, because, if the circuit is shorted (accidentally), the current drawn will exceed safety limits. This can be dangerous for human life and can cause fatal accidents.

Number of secondary cells, n = 6
Emf of each secondary cell, E = 2.0 V
Internal resistance of each cell, r = 0.015 Ω
series resistor is connected to the combination of cells.
Resistance of the resistor, R = 8.5 Ω
Current drawn from the supply = I, which is given by the relation,
\(I=\frac{n E}{R+n r}\)
\(=\frac{6 \times 2}{8.5+6 \times 0.015}\)
\(=\frac{12}{8.59}=1.39 A\)
Terminal voltage, V = IR = 1.39 x 8.5 = 11.87 A
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.

Given
\({ R }_{ 1 }=1\Omega ,{ R }_{ 2 }=2\Omega ,{ R }_{ 3 }=3\Omega \)
(a) Total resistance of series combination
\({ R }_{ s }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\)
\({ R }_{ s }\) = 1 + 2 + 3 = \(6\Omega \)

(b) Since E = I (R + r)
I = \(\frac { E }{ { R }_{ s }+0 } =\frac { E }{ { R }_{ s } } =2A\)
\({ V }_{ 1 }={ IR }_{ 1 }=2\times 1=2V\)
\({ V }_{ 2 }={ IR }_{ 2 }=2\times 2=4V\)
\({ V }_{ 3 }={ IR }_{ 3 }=2\times 3=6V\)

Reading of voltmeter = 4 V, Resistance of voltmeter  \(=10^{4} \Omega\)
Current drawn by the voltmeter, \(I=\frac{4}{10^{4}}=4 \times 10^{-4} \mathrm{~A}\)
Again, \(I=\frac{E}{X+10^{4}}\)
\( \Rightarrow \left(X+10^{4}\right) I =E \)
\(\therefore X I =E-I \times 10^{4} \)
\(\Rightarrow X =\frac{E}{I}-10^{4}=\frac{120}{4 \times 10^{-4}}-10^{4} \)
\(\therefore X =29 \times 10^{4} \Omega=290 \mathrm{k} \Omega \)