Current Electricity

Case 1 : When $\mathrm{S}_1$ and $\mathrm{S}_2$ are open

$ \left(R_{e q}\right)_1=1+\frac{5 \times \frac{1}{2}}{5+\frac{1}{2}}=1+\frac{5}{11}=\frac{16}{11} $

$ P_1=\frac{V^2}{(R_{e q})_1}=\frac{(6)^2}{16} \times 11=\frac{36 \times 11}{16}=24.75 \mathrm{~W} $

$ \left(R_{e q}\right)_2=\frac{6}{11} \Omega $

$ P_2=\frac{V^2}{(R_{e q})_2}=\frac{(6)^2}{6} \times 11=\frac{36 \times 11}{6}=66 \mathrm{~W} $

$ P_2>P_1 $

Option (A) is correct.

Case 2 : If I = $2 \mathrm{~A}$ source is used in both circuits, then

$ P_1=i^2\left(R_{e q}\right)_1=(2)^2 \times \frac{16}{11}=\frac{64}{11}=5.818 \mathrm{~W} $

$ P_2=i^2\left(R_{e q}\right)_2=(2)^2 \times \frac{6}{11}=\frac{24}{11}=2.1818 \mathrm{~W} $

$ P_1>P_2 $

Option (B) is correct.

Case 3 : For $Q_1$

$ \begin{aligned} & R_{e q}=\frac{5}{11} \Omega \\\\ & Q_1=\frac{V^2}{R_{e q}}=\frac{(6)^2}{\frac{5}{11}}=\frac{36 \times 11}{5}=79.2 \mathrm{~W} \\\\ & P_1=24.75 \mathrm{~W} \\\\ & Q_1>P_1 \end{aligned} $

Option (C) is correct.

Case 4 : For option (D)

$ \begin{aligned} & Q_1=i^2 R_{e q}=(2)^2 \times \frac{5}{11}=\frac{20}{11}=1.81 \mathrm{~W} \\\\ & Q_2=i^2 R_{e q}=(2)^2 \times \frac{1}{2}=\frac{4}{2}=2 \mathrm{~W} \\\\ & Q_2>Q_1 \end{aligned} $

Option (D) is incorrect.

Point A and B are at same potential so they can be merged/folded.

For loop 1

$ \begin{aligned} & -\frac{1}{2} x-\frac{1}{2}(x+y)-x+6=0 \\\\ &\Rightarrow -2 x-\frac{y}{2}=-6 \\\\ &\Rightarrow -4 x-y=-12 \\\\ &\Rightarrow 4 x+y=12 .......(1) \end{aligned} $

For loop 2

$ \begin{aligned} & \frac{1}{2} y+1 y+12+\frac{1}{2}(x+y)=0 \\\\ &\Rightarrow \frac{3}{2} y+\frac{y}{2}+\frac{x}{2}=-12 \\\\ &\Rightarrow 2 y+\frac{x}{2}=-12 \\\\ & \begin{array}{l} \Rightarrow4 y+x=-24 \\\\ \Rightarrow4 y+x-16 x-4 y =-24-48 \end{array} \\\\ &\Rightarrow -15 x=-72 \end{aligned} $

$ \Rightarrow $ $x=\frac{72}{15}$

$ \Rightarrow $ $4\left(\frac{72}{15}\right)+y=12$

$ \begin{aligned} \Rightarrow y & =12-\frac{288}{15} \\\\ & =\frac{180-288}{15} \\\\ & =\frac{-108}{15}=-7.2 \mathrm{~A} \end{aligned} $

Current in $R_1=7.2 \mathrm{~A}$

Current in $R_2=\frac{x+y}{2}=\left(\frac{72}{15}-\frac{108}{15}\right) \frac{1}{2}$

$ \begin{aligned} & =\frac{1}{2}\left(\frac{36}{15}\right)=\frac{2.4}{2} \mathrm{~A} \\\\ & =1.2 \mathrm{~A} \end{aligned} $

Current in $R_3=x=4.8 \mathrm{~A}$

Current in $R_5=\frac{1}{2} x=2.4 \mathrm{~A}$

All the elements are in parallel.

$ \therefore $ $\int {{l \over {dr}} = \int\limits_{{R_1}}^{{R_2}} {{{t\,dx} \over {\rho \,\pi x}}} } $

${l \over r} = {t \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$

Resistance = ${{\pi \rho } \over {t\ln \left( {{{{R_2}} \over {{R_1}}}} \right)}}$

$ \therefore $ $I = {V \over {{\mathop{\rm Re}\nolimits} sis\tan ce}}$

$ = {{{V_t}} \over {\pi \rho }}\ln \left( {{{{R_2}} \over {{R_1}}}} \right)$



$eE = {{m{v^2}} \over r}$

$I = neA{V_d}$

${{{V_0}tdr} \over {\rho \pi r}} = ne(drt)V$

$ \Rightarrow V = \sqrt {{{{V_0}} \over {\rho \pi ner}}} $

$ \therefore $ $E = {m \over {er}}{{V_0^2{t^2}} \over {{\rho ^2}{\pi ^2}{n^2}{e^2}{r^2}}}$

$ \Rightarrow E = {{mV_0^2} \over {{e^3}{r^2}{\rho ^2}{\pi ^2}{n^2}}}$

$dV = - Edr \Rightarrow dV = - {K \over {{r^3}}}dr$

$V = K\int\limits_{{R^1}}^{{R_2}} {{r^{ - 3}}dr} $

$\Delta V = {{mV_0^2} \over {{e^3}{\rho ^2}{\pi ^2}{n^2}}}\left( {{1 \over {r_1^2}} - {1 \over {r_2^2}}} \right)$

$\Delta V \propto V_0^2 $

$ \Rightarrow $ $\Delta $V $ \propto $ I² ( As $ I \propto {V_0}$ )

Just after closing of switch charge on any capacitor is zero.

$ \therefore $ Replace all capacitors by conducting wires.



Current flow in the circuit,

$i = {5 \over {70 + 100 + 30}} = {5 \over {200}} = 25$ mA

Now S₁ is kept closed for long time circuit is in steady state.



${q \over {10}} + {q \over {80}} + {q \over {80}} - 5 = 0 \Rightarrow {{10q} \over {80}} = 5$

$ \therefore $ $q = 40\mu C$

$ \therefore $ V across ${C_1} = {{40} \over {10}} = 4V$

Now just after closing S₂ charge on each capacitor remains same.



Applying KVL,

$ - 10 + x \times 30 + {{40} \over {10}} + y \times 70 = 0$

30x + 70y = 6 .....(1)

$ - {{40} \over {80}} + 5 + (x - y)30 - {{40} \over {80}} + (x + y) \times 100 - 10 + x \times 30 = 0$

$160x - 130y - 6 = 0$ .... (2)

$y = {{96} \over {1510}}$

x = 0.05 A

$V = 100 \times {10^{ - 3}} = {10^{ - 1}}V$

$V = {{\mathop{\rm l}\nolimits} _g}({R_g} + {R_v})$

${{{{10}^{ - 1}}} \over {2 \times {{10}^{ - 6}}}}={R_g} + {R_v}$

$5 \times {10^4}\Omega \approx {R_v}$ ($ \because $${R_v} < {10^5}\Omega $)



I_gR_g = (I - I_g)S


$S = {{2 \times {{10}^{ - 6}} \times 10} \over {{{10}^{ - 3}} - 2 \times {{10}^{ - 6}}}}$

$S = 2 \times {10^{ - 5}} \times {10^3} = 2 \times {10^{ - 2}} = 20m\Omega $

${R_A} = {{S{R_g}} \over {S + {R_g}}} = {{20 \times {{10}^{ - 3}} \times 10} \over {10 + 20 \times {{10}^{ - 3}}}}$

$ \approx 20 \times {10^{ - 3}}\Omega $

$i = {\varepsilon \over {\left( {{{1000 \times 50 \times {{10}^3}} \over {51 \times {{10}^3}}}} \right)}} = {{51\varepsilon } \over {5 \times {{10}^4}}}$

($ \because $ R_A $ \to $ 0)

$i' = i\left( {{{{R_v}} \over {51 \times {{10}^3}}}} \right) = {\varepsilon \over {1000}}$

Measured resistance,

$ \therefore $ ${R_m} = {{i' \times 1000} \over i} = {\varepsilon \over {51\varepsilon }} \times 5 \times 10 = 980.4\Omega $

If the voltmeter shows full scale deflection then

${\varepsilon \over {980}} \times \left( {{{1000} \over {51 \times {{10}^3}}}} \right) \times 5 \times {10^4} = {10^{ - 1}}$

$\varepsilon = 999.6$ mV

Since, i_A = 1 mA so maximum reading of R can be

${{999.6\,mV} \over {1\,mA}} = 999.6\Omega $

Let i_g be the current that gives full deflection of the galvanometer. When all components are connected in series (see figure), effective resistance of the circuit is ${R_e} = 2R + 2{R_c}$ and maximum current allowed in the circuit is i_g.

Thus, the voltage between A and B is

${V_{AB}} = {i_g}{R_e} = 2{i_g}(R + {R_c})$.

Consider the case when two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel.

The maximum current through each galvanometer is i_g and the maximum current through the resistors is 2i_g. Apply Kirchhoff's law to get the voltage between A and B as

$V{'_{AB}} = 2{i_g}R + 2{i_g}R + {i_g}{R_c}$

$ = 2{i_g}(R + {R_c}) + 2{i_g}(R - {R_c}/2)$

$ = {V_{AB}} + 2{i_g}(R - {R_c}/2)$

$ > {V_{AB}}$ ($\because$ ${R_c} < R/2$).

Consider the case when all four components are connected in parallel.

Let i and i_g be the currents through the resistors and the galvanometers. By Kirchhoff's law, $iR = {i_g}{R_c}$, which gives $i = {i_g}{R_c}/R$. The current between A and B is

${I_{AB}} = 2i + 2{i_g} = 2{i_g}(1 + {R_c}/R)$.

Consider the case when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

$I{'_{AB}} = {i_g} + 2i = {i_g} + 2{i_g}(2{R_c}/R)$

$ = 2{i_g}(1 + {R_c}/R) - (1 - 2{R_c}/R){i_g}$

$ = {I_{AB}} - (1 - 2{R_c}/R){i_g}$

$ < {I_{AB}}$ ($\because$ ${R_c} < R/2$).

When the key is just pressed, there is no charge on capacitors

$\therefore$ $5 = 25000{I_1} \Rightarrow {I_1} = {5 \over {25000}} = 0.2\,mA$

and $5 = 50000{I_2} \Rightarrow {I_2} = {5 \over {50000}} = 0.1\,mA$

Reading of voltmeter $ = {V_Q} - {V_P} = - 25000{I_1}$

$ = - 25000 \times 0.2 \times {10^{ - 3}}V = - 5V$

After a very long time, steady state is reached, i.e. ${I_1} - {I_2} = 0$

$\therefore$ $5 = {{{q_1}} \over {40 \times {{10}^{ - 6}}}}$ or ${q_1} = 200\mu C$

and $\therefore$ $5 = {{{q_2}} \over {20 \times {{10}^{ - 6}}}}$ or ${q_2} = 100\mu C$

Now, reading of voltmeter $ = {V_Q} - {V_P}$

$ = {{{q_2}} \over {20\mu F}} = {{100\mu C} \over {20\mu F}} = 5V$

For capacitor of 40$\mu$F,

${\tau _1} = RC = 25 \times {10^3} \times 40 \times {10^{ - 6}} = 1\,s$

and for capacitor of 20$\mu$F,

${\tau _2} = RC = 50 \times {10^3} \times 20 \times {10^{ - 6}} = 1\,s$

$\therefore$ At any instant t,

${q_1} = 200[1 - {e^{ - t}}]\mu C$, ${q_2} = 100[1 - {e^{ - t}}]\mu C$

${I_1} = 0.2\,{e^{ - t}}$ mA, ${I_2} = 0.1\,{e^{ - t}}$ mA

${V_Q} - {V_P} = {{100(1 - {e^{ - t}})\mu C} \over {20\mu F}} - 25k\Omega \times 0.2\,{e^{ - t}}$ mA

$ = 5(1 - {e^{ - t}}) - 5{e^{ - t}} = 5 - 10{e^{ - t}}$

At t = ln 2 s; ${V_Q} - {V_P} = 5 - 10{e^{ - \ln 2}} = 5 - {{10} \over 2} = 0$ V

Initially, I₀ = 0.1 mA + 0.2 mA = 0.3 mA

At t = 1 s, I = I₁ + I₂

$ = (0.2{e^{ - 1}} + 0.1{e^{ - 1}}) = {{0.3} \over e}\,mA = {{{I_0}} \over e}$

After a long time, i.e. t $\to$ $\infty$

$I = {I_1} + {I_2} = 0.2{e^{ - \infty }} + 0.1{e^{ - \infty }} = 0$

Because of non-uniform evaporation at different section, area of cross-section would be different at different sections.

Region of highest evaporation rate would have rapidly reduced area and would become break up cross-section.

Resistance of the wire as whole increases with time. Overall resistance increases hence power decreases.

($p = {{{V^2}} \over R}$ or $p \propto {1 \over R}$ as V is constant).

At break up junction temperature would be highest, thus light of highest band frequency would be emitted at those cross-section.

Resistance of a wire, $R = {{\rho l} \over A}$

For iron (Fe) bar,

$\rho$ = 10^$-$7 $\Omega$m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (2 mm) $\times$ (2 mm) = 4 mm² = 4 $\times$ 10^$-$6 m²

${R_1} = {{{{10}^{ - 7}} \times 50 \times {{10}^{ - 3}}} \over {4 \times {{10}^{ - 6}}}} = 1250 \times {10^{ - 6}}$ $\Omega$ = 1250 $\mu$W

For aluminium (Al) bar,

$\rho$ = 2.7 $\times$ 10^$-$8 $\Omega$ m, l = 50 mm = 50 $\times$ 10^$-$3 m

A = (7² $-$ 2²) mm² = 45 mm² = 45 $\times$ 10^$-$6 m²

$\therefore$ ${R_2} = {{2.7 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 3}}} \over {45 \times {{10}^{ - 6}}}}$

$ = {{27 \times 50} \over {45}} \times {10^{ - 6}} = 30 \times {10^{ - 6}}$ $\Omega$ = 30 $\mu$$\Omega$

Potential difference across both bars (resistors) is same so they are in parallel combination.

Equivalent resistance between P and Q is given by

$R = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {{1250 \times 30} \over {1250 + 30}} = {{125 \times 30} \over {128}} = {{1875} \over {64}}\mu \Omega $.

Let i₁ and i₂ be the currents as shown in the figure.


Apply Kirchhoff's law in the loop ABCDA and CEFDC to get

${i_1}{R_1} + ({i_1} - {i_2}){R_2} = {V_1}$ ..... (1)

${i_2}{R_3} - ({i_1} - {i_2}){R_2} = {V_2}$ ...... (2)

Multiply equation (1) by R₃ and (2) by R₁ and then subtract to get the current through R₂ as

$({i_1} - {i_2}) = {{{V_1}{R_3} - {V_2}{R_1}} \over {{R_1}{R_3} + {R_2}{R_3} - {R_1}{R_2}}}$.

The current through R₂ becomes zero when V₁R₃ = V₂R₁.

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown below :

${I_2} = {{12} \over {2 + 2 + 2}} = 2A$

${I_3} = {{12} \over {4 + 4 + 4}} = 1A$

$\therefore$ ${I_1} = {I_2} + {I_3} = 3A$

I_PQ = 0 because V_P = V_Q

Potential drop (from left to right) across each resistance is

${{12} \over 3} = 4V$

$\therefore$ ${V_{MS}} = 2 \times 4 = 8V$

${V_{NQ}} = 1 \times 4 = 4V$

or, ${V_S} < {V_Q}$

The current in the net circuit is

$I = {V \over {{R_{effective}}}}$

$ = {V \over {{R_1} + [({R_2}{R_L})/({R_1} + {R_L})]}}$

$ = {{24} \over {2 \times {{10}^{ - 3}} + [(6 \times 1.5 \times {{10}^{ - 6}})/(7.5 \times {{10}^{ - 3}})]}} = 7.5$ mA

The potential across $R_L$ is

${V_L} = I \times {{{R_2}{R_L}} \over {{R_2} + {R_L}}}$

$ = 7.5 \times {10^{ - 3}} \times 1.2 \times {10^3} = 9$ V

Therefore,

$I = {9 \over {{R_L}}} = 6$ mA

The ratio of power dissipated in $R_1$ and $R_2$ is

${{I{R_1}} \over {(I - i){R_2}}} = {{{{(7.5 \times {{10}^{ - 3}})}^2} \times 2 \times {{10}^3}} \over {{{(1.5 \times {{10}^{ - 3}})}^2}6 \times {{10}^3}}} = 0.75$ J

The power dissipated in $R_L$ is

${{V_L^2} \over {{R_L}}} = {{{9^2}} \over {1.5 \times {{10}^3}}} = 54$ mJ

If $R_1$ and $R_2$ are interchanged, then

$I' = {V \over {{R_{effective}}}}$

$ = {V \over {{R_2} + [({R_1}{R_L})/({R_L} + {R_1})]}}$

$ = {{24} \over {6 \times {{10}^{ - 3}} + [(2 \times 1.5 \times {{10}^{ - 6}})/3.5 \times {{10}^{ - 3}}]}} = 3.5$ mA

Hence, the voltage drop across the load is

$V{'_L} = I \times {{{R_1}{R_L}} \over {{R_1} + {R_L}}} = 1$ V

Therefore, $i' = 2$ mA. The magnitude of the power dissipated is the ratio between ${i^2}$ and ${(i')^2}$:

${{{P_1}} \over {{P_2}}} = {{{i^2}} \over {{{(i')}^2}}} = {{{6^2}} \over {{2^2}}} = 9$

Given, current carrying circular coil does not have an electric field and a gravitational field.

But it has a magnetic field and induced electric field if the magnetic field due to current in the coil is changing with time.

Let the charge per unit length on the axis be $\lambda$(t) at a time t. The electric field due to this line charge at a radial distance r is given by

$\overrightarrow E (r,t) = {{\lambda (t)} \over {2\pi \in r}}\widehat r$. ...... (1)

The free charges inside the conductor start moving radially due to the presence of electric field. The current density at a point is defined as the current flowing across a unit area placed perpendicular to the direction of current flow. The current density at a point is related to the electric field at that point by Ohm's law i.e.,

$\overrightarrow j (r,t) = \sigma \overrightarrow E (r,t)$, ....... (2)

where $\sigma$ is the electrical conductivity of the conductor. The current through a cylindrical shell of radius r and length l is given by

$I = \int_{surface} {\overrightarrow j (r,t)\,.\,d\overrightarrow A = j(r,t)(2\pi rl)} $. ...... (3)

By conservation of charge, the current I is equal to the rate of decrease of charge q on axial line segment of length l i.e.,

$I = - {{dq} \over {dt}} = - l{{d\lambda (t)} \over {dt}}$. ...... (4)

From equations (1) - (4)

${{d\lambda (t)} \over {dt}} = - {{\sigma \lambda (t)} \over \in }$.

Integrate with initial condition $\lambda$(t = 0) = $\lambda$₀ to get

$\lambda (t) = {\lambda _0}\exp {e^{ - {{\sigma t} \over \in }}}$,

Substitute $\lambda$(t) in equation (2) to get

$\overrightarrow j (r,t) = {{\sigma {\lambda _0}} \over {2\pi \in r}}{e^{ - {{\sigma t} \over \in }}}$.

Note that current density varies as 1/r with radial distance r. It decreases exponentially with time and becomes zero at t $\to$ $\infty$.

when a metre bridge is balanced, then

${{{R_1}} \over x} = {{{R_2}} \over {(100 - x)}}$ ..... (i)

From figure,

R₁ = R, R₂ = 90 $\Omega$, x = 40 cm

Then, ${R \over {40}} = {{90} \over {(100 - 40)}} = {{90} \over {60}}$

$\Rightarrow$ R = 60 $\Omega$

Now, for $\Delta$R taking natural log on both sides of eqn. (i),

$\ln {R_1} = \ln {R_2} + \ln x - \ln (100 - x)$

or, $\ln R = \ln x - \ln (100 - x) + \ln (90)$

On differentiating,

${{\Delta R} \over R} = {{\Delta x} \over x} - {{\Delta (100 - x)} \over {(100 - x)}}$

$ \Rightarrow {{\Delta R} \over R} = {{\Delta x} \over x} + {{\Delta x} \over {(100 - x)}}$

$ \Rightarrow \Delta R = \left( {{{0.1} \over {40}} + {{0.1} \over {60}}} \right) \times 60 = {{0.5} \over {120}} \times 60 = 0.25\Omega $

$\therefore$ Required value of R = (60 $\pm$ 0.25) $\Omega$

P = Vi

$\therefore$ $i = {P \over V} = {{600 \times {{10}^3}} \over {4000}} = 150$ A

Total resistance of cables,

R = 0.4 $\times$ 20 = 8$\Omega$

$\therefore$ Power loss in cables = i²R

= (150)² (8)

= 180000 W = 180 kW

This loss is 30% of 600 kW.

Corrected length L₁ (AJ) = 52 + 1 = 53 cm

Corrected length L₂ (BJ) = (100 $-$ 52) + 2 = 50 cm

For a balanced Wheatstone bridge,

${X \over {10}} = {{{L_1}} \over {{L_2}}} = {{53} \over {50}}$

$\Rightarrow$ X = 10.6 $\Omega$

The power of the bulb is

$P = {{{V^2}} \over R}$

Therefore,

$100 = {{{V^2}} \over {{R_{100}}}} \Rightarrow {1 \over {{R_{100}}}} = {{100} \over {{V^2}}}$

where R₁₀₀ is the resistance (at any temperature) corresponds to 100 W. Similarly,

$60 = {{{V^2}} \over {{R_{60}}}} \Rightarrow {1 \over {{R_{60}}}} = {{60} \over {{V^2}}}$ and $40 = {{{V^2}} \over {{R_{40}}}} \Rightarrow {1 \over {{R_{40}}}} = {{40} \over {{V^2}}}$

From these equations, we get

${P_{100}} > {P_{60}} > {P_{40}} \Rightarrow {1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$

To verify Ohm's law, we need to measure the voltage across the test resistance R_T and current passing through it. The voltage can be measured by connecting a high resistance R₁ in series with galvanometer. This combination becomes a voltmeter and should be connected in parallel to R_T. The current can be measured by connecting a low resistance R₂ (shunt) in parallel with galvanometer. This combination becomes an ammeter and should be connected in series to measure the current through R_T.

The sheet is of square shape with thickness t, width $\omega$ = L, and length l = L. The resistance between the two opposite faces is given by

$R = {{\rho l} \over A} = {{\rho l} \over {wt}} = {{\rho L} \over {Lt}} = {\rho \over t}$.

We know that $P = {{{V^2}} \over R}$

If potential is constant we have

$P \propto {1 \over R}$

Case I

Hence, this is a clear case of wheat stone bridge R$_1$ = 1 $\Omega$

Case II

Equivalent resistance (R$_2$)

$\therefore$ ${1 \over {{R_2}}} = {1 \over 2} + {1 \over 1} + {1 \over 2} = {{1 + 2 + 1} \over 2} = {4 \over 2} = 2\,\Omega $

Hence, ${R_2} = {1 \over 2} = 0.5\,\Omega $

It is clear that the equivalent resistance (R$_2$) will be less than 1 $\Omega$.

Case III

Hence, R$_3$ = 2 $\Omega$

Since, R$_2$ < R$_1$ < R$_2$

$\therefore$ P$_2$ > P$_1$ > P$_3$ ($ \because $ $P \propto {1 \over R}$)

When the temperature of a metal increases, its resistance will increase.

Therefore statement-2 is correct.

For meter bridge, when null point N is obtained, we get

${X \over l} = {R \over {100 - l}}$

When the unknown resistance is put inside an enclosure, maintained at a high temperature, then X increases. To maintain the ratio of null point, $l$ should also increase. But if we want to keep the null point at the initial position (i.e., if we want to change the value of $l$) to maintain the ratio, R should be increased.

Therefore, statement - 1 is false.

Given, that x is greater than 2 $\Omega$, when the bridge is balanced then,

$\frac{R}{l}=\frac{x}{100-l}$ ..... (i)

(from metre bridge balanced bridge formula)

$\Rightarrow 100R-Rl=lx$

$\Rightarrow 200-2l=lx~~(\because R=2\Omega)$

$l=\frac{200}{x+2}$ ..... (ii)

Where, $l$ = length of segment of wire from one end where null point is obtained.

Now, these resistances are interchanged, the Jockey shifts 20 cm. So,

$\frac{x}{l+20}=\frac{2}{80-l}$ ...... (iii)

From equation (iii) we have,

$80x=l(x+2)+40$

$\Rightarrow 80x=\left(\frac{200}{x+2}\right)(x+2)+40$

$\Rightarrow x=\frac{240}{80}=3\Omega$

$\therefore x=3\Omega$

Resistance of $1^{\text {st }}$ half part is $\mathrm{R}_1=\frac{\rho l_1}{\mathrm{~A}_1}=\frac{\rho(1 / 2)}{\pi(4 r)^2}$

Resistance of $2^{\text {nd }}$ part is, $\mathrm{R}_2=\frac{\rho l_2}{\mathrm{~A}_2}=\frac{\rho(1 / 2)}{(2 r)^2 \pi}$

$Thus, \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\frac{\rho(1 / 2)}{\pi 16 r^2}}{\frac{\rho(1 / 2)}{4 \pi r^2}}=\frac{1}{4} $

(A) Power loss : $P_1=I^2 R_1$ and $P_2=I^2 R_2$,

So $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4} \Rightarrow \mathrm{P}_1 .4=\mathrm{P}_2$

(B) Voltage drop : $\mathrm{V}_1=\mathrm{IR}_1$ and $\mathrm{V}_2=\mathrm{IR}_2$

So $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{1}{4}$

(C) Current density : $\mathrm{J}_1=\frac{\mathrm{I}}{\mathrm{A}_1}$ and $\mathrm{J}_2=\frac{\mathrm{I}}{\mathrm{A}_2}$

So $\frac{\mathrm{J}_1}{\mathrm{~J}_2}=\frac{\mathrm{A}_2}{\mathrm{~A}_1}=\frac{\pi r^2}{\pi(2 r)^2}=\frac{1}{4}$

(D) Electric field: $\frac{E_1}{E_2}=\frac{J_1}{J_2}=\frac{1}{4} \quad(J=\sigma E)$

At null points, the Wheatstone bridge will be balanced.

Therefore,

$\frac{\mathrm{X}}{r_{1}}=\frac{\mathrm{R}}{r_{2}} \text { (or) } \mathrm{X}=\mathrm{R} \frac{r_{1}}{r_{2}}$

Where $\mathrm{R}$ is a constant and $r_{1}$ and $r_{2}$ are variable. The maximum fraction error is

${{\Delta x} \over x} = {{\Delta {r_1}} \over {{r_1}}} + {{\Delta {r_2}} \over {{r_2}}}$

$\begin{array}{rr} & \mathrm{R}=\mathrm{R}_{1}, \mathrm{R}=\mathrm{R}_{2}, \mathrm{R}=\mathrm{R}_{3} \\ \text { Here, } & \Delta r_{1}=\Delta r_{2}=y \text { (say), }\end{array}$

$y$ can be considered as least count of the scale,

Then, $ \frac{\Delta x}{x}=y\left[\frac{r_{2}+r_{1}}{r_{1} r_{2}}\right]$

For $\frac{\Delta x}{x}$ to be minimum, $r_{1} \times r_{2}$ should be maximum [Since $r_{1}+r_{2}=\mathrm{c}$ (constant $)$ ]

Let,

$\begin{aligned} \mathrm{E} & =r_{1} \times r_{2}=r_{1} \times\left(c-r_{1}\right) \\ & =r_{1} c-r_{1}^{2} \\ \therefore \frac{d \mathrm{E}}{d r_{1}} & =c-2 r_{1}=0 \text { or } \frac{c}{2}=r_{1}\end{aligned}$

(or) $r_{2}=c / 2 \text { (or) } r_{1}=r_{2}$