Physics 5 Marks Questions

1. With rise of temperature, the collision of electrons with fixed lattice ions/ atoms increases so that relaxation time $(\tau)$ decreases. Consequently, the conductivity of metals decreases with rise of temperature.
2. Conductivity of ionic conductor increases with increase of temperature because with increase of temperature, the ionic bonds break releasing positive and negative ions which are charge carriers in ionic conductors.
3. In the case of a semiconductors, when temperature increases, covalent bonds break and charge carriers (electrons and holes) become free i.e., n increases, so conductivity increases with rise of temperature.

$\frac{10\times\text{R}'}{10+\text{R}}>8$

$\Rightarrow\ 2\text{R}'>80$

$\Rightarrow\ \text{R}'>40$

$\frac{10\times\frac{3}{4}\text{R}'}{10+\text{R}'}>8$

$\Rightarrow\ 7.5\text{R}'<80+8\text{R}'$

$\text{R}'>160$

$\Rightarrow\ 160<\text{R}'<200$

1. The change in potential in traversing a resistance in the direction of current is -iR while in the opposite direction +iR.

2. The change in potential in traversing an emf source from negative to positive terminal is +E while in the opposite direction -E irrespective of the direction of current in the circuit.

$\text{G}=25\Omega,\text{lg}=1\text{ma},\text{l}=2\text{A},\text{S}=?$

Potential across A B is same

$25\times10^{-3}=(2-10^{-3})\text{S}$

$\Rightarrow\text{S}=\frac{25\times10^{-3}}{2-10^{-3}}=\frac{25\times10^{-3}}{1.999}$

$=12.5\times10^{-3}=1.25\times10^{-2}.$

$\text{i}_1\times20=\text{i}_2\times10$

$\Rightarrow\frac{\text{i}_1}{\text{i}_2}=\frac{10}{20}=\frac{1}{2}$

i₁ = 4mA, i₂ = 8mA

Current in $20\text{K}\Omega$ resistor = 4mA

Current in $10\text{K}\Omega$ resistor = 8mA

Current in $100\text{K}\Omega$ resistor = 12mA

$\text{V}=\text{V}_1+\text{V}_2+\text{V}_3$

$=5\text{K}\Omega\times12\text{mA}+10\text{K}\Omega\times8\text{mA}\\+100\text{K}\Omega\times12\text{mA}$

$=60+80+1200=1340\text{ volts}.$

1. $\text{R}_\text{eff}=\frac{\frac{15\times5}{6}\times\frac{15}{6}}{\frac{15\times5}{6}+\frac{15}{6}}=\frac{\frac{15\times5\times15}{6\times6}}{\frac{75+15}{6}}$

$=\frac{15\times5\times15}{6\times90}=\frac{25}{12}=2.08\Omega$

2. Across AC,

$\text{R}_\text{eff}=\frac{\frac{15\times4}{6}\times\frac{15\times2}{6}}{\frac{15\times4}{6}+\frac{15\times2}{6}}=\frac{\frac{15\times4\times15\times2}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times4\times15\times2}{6\times90}=\frac{10}{3}=3.33\Omega$

3. Across AD,

$\text{R}_\text{eff}=\frac{\frac{15\times3}{6}\times\frac{15\times3}{6}}{\frac{15\times3}{6}+\frac{15\times3}{6}}=\frac{\frac{15\times3\times15\times3}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times3\times15\times3}{6\times90}=\frac{15}{4}=3.75\Omega.$

1. For same potential gradient of potentiometer wire, the formula for comparison of emfs of cells is,

$\frac{\varepsilon_2}{\varepsilon_1}=\frac{\text{l}_2}{\text{l}_1}\Rightarrow\frac{\varepsilon}{\varepsilon_\text{s}}=\frac{\text{l}}{\text{l}_\text{s}}$

$\varepsilon=\frac{\text{l}}{\text{l}_\text{s}}\varepsilon_\text{s}$

$\varepsilon_\text{s}=$ emf of standard cell = 1.02V

I_s = balancing with length standard cell = 67.3cm

l = balancing length with cell of unknown emf = 82.3cm

$\therefore$ Unknown emf $\varepsilon=\frac{(82.3\text{cm})}{(67.3\text{cm})}\times1.02\text{V}=1.25\text{V}$

2. The purpose of high resistance is to reduce the current through the galvanometer. When jockey is far from the balance point, this saves the standard cell from being damaged.
3. The balance point is not affected by the presence of high resistance because in balanced-position there is no current in cell-circuit (secondary circuit).
4. No, the balance point is not affected by the internal resistance of driver cell, because we have already set the constant potential gradient of wire.
5. No, since for the working of potentiometer the emf of driver cell must be greater than emf of secondary circuit.
6. No, the circuit will have to be modified by putting variable resistance (R) in series with the driver cell the value of R is so adjusted that potential drop across wire is slightly greater than emf of secondary cell, so that the balance point may be obtained at a longer length. This will reduce the error and increase the accuracy of measurement.

$\text{V}=250\text{V}$

$\text{P}=100\text{w}$

$\text{R}=\frac{\text{v}^2}{\text{P}}=\frac{(250)^2}{100}=625\Omega$

Resistance of wire $=\text{R}=\frac{\text{fl}}{\text{A}}=1.7\times10^{-8}\times\frac{10}{5\times10^{-6}}$

$=0.034\Omega$

$\therefore$ The effect in resistance $=625.034\Omega$

$\therefore$ The current in the conductor $=\frac{\text{V}}{\text{R}}=\Big(\frac{220}{625.034}\Big)\text{A}$

$\therefore$ The power supplied by one side of connecting wire $=\Big(\frac{220}{625.034}\Big)^2\times0.034$

$\therefore$ The total power supplied $=\Big(\frac{220}{625.034}\Big)^2\times0.034\times2$

$\Rightarrow0.0084\text{w}=8.4\text{mw}$

1. $\text{R}_\text{eff}=\frac{180}{3}=60\Omega$

$\text{i}=\frac{60}{60}=1\text{A}$

2. $\text{R}_\text{eff}=\frac{180}{2}=90\Omega$

$\text{i}=\frac{60}{90}=0.67\text{A}$

3. $\text{R}_\text{eff}=180\Omega\Rightarrow\text{i}=\frac{60}{180}=0.33$

1. When switch is open, no current passes through the ammeter. In the upper part of the circuit the Voltmenter has $\infty$ resistance. Thus current in it is 0.

$\therefore$ Voltmeter read the emf. (There is not Pot. Drop across the resistor).

2. When switch is closed current passes through the circuit and if its value of i.

The voltmeter reads

$\in-\text{ir}=1.45$

$\Rightarrow1.52-\text{ir}=1.45$

$\Rightarrow\text{ir}=0.07$

$\Rightarrow1\text{r}=0.07\Rightarrow\text{r}=0.07\Omega.$

1. 1eq. mass of the substance requires 96500 coulombs

Since the element is monoatomic, thus eq. mass = mol. Mass

6.023 × 10²³ atoms require 96500C

1 atoms require $\frac{96500}{6.023\times10^{23}}\text{C}=1.6\times10^{-19}\text{C}$

2. Since the element is diatomic eq.mass = $=\frac{1}{2}\text{mol.mass}$

$\therefore\Big(\frac{1}2{}\Big)\times6.023\times10^{23}$ atoms 2eq. 96500C

$\Rightarrow1\ \text{atom require}=\frac{96500\times2}{6.023\times10^{23}}$

$=3.2\times10^{-19}\text{C}$

$\text{i}_\text{min}=\frac{5.5\times3}{110}=0.15$

$\text{i}_\text{min}=\frac{5.5\times3}{20}=\frac{16.5}{20}=0.825.$

1. Current through (1) $4\Omega$ resistor = 0
2. Current through (2) and (3)

net E = 4V - 2V = 2V

(2) and (3) are in series,

$\text{R}_\text{eff}=4+6=10\Omega$

$\text{i}=\frac{2}{10}=0.2\text{A}$

Current through (2) and (3) are 0.2A.

1. Volume of water boiled = 4 × 200cc = 800cc

T₁ = 25^°C

T₂ = 100^°C

⇒ T₂ – T₁ = 75°C

Mass of water boiled = 800 × 1 = 800gm = 0.8kg(heat req.) $=\text{MS}\Delta\theta=0.8\times4200\times75=252000\text{J}.$

1000 watt-hour = 1000 × 3600 watt-sec = 1000 × 3600J

No. of units $=\frac{252000}{1000\times3600}=0.07=7\ \text{paise}$

2. $\text{Q}=\text{mS}\Delta\text{T}=0.8\times4200\times95\text{J}$

No. of units $=\frac{0.8\times4200\times95}{1000\times3600}=0.0886\approx0.09$

Money consumed = 0.09 Rs = 9 paise.

If the wire is connected to the potentiometer wire so that $\frac{\text{R}_\text{AD}}{\text{R}_\text{DB}}=\frac{8}{12},$ then according to wheat stone’s bridge no current will flow through galvanometer.

$\frac{\text{R}_\text{AB}}{\text{R}_\text{DB}}=\frac{\text{L}_\text{AB}}{\text{L}_\text{B}}=\frac{8}{12}=\frac{2}{3}$ (Acc. To principle of potentiometer).

$\text{I}_\text{AB}+\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\text{I}_\text{DB}\frac{2}{3}+\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\Big(\frac{2}{3}+1\Big)\text{I}_\text{DB}=40\text{cm}$

$\Rightarrow\frac{5}{3}\text{I}_\text{DB}=40$

$\Rightarrow\text{L}_\text{DB}=\frac{40\times3}{5}=24\text{cm}.$

$\text{I}_\text{AB}=(40-24)\text{cm}=16\text{cm}.$

$\text{R}_\text{eff}=(1150+50)\Omega=1200\Omega$

$\text{i}=\Big(\frac{12}{1200}\Big)\text{A}=0.01\text{A}.$

(The resistor of $50\Omega$ can tolerate)

Let R be the resistance of sheet used.

The potential across both the resistors is same.

$0.01\times50=1.99\times\text{R}$

$\Rightarrow\text{R}=\frac{0.01\times50}{1.99}=\frac{50}{199}=0.251\Omega$

1. Potential difference between A and B is 6V.

B is at 0 potential.

Thus potential of A point is 6V.

The potential difference between Ac is 4V.

$\text{V}_\text{A}-\text{V}_\text{C}=0.4$

$\text{V}_\text{C}=\text{V}_\text{A}-4=6-4=2\text{V}$

2. The potential at $\text{D}=2\text{V},\text{V}_\text{AD}=4\text{V};\text{V}_\text{BD}=\text{OV}$

Current through the resisters R₁ and R₂ are equal

Thus, $\frac{4}{\text{R}_1}=\frac{2}{\text{R}_2}$

$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=2$

$\Rightarrow\frac{\text{I}_1}{\text{I}_2}=2$ (Acc. to the law of potentiometer)

$\text{I}_1+\text{I}_2=100\text{cm}$

$\Rightarrow\text{I}_1+\frac{\text{I}_1}{2}=100\text{cm}\Rightarrow\frac{3\text{I}_1}{2}=100\text{cm}$

$\Rightarrow\text{I}_1=\frac{200}{3}\text{cm}=66.67\text{cm}.$

$\text{AD}=66.67\text{cm}$

3. When the points C and D are connected by a wire current flowing through it is 0 since the points are equipotential.

4. Potential at A = 6 v

Potential at C = 6 - 7.5 = -1.5V

The potential at B = 0 and towards A potential increases.

Thus -ve potential point does not come within the wire.

1. Total emf = n₁E

in 1 row

Total emf in all news = n₁E

Total resistance in one row = n₁r

Total resistance in all rows $=\frac{\text{n}_1\text{r}}{\text{n}_2}$

Net resistance $=\frac{\text{n}_1\text{r}}{\text{n}_2}+\text{R}$

Current $=\frac{\text{n}_1\text{E}}{\frac{\text{n}_1}{\text{n}_2\text{r}}+\text{R}}=\frac{\text{n}_1\text{n}_2\text{E}}{\text{n}_1\text{r}+\text{n}_2\text{R}}$

2. $\text{I}=\frac{\text{n}_1\text{n}_2\text{E}}{\text{n}_1\text{r}+\text{n}_2\text{R}}$

for I = max,

n₁r + n₂R = min

$\Rightarrow\Big(\sqrt{\text{n}_1\text{r}}-\sqrt{\text{n}_2\text{R}}\Big)^2+2\sqrt{\text{n}_1\text{rn}_2\text{R}}=\text{min}$

it is min, when

$\sqrt{\text{n}_1\text{r}}=\sqrt{\text{n}_2\text{R}}$

$\Rightarrow\text{n}_1\text{r}=\text{n}_2\text{R}$

I is max when n₁r = n₂R.

1. Potential difference = E across resistor

2. Current in the circuit $=\frac{\text{E}}{\text{R}}$

3. Pd. Across capacitor $=\frac{\text{E}}{\text{R}}$

4. Energy stored in capacitor $=\frac{1}{2}\text{CE}^2$

5. Power delivered by battery $=\text{E}\times\text{I}=\text{E}\times\frac{\text{E}}{\text{R}}=\frac{\text{E}^2}{\text{R}}$

6. Power converted to heat $=\frac{\text{E}^2}{\text{R}}$

1. In circuit ABFGA,

$\text{i}_150+2\text{i}+\text{i}-4.3=0$

$\Rightarrow50\text{i}_1+3\text{i}=4.3\ ...(1)$

In circuit BEDCB,

$50\text{i}-200\text{i}+200\text{i}_1=0$

$\Rightarrow250\text{i}_1-200\text{i}=0$

$\Rightarrow50\text{i}_1-40\text{i}=0\ ...(2)$

From (1) and (2)

$43\text{i}=4.3\Rightarrow\text{i}=0.1$

$5\text{i}_1=4\times\text{i}=4\times0.1$

$\Rightarrow\text{i}_1=\frac{4\times0.1}{5}=0.08\text{A}$

Ammeter reads a current $=\text{i}=0.1\text{A}$

Voltmeter reads a potential difference equal to $\text{i}_1\times50=0.08\times50=4\text{V}.$

2. In circuit ABEFA,

$50\text{i}_1+2\text{i}_1+1\text{i}-4.3=0$

$\Rightarrow52\text{i}_1+\text{i}=4.3$

$\Rightarrow200\times52\text{i}_1+200\text{i}=4.3\times200\ ...(1)$

In circuit BCDEB,

$(\text{i}-\text{i}_1)200-\text{i}_12-\text{i}_150=0$

$\Rightarrow200\text{i}-200\text{i}_1-2\text{i}_1-50\text{i}_1=0$

$\Rightarrow200\text{i}-252\text{i}_1=0\ ...(2)$

From (1) and (2)

$\text{i}_1(10652)=4.3\times2\times100$

$\Rightarrow\text{i}_1=\frac{4.3\times2\times100}{10652}=0.08$

$\text{i}=4.3-52\times0.08=0.14$

Reading of the ammeter = 0.08a

Reading of the voltmeter $=(\text{i}-\text{i}_1)200=(0.14-0.08)\times200=12\text{V}.$

1. The circuit is shown in figure.
2. Suppose I₁ are I₂ current drawn from cells $\varepsilon_1$ and $\varepsilon_2$ respectively, then

According to Kirchhoff’s junction law, current in $\text{R}=5\Omega,\text{I}=\text{I}_1+\text{I}_2.$

Applying Kirchhoff’s second law to mesh ABFEA

$1\times\text{I}_1+1.5-5(\text{I}_1+\text{I}_2)=0$

$\Rightarrow\text{I}_1+5\text{I}_2=1.5\dots\text{(i)}$

Applying Kirchhoff’s second law to mesh CDEFC

$-2\text{I}_2+2-5(\text{I}_1+\text{I}_2=0$

$\Rightarrow\text{5I}_1+7\text{I}_2=2\dots\text{(ii)}$

Solving equation (i) and (ii), we get

$\text{I}_1=\frac{1}{34}\text{A},\text{I}_2=\frac{9}{34}\text{A}$

$\text{I}=\text{I}_1+\text{I}_2=\frac{1}{34}+\frac{9}{34}=\frac{10}{34}$

Potential difference avross $\text{R}=5\Omega$ resistor

$(\text{I}_1+\text{I}_2)\text{R}=\frac{10}{34}\times5=\frac{25}{17}\text{Volt}$

1. $0.1\text{i}_1+1\text{i}_1-6+1\text{i}_1-6=0$

$\Rightarrow0.1\text{i}_1+1\text{i}_1+1\text{i}_1=12$

$\Rightarrow\text{i}_1=\frac{12}{2.1}$

$\text{ABCDA}$

$\Rightarrow0.1\text{i}_2+1\text{i}-6=0$

$\Rightarrow0.1\text{i}_2+1\text{i}$

$\text{ADEFA}$

$\Rightarrow\text{i}-6+6-(\text{i}_2-\text{i})1=0$

$\Rightarrow\text{i}-\text{i}_2+\text{i}=0$

$\Rightarrow2\text{i}-\text{i}_2=0\Rightarrow-2\text{i}\pm0.2\text{i}=0$

$\Rightarrow\text{i}_2=0.$

2. $1\text{i}_1+1\text{i}_1-6+1\text{i}_1=0$

$\Rightarrow3\text{i}_1=12\Rightarrow\text{i}_1=4$

$\text{DCFED}$

$\Rightarrow\text{i}_2+\text{i}-6=0\Rightarrow\text{i}_2+\text{i}=6$

$\text{ABCDA}$

$\text{i}_2+(\text{i}_2-\text{i})-6=0$

$\Rightarrow\text{i}_2+\text{i}_2-\text{i}=6\Rightarrow2\text{i}_2-\text{i}=6$

$\Rightarrow-2\text{i}_2\pm2\text{i}=6\Rightarrow\text{i}=-2\text{i}_2+\text{i}=6$

$\Rightarrow\text{i}_2-2=6\Rightarrow\text{i}_2=8$

$\frac{\text{i}_1}{\text{i}_2}=\frac{4}{8}=\frac{1}{2}.$

3. $10\text{i}_1+1\text{i}_1-6+1\text{i}_1-6=0$

$\Rightarrow12\text{i}_1=12\Rightarrow\text{i}_1=1$

$10\text{i}_2-\text{i}_1-6=0$

$\Rightarrow10\text{i}_2-\text{i}_1=6$

$\Rightarrow10\text{i}_2+(\text{i}_2-\text{i})1-6=0$

$\Rightarrow11\text{i}_2=6$

$\Rightarrow-\text{i}_2=0.$

1. $\text{R}_\text{eff}=\frac{100\times400}{500}+200=280$

$\text{i}=\frac{84}{280}=0.3$

$100\text{i}=(0.3-\text{i})400$

$\Rightarrow\text{i}=1.2-4\text{i}$

$\Rightarrow5\text{i}=1.2\Rightarrow\text{i}=0.24$

Voltage measured by the voltmeter $=\frac{0.24\times100}{24\text{V}}$

2. If voltmeter is not connected

$\text{R}_\text{eff}=(200+100)=300\Omega$

$\text{i}=\frac{84}{300}=0.28\text{A}$

Voltage across $100\Omega=(0.28\times100)=28\text{V}.$

1. $\text{C} = 12.0 \mu\text{F} = 12 \times 10^{-6}$

$\text{emf} = 6.00 \text{V, R} = 1\Omega$

$\text{t} = 12 \mu\text{c}, \text{ i} = \text{i}_0 \text{e}^{\frac{-\text{t}}{\text{RC}}}$

$=\frac{\text{CV}}{\text{T}}\times\text{e}^\frac{-\text{t}}{\text{RC}}=\frac{12\times10^{-6}\times6}{12\times10^{-6}}\times\text{e}^{-1}$

$=2.207=2.1\text{A}$

2. Power delivered by battery

We known, $\text{V}=\text{V}_0\text{e}^{\frac{-\text{t}}{\text{RC}}}$ (where V and V₀ are potential VI)

$\text{VI}=\text{V}_0\text{I}\text{e}^{\frac{-\text{t}}{\text{RC}}}$

$\Rightarrow\text{VI}=\text{V}_0\text{I}\times\text{e}^{-1}=6\times6\times\text{e}^{-1}=13.24\text{W}$

3. $\text{U}=\frac{\text{CV}^2}{\text{T}}\Big(\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)^2$ $\Big[\frac{\text{CV}^2}{\text{T}}$ = energy drawing per unit time$\Big]$

$=\frac{12\times10^{-6}\times36}{12\times10^{-6}}\times\big(\text{e}^{-1}\big)^2=4.872.$

1. $\text{R}_\text{eff}=\frac{\Big(\frac{2\text{r}}{2}\Big)\times\text{r}}{\Big(\frac{2\text{r}}{2}\Big)+\text{r}}$

$=\frac{\text{r}^2}{2\text{r}}=\frac{\text{r}}{2}$

2. At 0 current coming to the junction is current going from BO = Current going along OE.

Current on CO = Current on OD

Thus it can be assumed that current coming in OC goes in OB.

Thus the figure becomes

$\Bigg[\text{r}+\Big(\frac{2\text{r}\times\text{r}}{3\text{r}}\Big)+\text{r}\Bigg]=2\text{r}+\frac{2\text{r}}{3}=\frac{8\text{r}}{3}$

$\text{R}_\text{eff}=\frac{\Big(\frac{8\text{r}}{6}\Big)\times2\text{r}}{\Big(\frac{8\text{r}}{6}\Big)+2\text{r}}=\frac{\frac{8\text{r}^2}{3}}{\frac{20\text{r}}{6}}=\frac{8\text{r}^2}{3}\times\frac{6}{20}=\frac{8\text{r}}{10}=4\text{r}.$

$\text{R}_\text{eff}=\frac{12}{8}+1=\frac{5}{2}$

$\text{i}=\frac{6}{\frac{5}{2}}=\frac{12}{5}\text{Amp.}$

$\text{i}'6=({\text{i-i}'})2\Rightarrow\text{i}'6=\frac{12}{\frac{5}{2}}\times2-2\text{i}$

$8\text{i}'=\frac{24}{5}\Rightarrow\text{i}'=\frac{24}{5\times8}=\frac{3}{5}\text{Amp}$

$\text{i-i}'=\frac{12}{5}-\frac{3}{5}=\frac{9}{5}\text{Amp}$

1. Heat $=\text{i}^2\text{RT}=\frac{9}{5}\times\frac{9}{5}\times2\times15\times60=5832$

2000J of heat raises the temp. by 1K

5832J of heat raises the temp. by 2.916K.

2. When $6\Omega$ resistor get burnt $\text{R}_\text{eff}=1+2=3\Omega$

$\text{i}=\frac{6}{3}=2\text{Amp}$

Heat = 2 × 2 × 2 × 15 × 60 = 7200J

2000J raises the temp. by 1K

7200J raises the temp by 3.6k.

1. Let the equation resistance of the combination be R.

$\Big(\frac{2\text{R}}{\text{R}+2}\Big)+1=\text{R}$

$\Rightarrow\frac{2\text{R}+\text{R}+2}{\text{R}+2}=\text{R}\Rightarrow3\text{R}+2=\text{R}^2+2\text{R}$

$\Rightarrow\text{R}^2-\text{R}-2=0$

$\Rightarrow\text{R}=\frac{+1\pm\sqrt{1+4\times1\times2}}{2.1}$

$=\frac{1\pm\sqrt{9}}{2}=\frac{1\pm3}{2}=2\Omega.$

2. Total current sent by battery $=\frac{6}{\text{R}_\text{eff}}=\frac{6}{2}=3$

Potential between A and B

$3.1+2\text{i}=6$

$\Rightarrow3+2\text{i}=6\Rightarrow2\text{i}=3$

$\Rightarrow\text{i}=1.5\text{a}$

1.  $\text{I}=\frac{\varepsilon}{\text{R}+\frac{\rho_1\text{l}}{\text{A}_1}}$ for set A and $\text{I}=\frac{\in}{\text{R}+\frac{\rho_1\text{l}}{2\text{A}_1}+\frac{\rho_2\text{l}}{2\text{A}_2}}\text{for set B}$

Equating the above two expressions, we have,

$\frac{\varepsilon}{\text{R}+\frac{\rho_1\text{l}}{2\text{A}_1}}=\frac{\varepsilon}{\text{R}+\frac{\rho_1\text{l}}{2\text{A}_1}+\frac{\rho_2\text{l}}{2\text{A}_2}}$

$\Rightarrow\text{R}+\frac{\rho_1\text{l}}{\text{A}_1}=\text{R}+\frac{\rho_1\text{l}}{2\text{A}_1}+\frac{\rho_1\text{l}}{2\text{A}_2}$

$\Rightarrow\frac{\rho_1\text{l}}{\text{A}_1}-\frac{\rho_1\text{l}}{2\text{A}_1}=\frac{\rho_2\text{l}}{2\text{A}_2}\dots\text{(i})$

$\Rightarrow\frac{\rho_1}{\text{A}_1}=\frac{\rho_2}{\text{A}_2}$

2. Potential gradient of the potentiometer wire for set $\text{A},\text{K}=\text{I}\frac{\rho_1}{\text{A}_1}$

Potential drop across the potentiometer wire in set B

$\text{V}=\text{I}\Big(\frac{\rho_1\text{l}}{2\text{A}_1}+\frac{\rho_2\text{l}}{2\text{A}_2}\Big)\Rightarrow\text{V}=\frac{1}{2}\Big(\frac{\rho_1}{\text{A}_1}+\frac{\rho_2}{\text{A}_2}\Big)\text{I}$

$\text{K}'=\frac{\text{I}}{2}\Big(\frac{\text{r}_1}{\text{A}_1}+\frac{\text{r}_2}{\text{A}_2}\Big),$

Using the condition (i), we get

$\text{K}'=\text{I}\frac{\rho_1}{\text{A}_1},$ which is equal to K.

Therefore, balancing length obtained in the two sets is same.