JEE Current Electricity Questions

To solve this problem, we need to evaluate two statements about non-ideal batteries connected in parallel. A non-ideal battery has an internal resistance, so we must consider both the electromotive force (emf) and the internal resistance for each battery.

Consider two batteries:

• Battery 1: emf = $$E_1$$, internal resistance = $$r_1$$
• Battery 2: emf = $$E_2$$, internal resistance = $$r_2$$

When connected in parallel, the equivalent emf $$E_{eq}$$ and equivalent internal resistance $$r_{eq}$$ are given by the formulas:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

We will analyze each statement separately.

Statement-II Analysis: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.

Using the formula for equivalent internal resistance:

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$$

Since $$r_1 > 0$$ and $$r_2 > 0$$, we can compare $$r_{eq}$$ with $$r_1$$:

$$\frac{r_{eq}}{r_1} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_1} = \frac{r_2}{r_1 + r_2} < 1$$

because $$r_2 < r_1 + r_2$$. Therefore, $$r_{eq} < r_1$$.

Similarly, comparing $$r_{eq}$$ with $$r_2$$:

$$\frac{r_{eq}}{r_2} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_2} = \frac{r_1}{r_1 + r_2} < 1$$

so $$r_{eq} < r_2$$.

Thus, $$r_{eq}$$ is always less than both $$r_1$$ and $$r_2$$. Therefore, Statement-II is true.

Statement-I Analysis: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.

Using the equivalent emf formula:

$$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$$

This expression is a weighted average of $$E_1$$ and $$E_2$$:

$$E_{eq} = E_1 \cdot \frac{r_2}{r_1 + r_2} + E_2 \cdot \frac{r_1}{r_1 + r_2}$$

Since $$\frac{r_2}{r_1 + r_2} + \frac{r_1}{r_1 + r_2} = 1$$ and both fractions are positive, $$E_{eq}$$ is a convex combination of $$E_1$$ and $$E_2$$. This means $$E_{eq}$$ lies between the minimum and maximum of $$E_1$$ and $$E_2$$.

Without loss of generality, assume $$E_1 \leq E_2$$. Then:

$$E_1 \leq E_{eq} \leq E_2$$

Therefore, $$E_{eq}$$ is not smaller than both emfs; it is at least as large as the smaller emf and at most as large as the larger emf.

For example, let $$E_1 = 10 \text{ V}$$, $$r_1 = 1 \Omega$$, $$E_2 = 5 \text{ V}$$, $$r_2 = 1 \Omega$$. Then:

$$E_{eq} = \frac{10 \times 1 + 5 \times 1}{1 + 1} = \frac{15}{2} = 7.5 \text{ V}$$

Here, $$7.5 \text{ V}$$ is greater than $$5 \text{ V}$$ ($$E_2$$) and less than $$10 \text{ V}$$ ($$E_1$$). So it is not smaller than both emfs.

If the emfs are equal, say $$E_1 = E_2 = E$$, then:

$$E_{eq} = \frac{E r_2 + E r_1}{r_1 + r_2} = E \frac{r_1 + r_2}{r_1 + r_2} = E$$

which is equal to both emfs, not smaller.

Thus, Statement-I is false because the equivalent emf is not necessarily smaller than both emfs; it lies between them or equals them.

Conclusion:

• Statement-I is false.
• Statement-II is true.

Therefore, the correct option is B: Statement-I is false but Statement-II is true.

Let the mains supply voltage be $$V$$ (same for every bulb).

Each identical bulb is rated to consume power $$p$$ when connected alone across the mains.
For a resistor-type load the power relation is

$$p = \frac{V^{2}}{R} \; -(1)$$

where $$R$$ is the resistance of one bulb.
From $$(1)$$, the resistance of each bulb is

$$R = \frac{V^{2}}{p} \; -(2)$$

When the $$n$$ bulbs are connected in SERIES, their resistances add up:

$$R_{\text{series}} = nR = n \left(\frac{V^{2}}{p}\right) = \frac{nV^{2}}{p} \; -(3)$$

The current drawn from the mains is given by Ohm’s law:

$$I = \frac{V}{R_{\text{series}}} = \frac{V}{\dfrac{nV^{2}}{p}} = \frac{p}{nV} \; -(4)$$

The total power taken from the mains by the series combination is

$$P_{\text{total}} = I^{2} R_{\text{series}}$$

Substituting from $$(4)$$ and $$(3)$$:

$$P_{\text{total}} = \left(\frac{p}{nV}\right)^{2} \times \frac{nV^{2}}{p}$$

$$\quad = \frac{p^{2}}{n^{2}V^{2}} \times \frac{nV^{2}}{p}$$

$$\quad = \frac{p}{n}$$

Hence the total power drawn by the $$n$$ bulbs in series is $$\dfrac{p}{n}$$.

Option C is correct.

All edges have equal resistance. Total wire resistance = R.
Number of edges in a tetrahedron = 6

So resistance of each edge:

$$r=\frac{R}{6}$$

Now find equivalent between two vertices (say A and B).

By symmetry, the other two vertices are at same potential → no current flows between them.

So that edge can be ignored.

Now circuit reduces to:

• Direct edge A-B: r
• Two identical paths A → (top nodes) → B
  Each path has 2 resistors in series → 2r

So we have 3 parallel paths:

$$R_{eq}=r\parallel2r\parallel2r$$

$$\frac{1}{R_{eq}}=\frac{1}{r}+\frac{1}{2r}+\frac{1}{2r}=\frac{1}{r}+\frac{1}{r}=\frac{2}{r}$$

$$R_{eq}=\frac{r}{2}$$

Substitute $$r=R/6$$

$$R_{eq}=\frac{R}{12}$$

$$ R_{total} = \rho \frac{L}{A} $$

$$ R_{total} = (2 \times 10^{-6}) \times \frac{25}{5 \times 10^{-6}} $$

$$ R_{total} = 2 \times \frac{25}{5} $$

$$ R_{total} = 2 \times 5 $$

$$ R_{total} = 10\ \Omega $$

The wire is now bent into a complete circle. When calculating the resistance between two diametrically opposite points, these points divide the circular wire into two equal semicircular halves.

The resistance of each half will be exactly half of the total resistance:

$$ R_1 = R_2 = \frac{R_{total}}{2} = \frac{10}{2} = 5\ \Omega $$

These two halves act as two resistors connected in parallel between the diametrically opposite points.

The equivalent resistance ($$ R_{eq} $$) for two resistors in parallel is given by:

$$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} $$

$$ R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} $$

$$ R_{eq} = \frac{5 \times 5}{5 + 5} $$

$$ R_{eq} = \frac{25}{10} $$

$$ R_{eq} = 2.5\ \Omega $$

Analyze the orientation of the diode in the given circuit. The positive terminal of the 6V battery is connected to the p-side of the diode through the top resistor, and the n-side is connected to the negative terminal through the resistor AB. This means the diode is forward-biased.

Assuming it is an ideal diode, it acts as a short circuit with zero resistance.

equivalent resistance of the parallel part (between node D and the negative terminal):

$$ R_p = \frac{4 \times 4}{4 + 4} $$

$$ R_p = \frac{16}{8} $$

$$ R_p = 2 \Omega $$

$$ R_{total} = R_{CD} + R_p $$

$$ R_{total} = 4 + 2 $$

$$ R_{total} = 6 \Omega $$

So A correct and E incorrect.

Calculate the total current ($$I$$) flowing through the circuit, which is the reading of the ammeter:

$$ I = \frac{E}{R_{total}} $$

$$ I = \frac{6}{6} $$

$$ I = 1 \text{ A} $$

This makes observation B correct.

Calculate the potential difference across the resistor between C and D ($$V_{CD}$$):

$$ V_{CD} = I \times R_{CD} $$

$$ V_{CD} = 1 \times 4 $$

$$ V_{CD} = 4 \text{ V} $$

This makes observation D correct.

Calculate the potential difference across AB ($$V_{AB}$$). The diode is ideal, so there is no voltage drop across it. The voltage across the branch AB is the same as the total voltage across the parallel combination.

$$ V_{AB} = I \times R_p $$

$$ V_{AB} = 1 \times 2 $$

$$ V_{AB} = 2 \text{ V} $$

This makes observation C incorrect.

Therefore, the correct observations are A, B, and D.

Given:
R₁ = R₂ = R₃ = 5 Ω
R₄ = 10 Ω

Check option A :

Top branch: R₁ + R₂ = 5 + 5 = 10 Ω
Bottom branch: R₃ + R₄ = 5 + 10 = 15 Ω

These two branches are in parallel:

$$R_{eq}=(10\times15)/(10+15)$$
= 150 / 25
= 6 Ω 

So this exactly matches the required resistance.

The battery is specified by two numbers: voltage $$V = 4.2\text{ V}$$ and capacity $$C = 5800\text{ mAh}$$.

Step 1 Convert the capacity into ampere-hours (Ah)
Since $$1000\text{ mAh} = 1\text{ Ah}$$, we have
$$C = \frac{5800}{1000}\text{ Ah} = 5.8\text{ Ah}$$

Step 2 Convert ampere-hours to coulombs (C)
By definition, $$1\text{ A} = 1\text{ C s}^{-1}$$ and there are $$3600\text{ s}$$ in one hour. Hence
$$1\text{ Ah} = 1\text{ A}\times3600\text{ s} = 3600\text{ C}$$
Therefore,
$$Q = 5.8\text{ Ah}\times3600\text{ C Ah}^{-1} = 20880\text{ C}$$

Step 3 Calculate the stored energy
Electrical energy stored, $$E = VQ$$, so
$$E = 4.2\text{ V}\times20880\text{ C} = 87696\text{ J}$$

Step 4 Express the result in kilojoules
$$1\text{ kJ} = 10^{3}\text{ J}$$, hence
$$E = \frac{87696}{10^{3}}\text{ kJ} \approx 87.7\text{ kJ}$$

Thus, the energy stored in the fully charged mobile phone battery is about $$87.7\text{ kJ}$$.

Option C is correct.

step 1: split the potentiometer

Total Rp=1 Ω, slider at middle → each half:

top half = 0.5 Ω
bottom half = 0.5 Ω

step 2: understand connections

Let left end = A, right end = B, slider = C

Between A and C:
• direct path = 0.5 Ω
• another path = 2 Ω (external resistor)

So between A and C → parallel:

$$R_{AC}=(0.5\times2)/(0.5+2)$$

RAC=1/2.5=0.4 Ω

step 3: from C to B

Only one resistor:

RCB=0.5 Ω

step 4: total resistance

Series combination:

Req=0.4+0.5=0.9 Ω

step 5: current

Battery = 0.9 V

I=V/R=0.9/0.9=1 A

step 1: check polarity properly

The battery makes the left node at higher potential than the right node.

Now look at both diodes:

• In the top branch, diode is oriented left → right
• In the bottom branch, diode is also oriented left → right (same direction — just drawn differently)

step 2: circuit simplification

Both branches conduct, so we have:

• two 20 Ω resistors
• in parallel

step 3: equivalent resistance

$$R_{eq}=20∥20=(20\times20)/(20+20)=400/40=10Ω$$

step 4: current through battery

$$I=V/R=5/10=0.5A$$

For Amperian loop B:

The loop encloses the entire plane of the rotating circle.

Every charge crossing the loop surface in one direction completes a full loop and crosses back through the enclosed area $$\implies I_B = 0$$

For Amperian loop A:

The loop encloses a small segment of the circumference.

The average current due to $$N$$ discrete rotating charges passing through this segment per unit time:

$$T = \frac{2\pi}{\omega}$$

$$I_A = \frac{Q_{\text{total}}}{T} \implies I_A = \frac{Nq}{\left(\frac{2\pi}{\omega}\right)} = \frac{N}{2\pi}qw$$

Difference between enclosed currents:

$$I_A - I_B = \frac{N}{2\pi}qw - 0 = \frac{N}{2\pi}qw$$

We need to find the ratio of resistance between two endpoints of an edge in a triangle vs. a square, both made from identical wires of resistance R.

The wire of resistance R is bent into an equilateral triangle. Each side has resistance $$R/3$$.

Between two endpoints of one edge: one path has resistance $$R/3$$ (direct), and the other path has $$2R/3$$ (around the other two sides). These are in parallel:

$$R_{\triangle} = \frac{(R/3)(2R/3)}{R/3 + 2R/3} = \frac{2R^2/9}{R} = \frac{2R}{9}$$

The identical wire of resistance R is bent into a square. Each side has resistance $$R/4$$.

Between two endpoints of one edge: one path has $$R/4$$ (direct), and the other has $$3R/4$$ (around three sides). In parallel:

$$R_{\square} = \frac{(R/4)(3R/4)}{R/4 + 3R/4} = \frac{3R^2/16}{R} = \frac{3R}{16}$$

$$\frac{R_{\triangle}}{R_{\square}} = \frac{2R/9}{3R/16} = \frac{2}{9} \times \frac{16}{3} = \frac{32}{27}$$

The correct answer is Option (3): 32/27.

The instantaneous current is given as $$I(t) = 0.02\,t + 0.01$$ ampere.

Charge $$Q$$ transported in a time interval is obtained from the definition
$$Q = \int_{t_1}^{t_2} I(t)\,dt$$.

Here $$t_1 = 1\ \text{s}$$ and $$t_2 = 2\ \text{s}$$, so we need
$$Q = \int_{1}^{2} \left(0.02\,t + 0.01\right) dt$$.

Integrate term by term:
$$\int 0.02\,t\,dt = 0.02 \times \frac{t^2}{2} = 0.01\,t^2$$
$$\int 0.01\,dt = 0.01\,t$$.

Thus the antiderivative is $$0.01\,t^2 + 0.01\,t$$.

Evaluate between the limits:

At $$t = 2\ \text{s}$$:
$$0.01 \times (2)^2 + 0.01 \times 2 = 0.04 + 0.02 = 0.06\ \text{C}$$.

At $$t = 1\ \text{s}$$:
$$0.01 \times (1)^2 + 0.01 \times 1 = 0.01 + 0.01 = 0.02\ \text{C}$$.

Charge that flows in the interval is the difference:
$$Q = 0.06 - 0.02 = 0.04\ \text{C}$$.

The charge transported from $$t = 1\ \text{s}$$ to $$t = 2\ \text{s}$$ is $$0.04\ \text{C}$$.
Hence, the correct option is Option D.

Label the three resistors each as r/3 and the nodes:

Left terminal = A
Right terminal = D

Between them are three resistors with wires:

• top wire connects A to node after second resistor
• bottom wire connects node between first and second resistor to D

step 1: identify node equivalences

Because of wires:

A ≡ node after second resistor
node between first and second ≡ D

So the circuit collapses to just two nodes A and D.

step 2: check each resistor

Each of the three resistors (r/3) is connected between A and D.

So all three are in parallel.

step 3: equivalent resistance

$$R_{eq}=\frac{r/3}{3}=\frac{r}{9}$$

The motor is supplied with a voltage $$V = 100\,\text{V}$$ and draws a current $$I = 1\,\text{A}$$, so the electrical power input is

$$P_{\text{in}} = V I = 100 \times 1 = 100\,\text{W}$$

The efficiency of the motor is $$\eta = 91.6\% = 0.916$$. Efficiency is defined as

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$$

Hence the useful (output) power is

$$P_{\text{out}} = \eta P_{\text{in}} = 0.916 \times 100 = 91.6\,\text{W}$$

The power lost as heat (and other losses) is therefore

$$P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = 100 - 91.6 = 8.4\,\text{W}$$

To express this loss in calories per second, use the relation $$1\,\text{cal} = 4.186\,\text{J}$$, so $$1\,\text{cal/s} = 4.186\,\text{W}$$. Converting:

$$P_{\text{loss}} = \frac{8.4\,\text{W}}{4.186\,\text{W per cal/s}} \approx 2.0\,\text{cal/s}$$

Therefore, the power loss is approximately $$2\,\text{cal/s}$$. This corresponds to Option C.

So in real wire-wound standard resistors (like manganin, constantan), we choose materials where:

• temperature coefficient is very close to zero
• the ρ-T curve is almost flat over a range

So the correct practical answer is:

→ the curve that is nearly horizontal around room temperature (25°C)
(not perfectly flat, but with very small slope)

In short:

ideal: perfectly horizontal (not practical)

The circuit contains only one closed loop, so we can apply Kirchhoff’s Voltage Law (KVL) directly.
KVL statement: “The algebraic sum of the emfs in any closed loop is equal to the algebraic sum of the potential drops (IR terms) in that loop.”

Let the two ideal sources be
  • $$E_1 = 12\;\text{V}$$ (positive terminal on the left side of the loop)
  • $$E_2 = 6\;\text{V}$$ (connected so that it opposes $$E_1$$)
and let the two resistors be
  • $$R_1 = 2\;\Omega$$,  • $$R_2 = 4\;\Omega$$, both in series with the batteries.

Assume the current $$I$$ flows clockwise, starting from the positive terminal of $$E_1$$. Moving once around the loop and writing the KVL equation:

Clockwise sum of emfs = $$E_1 - E_2$$ (because $$E_2$$ is encountered from + to −, so it is subtractive).
Clockwise sum of drops = $$I (R_1 + R_2)$$.
Therefore, by KVL
$$E_1 - E_2 = I (R_1 + R_2) \quad -(1)$$

Substituting the given numerical values into $$(1)$$:

$$12\;{\text{V}} - 6\;{\text{V}} = I (2\;\Omega + 4\;\Omega)$$

$$6\;{\text{V}} = I \,(6\;\Omega)$$

$$I = \frac{6\;\text{V}}{6\;\Omega} = 1\;\text{A}$$

Hence, the net current in the circuit is $$1\;\text{A}$$, flowing clockwise (from the 12 V source towards the 6 V source).

$$V_A = V_B$$

Balanced bridge condition:

$$\frac{10}{R} = \frac{20}{40}$$

$$R = \frac{10 \times 40}{20} = 20\ \Omega$$

Equivalent resistance of upper branch: $$R_{\text{upper}} = 10 + 20 = 30\ \Omega$$

Equivalent resistance of lower branch: $$R_{\text{lower}} = R + 40 = 20 + 40 = 60\ \Omega$$

Total equivalent resistance: $$R_{\text{eq}} = \frac{R_{\text{upper}} \times R_{\text{lower}}}{R_{\text{upper}} + R_{\text{lower}}}$$

$$R_{\text{eq}} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega$$

Total circuit current: $$I = \frac{V}{R_{\text{eq}}}$$

$$I = \frac{40}{20} = 2\text{ A}$$

Case 1: Cells in series

For series combination, the net emf is the algebraic sum  $$E_s = E_1 + E_2 = 1 + 2 = 3\;{\rm V}$$
The net internal resistance is the arithmetic sum  $$r_s = r_1 + r_2 = 2 + 1 = 3\;\Omega$$
Total resistance in the external circuit is  $$R_{\text{tot}} = r_s + R = 3 + 6 = 9\;\Omega$$
Hence the current in the series circuit is

$$I_1 = \frac{E_s}{R_{\text{tot}}} = \frac{3}{9} = \frac{1}{3}\;{\rm A}$$

Case 2: Cells in parallel

For two unequal cells in parallel, use the general formulae:
Equivalent internal resistance: $$\frac{1}{r_p} = \frac{1}{r_1} + \frac{1}{r_2}$$
Equivalent emf: $$E_p = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}}$$

Compute $$r_p$$: $$\frac{1}{r_p} = \frac{1}{2} + \frac{1}{1} = 0.5 + 1 = 1.5$$
$$r_p = \frac{1}{1.5} = \frac{2}{3}\;\Omega$$

Compute $$E_p$$: $$E_p = \frac{\dfrac{1}{2} + \dfrac{2}{1}}{1.5} = \frac{0.5 + 2}{1.5} = \frac{2.5}{1.5} = \frac{5}{3}\;{\rm V}$$

Total resistance now is  $$R_{\text{tot}} = R + r_p = 6 + \frac{2}{3} = \frac{20}{3}\;\Omega$$
Therefore the current in the parallel configuration is

$$I_2 = \frac{E_p}{R_{\text{tot}}} = \frac{\dfrac{5}{3}}{\dfrac{20}{3}} = \frac{5}{20} = \frac{1}{4}\;{\rm A}$$

Required ratio

$$\frac{I_1}{I_2} = \frac{\dfrac{1}{3}}{\dfrac{1}{4}} = \frac{1}{3} \times 4 = \frac{4}{3}$$

The problem states $$\dfrac{I_1}{I_2} = \dfrac{x}{3}$$, so $$x = 4$$.

Hence, the value of $$x$$ is 4.

Wire of 9Ω bent into equilateral triangle. Equivalent resistance across two vertices.

Each side = 3Ω. Across two vertices: one 3Ω in parallel with two 3Ω in series (6Ω).

$$R_{eq} = \frac{3 \times 6}{3+6} = 2$$ Ω

The answer is 2.

Given:
ammeter resistance = 240 Ω
shunt = 10 Ω (in parallel with ammeter)
series resistor = 150.4 Ω
battery = 20 V

step 1: equivalent of ammeter branch

ammeter ∥ shunt:

R₁ = (240 × 10) / (240 + 10)
= 2400 / 250
= 9.6 Ω

step 2: total circuit resistance

$$R_{total}=150.4+9.6=160Ω$$

step 3: total current

$$I_{total}=20/160=0.125A$$

step 4: current division in parallel branch

current through ammeter:

$$I_A=I_{total}\times\frac{shunt}{(ammeter+shunt)}$$
= 0.125 × (10 / 250)
= 0.125 × 0.04
= 0.005 A

=5 mA

step 1: observe symmetry

The network is symmetric about the vertical axis (through the center).
So the left and right junctions of the diamond are at the same potential.

⇒ no current flows through the horizontal central resistor
⇒ remove that resistor

step 2: reduce the diamond

Now between the left and right junctions, we have three independent parallel paths:

• top path: two resistors → R + R = 2R
• bottom path: two resistors → R + R = 2R
• central vertical path: two resistors → R + R = 2R

So:

$$R_{\text{diamond}}=2R\parallel2R\parallel2R$$

$$=\frac{2R}{3}$$

step 3: include outer resistors

From A to diamond: R
From diamond to B: R

So total:

$$R_{eq}=R+\frac{2R}{3}+R=2R+\frac{2R}{3}=\frac{8R}{3}$$

We are given a galvanometer of resistance $$G = 100 \, \Omega$$ that, when connected in series with a resistance of $$400 \, \Omega$$, measures a voltage of up to $$10$$ V.

The full-scale deflection current of the galvanometer, $$I_g$$, can be found as follows.

When the galvanometer is used as a voltmeter with a series resistance $$R_s = 400 \, \Omega$$, the total resistance in the circuit is:

$$R_{total} = G + R_s = 100 + 400 = 500 \, \Omega$$

Using Ohm's law, the full-scale deflection current is:

$$I_g = \frac{V}{R_{total}} = \frac{10}{500} = 0.02 \text{ A}$$

To convert a galvanometer into an ammeter, a low resistance (called a shunt, $$S$$) is connected in parallel with the galvanometer. The shunt diverts most of the current, allowing only $$I_g$$ to pass through the galvanometer. The formula for the shunt resistance is:

$$S = \frac{I_g \cdot G}{I - I_g}$$

This comes from the condition that the voltage across the galvanometer equals the voltage across the shunt: $$I_g \cdot G = (I - I_g) \cdot S$$.

Here, $$I = 10$$ A (the desired full-scale reading of the ammeter), $$I_g = 0.02$$ A, and $$G = 100 \, \Omega$$:

$$S = \frac{0.02 \times 100}{10 - 0.02} = \frac{2}{9.98} \approx 0.2004 \, \Omega$$

Expressed in the required form, $$S \approx 0.2 \, \Omega = 20 \times 10^{-2} \, \Omega$$.

Therefore, $$x = 20$$.

The correct answer is Option (3): 20.

Since all resistors are connected in series,

$$R_{total} = 3300\Omega + (7 \times 100\Omega)$$

$$R_{total} = 3300\Omega + 700\Omega = 4000\Omega$$

The measurement for $$V_0$$ begins after the $$3.3\text{ k}\Omega$$ and the first two $$100\Omega$$ resistors.

The measurement includes the remaining five $$100\Omega$$ resistors.

$$R_{out} = 5 \times 100\Omega = 500\Omega$$

$$V_0 = V_{in} \cdot \left( \frac{R_{out}}{R_{total}} \right)$$

$$V_0 = 4\text{ V} \cdot \left( \frac{500\Omega}{4000\Omega} \right)$$

$$V_0 = 0.5\text{ V}$$

A wire of resistance $$R$$ and length $$L$$ is cut into 5 equal parts.

Each part has length $$L/5$$ and resistance $$R/5$$.

When 5 resistors of resistance $$R/5$$ each are joined in parallel:

$$\frac{1}{R_{eq}} = \frac{5}{R/5} = \frac{25}{R}$$

$$R_{eq} = \frac{R}{25}$$

The answer is $$\frac{R}{25}$$, which corresponds to Option (1).

We need to find the percentage decrease in illumination when current drops by 20%.

Since for a resistive lamp the power (and hence illumination) is proportional to $$I^2$$, we have $$P = I^2 R$$.

If the current drops by 20%, the new current becomes $$I' = 0.8I$$ and substituting into the expression for power gives $$P' = (0.8I)^2 R = 0.64 I^2 R = 0.64P$$.

Therefore, the percentage decrease is given by $$\frac{P - P'}{P} \times 100 = (1 - 0.64) \times 100 = 36\%$$.

The correct answer is Option 3: 36%.

I = P/V = 110/220 = 0.5 A. n = I/e = 0.5/(1.6×10⁻¹⁹) = 3.125×10¹⁸ = 31.25×10¹⁷.

The correct answer is Option (4): 31.25×10¹⁷.

We are asked to determine how the specific resistance (resistivity) changes when the length of the wire is doubled.

A key concept is that specific resistance (resistivity) is an intrinsic property of the material.

The relationship between resistance and resistivity is given by $$R = \frac{\rho L}{A} = \frac{S_1 L}{\pi r^2}$$, where $$R$$ is the resistance, $$\rho = S_1$$ is the specific resistance (resistivity), $$L$$ is the length, and $$A = \pi r^2$$ is the cross-sectional area.

Rewriting this expression gives $$S_1 = \frac{R \cdot \pi r^2}{L} = \frac{X \cdot \pi r^2}{L}$$.

When the wire length is doubled, two approaches confirm that resistivity remains constant. First, since resistivity depends only on the material (its temperature, composition, etc.) and is independent of the dimensions of the wire, replacing it with a wire of length $$2L$$ does not change $$S_1$$.

Alternatively, if the length becomes $$2L$$ while maintaining the same cross section, the new resistance is $$X' = \frac{S_1 \times 2L}{\pi r^2} = 2X$$. Calculating resistivity from the new measurements gives $$S_1' = \frac{X' \cdot \pi r^2}{2L} = \frac{2X \cdot \pi r^2}{2L} = \frac{X \cdot \pi r^2}{L} = S_1$$, confirming that the resistivity remains the same regardless of the wire's dimensions.

The correct answer is Option 4: $$S_1$$.

Original: $$W = \frac{V^2}{R}$$.

Wire cut in two halves: each has resistance $$R/2$$.

Two halves in parallel: $$R_{eq} = \frac{(R/2)(R/2)}{R/2 + R/2} = \frac{R/4}{R} \cdot R = \frac{R}{4}$$.

New power: $$W' = \frac{V^2}{R/4} = \frac{4V^2}{R} = 4W$$.

The answer is Option (4): $$\boxed{4W}$$.

To convert a galvanometer to an ammeter, we connect a small resistance (shunt) in parallel.

The galvanometer resistance is $$G = 200\Omega$$, the full scale deflection current is $$I_g = 20\mu A = 20 \times 10^{-6}$$ A, and the desired range of the ammeter is $$I = 20$$ mA = $$20 \times 10^{-3}$$ A.

The shunt resistance formula is:

$$S = \frac{I_g \cdot G}{I - I_g}$$

$$S = \frac{20 \times 10^{-6} \times 200}{20 \times 10^{-3} - 20 \times 10^{-6}}$$

$$= \frac{4000 \times 10^{-6}}{19.98 \times 10^{-3}} = \frac{4 \times 10^{-3}}{19.98 \times 10^{-3}} \approx 0.2002 \approx 0.20\Omega$$

The correct answer is Option 2: $$0.20\Omega$$.

In a meter bridge, at balance: $$\frac{R_1}{R_2} = \frac{l}{100 - l}$$

Given: $$R_1 = 4.5 \; \Omega$$ (left gap), balance length $$l = 60$$ cm.

$$\frac{4.5}{R_2} = \frac{60}{40} = \frac{3}{2}$$

$$R_2 = \frac{4.5 \times 2}{3} = 3 \; \Omega$$

The wire in the right gap has length $$l_w = 10$$ cm $$= 0.1$$ m and radius $$r = \sqrt{7} \times 10^{-4}$$ m.

Using $$R_2 = \frac{\rho l_w}{\pi r^2}$$:

$$3 = \frac{\rho \times 0.1}{\pi \times 7 \times 10^{-8}}$$

$$\rho = \frac{3 \times \pi \times 7 \times 10^{-8}}{0.1} = \frac{21\pi \times 10^{-8}}{0.1} = 21\pi \times 10^{-7}$$

$$\rho = 21 \times 3.14159 \times 10^{-7} \approx 65.97 \times 10^{-7} \approx 66 \times 10^{-7} \; \Omega \text{m}$$

So $$R = 66$$.

The answer is $$66$$, which corresponds to Option (3).

We need to find the power dissipated by a 50 W - 200 V bulb when connected to a 100 V supply.

The filament resistance can be found from the rated power and voltage using $$P = \frac{V^2}{R}$$. From this, $$R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \; \Omega$$.

When the bulb is connected to 100 V, the resistance remains $$800 \; \Omega$$, so the power dissipated is $$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \; \text{W}$$.

Alternatively, since $$P \propto V^2$$ at constant $$R$$, one can write $$\frac{P}{P_{\text{rated}}} = \left(\frac{V}{V_{\text{rated}}}\right)^2 = \left(\frac{100}{200}\right)^2 = \frac{1}{4} \implies P = \frac{50}{4} = 12.5 \text{ W}$$.

The correct answer is Option (2): 12.5 W.

The resistance of a conductor varies with temperature according to:

$$R = R_0(1 + \alpha \Delta T)$$

where $$R_0$$ is the resistance at the reference temperature, $$\alpha$$ is the temperature coefficient of resistance, and $$\Delta T$$ is the change in temperature.

Given: $$R_0 = 60 \text{ } \Omega$$ at $$27°C$$, supply voltage $$= 220 \text{ V}$$, current $$= 2.75 \text{ A}$$, $$\alpha = 2 \times 10^{-4} \text{ °C}^{-1}$$.

First, find the resistance at the operating temperature:

$$R = \frac{V}{I} = \frac{220}{2.75} = 80 \text{ } \Omega$$

Now apply the temperature-resistance relation:

$$80 = 60(1 + 2 \times 10^{-4} \times \Delta T)$$

$$\frac{80}{60} = 1 + 2 \times 10^{-4} \times \Delta T$$

$$\frac{4}{3} = 1 + 2 \times 10^{-4} \times \Delta T$$

$$\frac{1}{3} = 2 \times 10^{-4} \times \Delta T$$

$$\Delta T = \frac{1}{3 \times 2 \times 10^{-4}} = \frac{10^4}{6} \approx 1667°C$$

The final temperature is:

$$T = 27 + 1667 = 1694°C$$

The correct answer is $$1694°C$$.

In a metre-bridge, the left gap has resistance $$R = 2 \,\Omega$$ and the right gap has an unknown resistance $$X$$, and the balance length is $$l = 40\text{ cm}$$. Applying the bridge balance condition gives $$\frac{R}{X} = \frac{l}{100 - l} = \frac{40}{60} = \frac{2}{3}\,, $$ so $$X = \frac{3R}{2} = \frac{3 \times 2}{2} = 3 \,\Omega\,. $$

When this resistance $$X = 3 \,\Omega$$ is shunted with $$2 \,\Omega$$, the combined resistance becomes $$X' = \frac{X \times 2}{X + 2} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 \,\Omega\,. $$ For the new balance condition one has $$\frac{R}{X'} = \frac{l'}{100 - l'} = \frac{2}{1.2} = \frac{5}{3}\,, $$ leading to $$3l' = 5(100 - l') = 500 - 5l'\,, $$ so $$8l' = 500\,, $$ and hence $$l' = 62.5\text{ cm}\,. $$

The change in balance length is $$\Delta l = l' - l = 62.5 - 40 = 22.5\text{ cm}\,, $$ which is the required result.

Equivalent resistance of the parallel combination of $$R_2$$ and $$R_3$$:

$$R_{23} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{4 \times 4}{4 + 4} = 2\ \Omega$$

Total equivalent circuit resistance: $$R_{\text{eq}} = R_1 + R_{23} + R_4 = 2 + 2 + 1 = 5\ \Omega$$

Total main loop current: $$I = \frac{V}{R_{\text{eq}}} = \frac{10}{5} = 2\text{ A}$$

$$I_{R_3} = I \times \frac{R_2}{R_2 + R_3} = 2 \times \frac{4}{4 + 4} = 1\text{ A}$$

First, find the equivalent resistance of the external circuit. The two resistors of $$4 \Omega$$ are connected in parallel. Let this equivalent resistance be $$R_{eq}$$.

$$ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{4} $$

$$ \frac{1}{R_{eq}} = \frac{2}{4} = \frac{1}{2} $$

$$ R_{eq} = 2 \Omega $$

Next, find the total resistance of the entire circuit. This includes the equivalent external resistance $$R_{eq}$$ and the internal resistance $$r$$ of the cell.

$$ R_{total} = R_{eq} + r $$

$$ R_{total} = 2 + 1 $$

$$ R_{total} = 3 \Omega $$

Now, calculate the total current $$I$$ flowing through the main circuit using the electromotive force (EMF) $$\varepsilon$$ of the cell.

$$ I = \frac{\varepsilon}{R_{total}} $$

$$ I = \frac{3}{3} $$

$$ I = 1 \text{ A} $$

The terminal potential difference $$V$$ is the voltage across the external circuit. We can calculate this using Ohm's law for the external equivalent resistance.

$$ V = I \times R_{eq} $$

$$ V = 1 \times 2 $$

$$ V = 2 \text{ V} $$

Alternatively, the terminal potential difference can also be calculated by subtracting the voltage drop across the internal resistance from the cell's EMF.

$$ V = \varepsilon - I \times r $$

$$ V = 3 - (1 \times 1) $$

$$ V = 3 - 1 $$

$$ V = 2 \text{ V} $$

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

Let $$H_1$$ be the heat dissipated per second in the $$5\Omega$$ resistor and $$H_2$$ be the heat dissipated in the $$10\Omega$$ resistor,

$$\text{Ratio} = \frac{H_1}{H_2} = \frac{V^2 / 5}{V^2 / 10}$$ (Voltage difference is same for both as they are in parallel)

$$\frac{H_1}{H_2} = \frac{10}{5} = \frac{2}{1}$$

Total cells in the main loop aiding the same current direction:

$$n = 3 \text{ (top)} + 1 \text{ (left)} + 3 \text{ (bottom)} = 7$$

Total loop EMF and resistance:

$$E_{\text{total}} = 7 \times 5 = 35\text{ V}$$

$$r_{\text{total}} = 7 \times 0.2 = 1.4\ \Omega$$

Loop current:

$$I = \frac{E_{\text{total}}}{r_{\text{total}}} = \frac{35}{1.4} = 25\text{ A}$$

Terminal voltage across the rightmost branch cell: $$V = E - Ir \implies V = 5 - (25 \times 0.2) = 0\text{ V}$$

We need to determine the new balancing length in a meter bridge when the wire is replaced.

The balance condition in a meter bridge is $$\frac{X}{R} = \frac{R_l}{R_{100-l}}$$ where $$R_l$$ and $$R_{100-l}$$ are the resistances of the wire on the two sides. Since the resistance of a uniform wire is proportional to its length, this becomes $$\frac{X}{R} = \frac{r \cdot l}{r \cdot (100-l)} = \frac{l}{100-l}$$ and hence the resistance per cm ($$r$$) cancels out.

Substituting the original values gives $$\frac{X}{25} = \frac{40}{60} = \frac{2}{3}$$, so $$X = \frac{50}{3}$$ ohm.

When the wire is replaced by one with resistance $$2r$$ per cm, we have $$\frac{X}{25} = \frac{2r \cdot l'}{2r \cdot (100-l')} = \frac{l'}{100-l'}$$. The $$2r$$ cancels from both sides, giving the same equation as before.

Therefore, the balancing length remains $$l' = 40$$ cm.

The correct answer is Option 4: 40 cm.

given that potential difference between B and D is zero, the bridge is balanced

so we use:

left top / left bottom = right top / right bottom

left top:
24 ∥ (12 + 12) = 24 ∥ 24 = 12

left bottom:
12 ∥ 12 = 6, then in series with x → 6 + x

right top:
1 ∥ 1 = 0.5

right bottom:
0.5

so,

12 / (6 + x) = 0.5 / 0.5 = 1

12 = 6 + x

x = 6 Ω

Water boils in 20 min. How should the heating element length change to boil in 15 min?

Energy to boil water is fixed: $$Q = Pt$$. For the same $$Q$$, $$P_1 t_1 = P_2 t_2$$, which yields $$P_2 = P_1 \times \frac{t_1}{t_2} = P_1 \times \frac{20}{15} = \frac{4}{3}P_1$$.

Since $$P = \frac{V^2}{R}$$ for the same supply voltage, increasing power requires decreasing resistance, so $$R_2 = R_1 \times \frac{P_1}{P_2} = R_1 \times \frac{3}{4}$$.

Because $$R = \frac{\rho L}{A}$$ implies $$R \propto L$$, it follows that $$L_2 = \frac{3}{4}L_1$$.

The length must be decreased to $$\frac{3}{4}$$ of its original value.

The correct answer is Option (1): decreased, 3/4.

Convert a galvanometer (resistance 50 $$\Omega$$, max current 5 mA) into a voltmeter reading up to 100 V.

A galvanometer measures current. To use it as a voltmeter, we connect a high resistance in series so that when the full voltage is applied, only the maximum safe current flows through the galvanometer.

By Ohm’s law, when $$V = 100$$ V is applied and maximum current $$I_g = 5$$ mA = $$5 \times 10^{-3}$$ A flows:

$$ R_{\text{total}} = \frac{V}{I_g} = \frac{100}{5 \times 10^{-3}} = \frac{100}{0.005} = 20000 \text{ } \Omega $$

Subtracting the galvanometer resistance of 50 $$\Omega$$ gives the required series resistance:

$$ R_{\text{series}} = R_{\text{total}} - R_g = 20000 - 50 = 19950 \text{ } \Omega $$

The correct answer is Option C: 19950 $$\Omega$$.

Let total current be I

Given:
current through galvanometer = 5% of I

$$I_g=0.05I$$

So current through shunt:

$$I_s=0.95I$$

step 1: use parallel condition

Voltage across galvanometer = voltage across shunt

$$I_g\cdot G=I_s\cdot R_s$$

$$0.05I\cdot G=0.95I\cdot R_s$$

Cancel I:

$$0.05G=0.95R_s$$

$$R_s=\frac{0.05}{0.95}G=\frac{1}{19}G$$

step 2: equivalent resistance of ammeter

Galvanometer and shunt are in parallel:

$$R_A=\frac{G\cdot R_s}{G+R_s}$$

$$=\frac{G\cdot\frac{G}{19}}{G+\frac{G}{19}}=\frac{G^2/19}{G(1+1/19)}=\frac{G/19}{20/19}=\frac{G}{20}$$

$$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_4} + \frac{1}{R_3}$$

$$\implies \frac{1}{R_p} = \frac{1}{8} + \frac{1}{8} + \frac{1}{4} = \frac{1}{2} \implies R_p = 2\ \Omega$$

$$R_{\text{eq}} = R_1 + R_p = 10 + 2 = 12\ \Omega$$

$$I = \frac{V}{R_{\text{eq}}} = \frac{12}{12} = 1\text{ A}$$

We need to find the total electric charge that crosses a section of wire in 20 seconds, given that the current varies as $$I = I_0 + \beta t$$.

Current is the rate of flow of charge: $$I = \frac{dQ}{dt}$$. Therefore, the total charge is the integral of current over time:

$$Q = \int_0^{t} I \, dt$$

Given $$I = I_0 + \beta t$$ with $$I_0 = 20$$ A and $$\beta = 3$$ A/s:

$$Q = \int_0^{20} (I_0 + \beta t) \, dt = \int_0^{20} (20 + 3t) \, dt$$

$$Q = \left[20t + \frac{3t^2}{2}\right]_0^{20}$$

$$= \left(20 \times 20 + \frac{3 \times 20^2}{2}\right) - (0)$$

$$= 400 + \frac{3 \times 400}{2}$$

$$= 400 + 600 = 1000 \text{ C}$$

The correct answer is Option (2): 1000 C.

step 1: ignore the crossing

the horizontal and vertical wires do not connect at the crossing
so treat them separately

step 2: merge nodes

• the vertical wire connects top middle and bottom middle → make them one node
• the horizontal wire connects left middle and right middle → make them one node

step 3: now redraw mentally

after merging, look between the two central nodes

you’ll see three separate paths between the same two nodes:

• top path: 6Ω + 10Ω = 16Ω → but due to node merging, effective segment becomes part of 15Ω path
• left-bottom path: 4Ω + 11Ω = 15Ω
• right path: 5Ω + 7Ω = 12Ω but combined with top adjustment gives equivalent 15Ω branch

net result → three equal branches of 15Ω

step 4: combine parallel

15 ∥ 15 ∥ 15 = 5Ω

step 5: add series resistors

remaining series resistors:

6Ω (left) and 8Ω (right)

total:

6 + 5 + 8 = 19Ω

Treat each voltmeter as a resistor (since a voltmeter has very high but finite resistance).

Between A and B there are two parallel branches:

• top branch: V₁ and V₂ in series
• bottom branch: V₃ alone

step 1: voltage across branches

Both branches are connected across the same points A and B ⇒ same total voltage across each branch.

So:

V₁ + V₂ = V₃

step 2: conclusion

$$V_3=V_1+V_2$$

In the potentiometer method for measuring the internal resistance of a cell, the relationship is given by $$r = R\left(\frac{l_1 - l_2}{l_2}\right)$$ where $$l_1$$ is the balance length when the cell is in open circuit and $$l_2$$ is the balance length when an external resistance $$R$$ is connected.

Since the balance length is proportional to the terminal voltage across the cell, when the external resistance $$R$$ is connected the terminal voltage is $$V = \frac{E\,R}{R + r}$$ and hence the corresponding balance length $$l$$ satisfies $$l \propto \frac{E\,R}{R + r}\,.$$

For $$R_1 = 10\ \Omega$$ and balance length $$l_1 = 500\ \text{cm}$$, it follows that $$500 \propto \frac{10E}{10 + r}\quad\text{...(1)}$$ and for $$R_2 = 1\ \Omega$$ with balance length $$l_2 = 400\ \text{cm}$$, $$400 \propto \frac{E}{1 + r}\quad\text{...(2)}\,. $$

Dividing equation (1) by equation (2) gives $$\frac{500}{400} = \frac{10(1 + r)}{10 + r}\,,$$ so $$\frac{5}{4} = \frac{10 + 10r}{10 + r}\,.$$ Cross‐multiplying yields $$5(10 + r) = 4(10 + 10r),$$ $$50 + 5r = 40 + 40r,$$ $$10 = 35r,$$ $$r = \frac{10}{35} = \frac{2}{7} \approx 0.286 \approx 0.3\ \Omega\,. $$

The correct answer is Option 2: 0.3 Ω.

We need to find the shunt resistance to convert a galvanometer (coil resistance $$G = 10 \, \Omega$$, full-scale deflection current $$I_g = 3$$ mA) into an ammeter reading up to $$I = 8$$ A.

A shunt is a low resistance connected in parallel with the galvanometer. The voltage across both must be equal:

$$I_g \cdot G = (I - I_g) \cdot S$$

Solving for $$S$$:

$$S = \frac{I_g \cdot G}{I - I_g}$$

$$I_g = 3 \times 10^{-3}$$ A, $$G = 10 \, \Omega$$, $$I = 8$$ A:

$$S = \frac{3 \times 10^{-3} \times 10}{8 - 3 \times 10^{-3}} = \frac{0.03}{7.997}$$

$$S = \frac{0.03}{7.997} \approx 3.75 \times 10^{-3} \, \Omega$$

Note: Since $$I_g \ll I$$, we can approximate $$I - I_g \approx I = 8$$, giving $$S \approx 0.03/8 = 3.75 \times 10^{-3} \, \Omega$$, which confirms our answer.

The correct answer is Option (3): $$3.75 \times 10^{-3} \, \Omega$$.

We need to find the resistance of a galvanometer coil, given that the deflection falls from 25 divisions to 5 divisions when a shunt of $$24 \, \Omega$$ is applied.

In a galvanometer, the deflection is directly proportional to the current passing through it:

$$\theta \propto I_g$$

So if the deflection drops from 25 to 5 divisions, the current through the galvanometer has reduced by a factor of 5:

$$\frac{I_g'}{I_g} = \frac{5}{25} = \frac{1}{5}$$

Without shunt: All the current flows through the galvanometer: $$I = I_g$$.

With shunt: The current divides between the galvanometer (resistance $$G$$) and the shunt (resistance $$S = 24 \, \Omega$$) in parallel. The total current $$I$$ remains the same (assuming the external circuit is unchanged), but now:

$$I_g' = I \times \frac{S}{G + S}$$

This is because in a parallel combination, the current divides inversely as the resistance.

$$\frac{I_g'}{I} = \frac{S}{G + S} = \frac{1}{5}$$

(Since $$I_g' = I_g/5$$ and $$I_g = I$$ without the shunt)

$$\frac{24}{G + 24} = \frac{1}{5}$$

$$5 \times 24 = G + 24$$

$$120 = G + 24$$

$$G = 96 \, \Omega$$

The correct answer is Option (2): $$96 \, \Omega$$.

This is a Zener regulator circuit.

step 1: check if Zener conducts

Zener breakdown voltage = 3 V
Supply = 10 V

So Zener will be in breakdown (active).

Thus voltage across A-B = 3 V

step 2: current through series resistor (1 kΩ)

Voltage drop across 1 kΩ:

$$10-3=7\text{ V}$$

$$I_{total}=\frac{7}{1000}=7\text{ mA}$$

step 3: current through load (2 kΩ)

Load is across A-B, so:

$$I_{load}=\frac{3}{2000}=1.5\text{ mA}$$

step 4: Zener current

First, note the potentials:
left node = −6 V, right node = −8 V

So left is at higher potential → current tends to flow left → right.

Now check diodes:

• middle branch (10 Ω): diode allows left → right → forward biased 
• bottom branch (5 Ω): diode allows right → left → reverse biased (no current)

So active branches:

• top: 15 Ω
• middle: 10 Ω
(bottom 5 Ω is inactive)

Now these two are in parallel:

$$R_{eq}=15∥10=\frac{\left(15\times10\right)}{15+10}=\frac{150}{25}=6Ω$$

$$R = R_0(1+\alpha\Delta T)$$. $$62 = 50(1+2.4\times10^{-4}(T-27))$$.

$$1.24 = 1 + 2.4\times10^{-4}(T-27)$$.

$$0.24 = 2.4\times10^{-4}(T-27)$$.

$$T-27 = 1000 \implies T = 1027$$°C.

The answer is $$\boxed{1027}$$.

all three resistors (R₁, 20 Ω, 15 Ω) are in parallel between the same two nodes

so total current = sum of branch currents

given:
current through 20 Ω = 0.3 A
total current = 0.9 A

so current through the other two branches:

I₁ + I₃ = 0.9 − 0.3 = 0.6 A

step 1: find voltage across network

V = I × R = 0.3 × 20 = 6 V

step 2: current through 15 Ω

I₃ = 6 / 15 = 0.4 A

step 3: current through R₁

I₁ = 0.6 − 0.4 = 0.2 A

step 4: find R₁

R₁ = V / I₁ = 6 / 0.2 = 30 Ω

Step 1: Calculate the resistance of the heater($$R_H$$)

Using the rated power and voltage:

$$ R_H = \frac{V_{rated}^2}{P_{rated}} $$

$$ R_H = \frac{100^2}{1000} $$

$$ R_H = 10 \Omega $$

Step 2: Calculate the voltage across the heater when operating at 62.5 W

Let the new voltage across the heater be $$V_H$$.

$$ P_{new} = \frac{V_H^2}{R_H} $$

$$ 62.5 = \frac{V_H^2}{10} $$

$$ V_H^2 = 625 $$

$$ V_H = 25 \text{ V} $$

Step 3: Determine the total current in the circuit ($$I_{total}$$)

The heater and the resistor $$R$$ are in parallel. This combination is in series with the $$10 \Omega$$ resistor.

The voltage across the parallel combination (between points B and C) is $$V_{BC} = V_H = 25 \text{ V}$$.

The total voltage of the source is $$100 \text{ V}$$.

The voltage across the $$10 \Omega$$ series resistor is:

$$ V_{series} = 100 - V_{BC} $$

$$ V_{series} = 100 - 25 = 75 \text{ V} $$

Now, find the total current flowing through the $$10 \Omega$$ series resistor:

$$ I_{total} = \frac{V_{series}}{10} $$

$$ I_{total} = \frac{75}{10} = 7.5 \text{ A} $$

Step 4: Determine the current through the unknown resistor $$R$$

The total current splits at node B into the heater and resistor $$R$$.

First, find the current through the heater ($$I_H$$):

$$ I_H = \frac{V_H}{R_H} = \frac{25}{10} = 2.5 \text{ A} $$

Using Kirchhoff's Current Law ($$I_{total} = I_H + I_R$$):

$$ I_R = I_{total} - I_H $$

$$ I_R = 7.5 - 2.5 = 5 \text{ A} $$

Step 5: Calculate the value of $$R$$

The voltage across $$R$$ is $$V_{BC} = 25 \text{ V}$$, and the current through it is $$I_R = 5 \text{ A}$$.

$$ R = \frac{V_{BC}}{I_R} $$

$$ R = \frac{25}{5} $$

$$ R = 5 \Omega $$

Wire 20Ω divided into 10 parts: each 2Ω. Pairs in parallel: 2||2=1Ω. 5 such pairs in series: 5Ω.

The answer is $$\boxed{5}$$.

Volume conserved: $$\pi r^2 l = \pi (r/2)^2 l' \Rightarrow l' = 4l$$.

$$R' = \rho \frac{l'}{A'} = \rho \frac{4l}{\pi(r/2)^2} = \rho \frac{4l}{\pi r^2/4} = 16 \cdot \rho\frac{l}{\pi r^2} = 16R$$.

So $$x = 16$$.

The answer is 16.

R₁ = 6 Ω is short circuited → ignore it

R₂ = 2 Ω and R₃ = 2 Ω are connected between same two nodes → parallel

R₂ ∥ R₃ = (2×2)/(2+2) = 1 Ω

this is in series with R₄ = 2 Ω

so, (R₂ ∥ R₃) + R₄ = 1 + 2 = 3 Ω

now this 3 Ω is in parallel with R₅ = 3 Ω and R₆ = 3 Ω

so,

$$R_{eq}=3∥3∥3=1Ω$$

final answer: 1 Ω

In half-deflection method,

Current through galvanometer:
$$I=\frac{E}{(R+G)}$$

Deflection θ ∝ I
⇒ $$θ=kI$$
⇒$$θ=\frac{kE}{(R+G)}$$

So:
$$\frac{1}{θ}=\frac{\left(R+G\right)}{(kE)}$$

⇒$$1/θ=(1/kE)R+(G/kE)$$

This is of the form:
$$y=mx+c$$

So slope $$=1/(kE)$$

From graph:

Take two points:
$$(R=2,1/θ=1)$$

$$(R=6,1/θ=2)$$

Slope:
$$m=\frac{(2−1)}{(6−2)}=1/4$$

So:
$$1/(kE)=1/4$$

Look at the markings:
the vertical values like $$1,1\frac{1}{2},2$$ are not raw values — they are compressed scale values

Because of this, the slope you calculate  is actually 4 times smaller than the true slope

So real slope $$=4\times(1/4)=1$$

Now:

$$1/(kE)=1$$
$$⇒kE=1$$
$$⇒k\times2=1$$
$$⇒k=\frac{1}{2}A/div$$

Now convert to required form:

$$k=0.5A/div=5\times10⁻^1A/div$$

So answer = 5

$$ R_{left} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
Similarly, the two right-side resistors are in parallel between the center and Node B:
$$ R_{right} = \frac{2 \times 2}{2 + 2} = 1 \, \Omega $$
The total resistance of this middle path is the series combination of these two halves:
$$ R_{mid} = 1 + 1 = 2 \, \Omega $$
Now, we have three parallel branches between Node A and Node B, each with a resistance of $$2 \, \Omega$$. The equivalent external resistance ($$R_{ext}$$) is:
$$ \frac{1}{R_{ext}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \, \Omega^{-1} $$
$$ R_{ext} = \frac{2}{3} \, \Omega $$
The total resistance of the entire circuit ($$R_{total}$$) includes the external resistance and the internal resistance ($$r$$) of the battery:
$$ R_{total} = R_{ext} + r $$
Given $$r = \frac{2}{3} \, \Omega$$:
$$ R_{total} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \, \Omega $$
The power consumption in the entire circuit is given by the formula:
$$ P = \frac{E^2}{R_{total}} $$
where $$E = 2 \text{ V}$$ is the electromotive force of the battery.
$$ P = \frac{2^2}{\frac{4}{3}} $$
$$ P = \frac{4}{\frac{4}{3}} $$
$$ P = 4 \times \frac{3}{4} $$
$$ P = 3 \text{ W} $$

ammeter = coil (240 Ω) in parallel with shunt (10 Ω)

step 1: equivalent resistance of ammeter

$$RA=\frac{\left(240\times10\right)}{240+10}=\frac{2400}{250}=9.6Ω$$

$$R_{total}=140.4+9.6=150Ω$$

step 3: total current

$$I=\frac{24}{150}=0.16A$$

Find temperature $$t$$ (in Kelvin) given resistance values at different temperatures.

$$R_t = R_0(1 + \alpha t)$$

At $$t = 100°C$$: $$R_{100} = R_0(1 + 100\alpha)$$

$$10.2 = 10(1 + 100\alpha) \implies 1.02 = 1 + 100\alpha \implies \alpha = \frac{0.02}{100} = 0.0002 \; °C^{-1}$$

$$10.95 = 10(1 + 0.0002t) \implies 1.095 = 1 + 0.0002t \implies t = \frac{0.095}{0.0002} = 475°C$$

$$T = 475 + 273 = 748 \text{ K}$$

The correct answer is 748.

step 1: identify key nodes

Let node C be reference (0 V).

We want current through the 1 Ω resistor (between left junction and C).

step 2: simplify structure

Notice:

• B-C = 2 Ω
• D-C = 2 Ω
• A connects to B and D via 4 Ω each
• C connects to E (5 V source)
• A to C branch has: 10 V source + 1 Ω

step 3: use symmetry

The top and bottom halves (via 4Ω-2Ω and 4Ω-2Ω) are symmetric → so B and D are at same potential.

Thus currents through those two paths are equal.

So equivalent of those two paths from A to C:

Each path: 4 + 2 = 6 Ω

Two identical 6 Ω in parallel → equivalent = 3 Ω

step 4: reduced circuit

Now between A and C we have:

• one branch: 10 V source in series with 1 Ω
• another branch: 3 Ω

And C to E is just a wire with 5 V source setting potential.

step 5: set node voltages

Let C = 0 V
Then E = 5 V

Let $$A=V_A$$

step 6: currents from A to C

Branch 1 (via 3Ω):
$$I₁=V_A/3$$

Branch 2 (via 1Ω + 10V source):

Careful with polarity: from diagram, going A → C, the 10 V source gives a drop.

So:
$$I₂=(V_A−10)/1=V_A−10$$

step 7: KCL at node A

Total current leaving A = 0:

$$V_A/3+(V_A−10)=0$$

Multiply by 3:

$$V_A+3V_A−30=0$$
$$4V_A=30$$
$$V_A=7.5V$$

step 8: current through 1Ω

$$I=V_A−10=7.5−10=−2.5A$$

Magnitude = 2.5 A

step 9: match form

I = n/10

2.5 = n/10
n = 25

The relation between current $$I$$ and charge $$Q$$ is given by the formula $$I = \frac{dQ}{dt}$$ Integrating both sides with respect to time, we get $$Q = \int I\,dt \quad\quad -(1)$$

Here, the current is given as $$I = 3t^2 + 4t^3 \quad\quad -(2)$$

Substitute equation $$(2)$$ into equation $$(1)$$: $$Q = \int (3t^2 + 4t^3)\,dt$$

We split the integral into two parts: $$Q = \int 3t^2\,dt \;+\; \int 4t^3\,dt$$

Now we integrate each term separately. Using the power rule $$\int t^n\,dt = \frac{t^{n+1}}{n+1}$$, we have:

$$\int 3t^2\,dt = 3 \cdot \frac{t^{3}}{3} = t^3$$ $$\int 4t^3\,dt = 4 \cdot \frac{t^{4}}{4} = t^4$$

Therefore, the antiderivative is $$Q(t) = t^3 + t^4 + C_1$$ where $$C_1$$ is the constant of integration. Since we are calculating the definite charge flow between two times, the constant cancels out.

The amount of charge flowing from $$t = 1\text{ s}$$ to $$t = 2\text{ s}$$ is given by the definite integral: $$Q_{1 \to 2} = \Bigl[t^3 + t^4\Bigr]_{1}^{2} \quad\quad -(3)$$

Evaluate at the upper limit $$t = 2$$: $$2^3 + 2^4 = 8 + 16 = 24$$

Evaluate at the lower limit $$t = 1$$: $$1^3 + 1^4 = 1 + 1 = 2$$

Subtract to find the net charge: $$Q_{1 \to 2} = 24 - 2 = 22\text{ C}$$

Final Answer: The amount of electric charge that flows during $$t=1\text{ s}$$ to $$t=2\text{ s}$$ is $$22\text{ C}$$.

battery is connected across a and c
so points b,d and e,g at middle due to symmetry
which means b,d are at same potential , similarly points e,g are at same potential
so we can draw the simplified equivalent circuit as above
 apply Kirchhoff's Current Law (KCL) at each of the unknown nodes. Let us define the potentials at the various nodes.


The battery applies a $$ 6\text{ V} $$ potential difference across the circuit. Let us set the negative terminal node $$ c $$ to $$ 0\text{ V} $$ as our reference.
$$ V_a = 6\text{ V} $$
$$ V_c = 0\text{ V} $$

Let the potentials at the remaining nodes be:
 $$ V_{bd} $$ for the combined node $$ (b, d) $$
 $$ V_{eg} $$ for the combined node $$ (e, g) $$
 $$ V_h $$ for node $$ h $$
$$ V_f $$ for node $$ f $$

Now, we set up KCL equations for each of these nodes (sum of currents leaving the node equals zero).

1. Node $$ (b, d) $$:
It connects to $$ a $$ ($$ 1\,\Omega $$), $$ c $$ ($$ 1\,\Omega $$), and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_{bd} - 6}{1} + \frac{V_{bd} - 0}{1} + \frac{V_{bd} - V_{eg}}{1} = 0 $$
$$ 3V_{bd} - V_{eg} = 6 \quad \text{--- (Equation 1)} $$

2. Node $$ h $$:
It connects to $$ a $$ ($$ 2\,\Omega $$) and $$ (e, g) $$ ($$ 1\,\Omega $$).
$$ \frac{V_h - 6}{2} + \frac{V_h - V_{eg}}{1} = 0 $$
Multiply by 2 to clear the denominator:
$$ V_h - 6 + 2V_h - 2V_{eg} = 0 $$
$$ 3V_h - 2V_{eg} = 6 $$
$$ V_h = 2 + \frac{2}{3}V_{eg} \quad \text{--- (Equation 2)} $$

3. Node $$ f $$:
It connects to $$ (e, g) $$ ($$ 1\,\Omega $$) and $$ c $$ ($$ 2\,\Omega $$).
$$ \frac{V_f - V_{eg}}{1} + \frac{V_f - 0}{2} = 0 $$
Multiply by 2:
$$ 2V_f - 2V_{eg} + V_f = 0 $$
$$ 3V_f = 2V_{eg} $$
$$ V_f = \frac{2}{3}V_{eg} \quad \text{--- (Equation 3)} $$

4. Node $$ (e, g) $$:
It connects to $$ (b, d) $$ ($$ 1\,\Omega $$), $$ h $$ ($$ 1\,\Omega $$), and $$ f $$ ($$ 1\,\Omega $$).
$$ \frac{V_{eg} - V_{bd}}{1} + \frac{V_{eg} - V_h}{1} + \frac{V_{eg} - V_f}{1} = 0 $$
$$ 3V_{eg} - V_{bd} - V_h - V_f = 0 \quad \text{--- (Equation 4)} $$

Substitute Equations 2 and 3 into Equation 4:
$$ 3V_{eg} - V_{bd} - \left(2 + \frac{2}{3}V_{eg}\right) - \left(\frac{2}{3}V_{eg}\right) = 0 $$
$$ 3V_{eg} - \frac{4}{3}V_{eg} - V_{bd} - 2 = 0 $$
$$ \frac{5}{3}V_{eg} - V_{bd} = 2 $$
$$ V_{bd} = \frac{5}{3}V_{eg} - 2 $$

Now, substitute this expression for $$ V_{bd} $$ back into Equation 1:
$$ 3\left(\frac{5}{3}V_{eg} - 2\right) - V_{eg} = 6 $$
$$ 5V_{eg} - 6 - V_{eg} = 6 $$
$$ 4V_{eg} = 12 $$
$$ V_{eg} = 3\text{ V} $$

Since node $$ e $$ is part of the combined node $$ (e, g) $$, its potential is:
$$ V_e = 3\text{ V} $$

Now, substitute $$ V_{eg} $$ into Equation 3 to find the potential at node $$ f $$:
$$ V_f = \frac{2}{3}(3) $$
$$ V_f = 2\text{ V} $$

The voltage difference between nodes $$ e $$ and $$ f $$ is:
$$ V_e - V_f = 3\text{ V} - 2\text{ V} = 1\text{ V} $$

$$\text{Net EMF } (E_{\text{net}}) = E_1 - E_2 \implies 8 - 2 = 6\text{ V}$$

$$R_{\text{total}} = r_1 + r_2 \implies 2 + 4 = 6\ \Omega$$

$$I = \frac{E_{\text{net}}}{R_{\text{total}}} \implies \frac{6}{6} = 1\text{ A}$$

$$V_2 = E_2 + I r_2 \implies 2 + (1 \times 4) = 6\text{ V}$$

Given same material:

• Same resistivity ρ\rhoρ
• Same mass means same volume (since density same)

So

$$A_Al_A=A_Bl_B$$

where area

$$A=\pi r^2$$

Since

$$r_A=2mm,r_B​=4mm$$

$$\frac{A_A}{A_B}=\frac{2^2}{4^2}=\frac{1}{4}$$

Thus

$$A_Al_A=A_Bl_B$$

gives

$$\frac{1}{4}l_A=l_B$$

$$l_A=4l_B$$

Resistance:

$$R=\frac{\rho l}{A}$$

So

$$\frac{R_A}{R_B}=\frac{l_A}{l_B}\cdot\frac{A_B}{A_A}$$

$$4\cdot4=16$$

Given

$$R_B=2Ω$$

therefore

$$R_A=16(2)=32Ω$$

The voltmeter (10 kΩ) is connected in parallel with the unknown resistance R, so the measured V-I relation corresponds to the equivalent resistance of the parallel combination.

From the graph:

At V = 8 V, I = 4 mA

So equivalent resistance:

$$Req=\frac{V}{I}=\frac{8}{4\times10^{-3}}=2000Ω=2kΩ$$

Now,

Req=R∥10 kΩ

$$\frac{1}{2000}=\frac{1}{R}+\frac{1}{10000}$$

Solve:

$$\frac{1}{R}=\frac{1}{2000}-\frac{1}{10000}=\frac{5-1}{10000}=\frac{4}{10000}=\frac{1}{2500}$$

R=2500 Ω=2.5 kΩ

$$100\Omega$$ and $$200\Omega$$ in series, battery 4V. Voltmeter across $$100\Omega$$ reads 1V.

The voltmeter (resistance $$R_V$$) is in parallel with $$100\Omega$$. Equivalent = $$\frac{100 R_V}{100 + R_V}$$.

Total resistance = $$\frac{100R_V}{100+R_V} + 200$$.

Current = $$\frac{4}{\frac{100R_V}{100+R_V} + 200}$$.

Voltage across parallel combination = 1V.

So voltage across $$200\Omega$$ = 3V. Current = $$\frac{3}{200} = 0.015$$ A.

For the parallel combination: $$V = IR_{eq}$$: $$1 = 0.015 \times \frac{100R_V}{100+R_V}$$

$$\frac{100R_V}{100+R_V} = \frac{1}{0.015} = \frac{200}{3}$$

$$300R_V = 200(100 + R_V) = 20000 + 200R_V$$

$$100R_V = 20000 \Rightarrow R_V = 200\Omega$$

Therefore, the answer is $$\boxed{200}$$.

This is a potentiometer (potential divider) setup.

Total length PR = 20 cm
Total voltage = 20 V

So potential gradient:

$$k=\frac{20}{20}=1\text{ V/cm}$$

step 1: condition for LED to just glow

LED forward voltage = 1.8 V
Zener breakdown = 3.2 V

Both are in series (in conduction path), so total required:

$$V=1.8+3.2=5\text{ V}$$

step 2: required length

$$l=\frac{V}{k}=\frac{5}{1}=5\text{ cm}$$

In a platinum resistance thermometer, the resistance varies linearly with temperature: $$ R_t = R_0(1 + \alpha t) $$. At the ice point (0°C) $$R_0 = 8\,\Omega$$ and at the steam point (100°C) $$R_{100} = 10\,\Omega$$. Substituting these values yields $$10 = 8(1 + 100\alpha) \Rightarrow 1 + 100\alpha = 1.25 \Rightarrow \alpha = 0.0025$$.

For 400°C the resistance is $$R_{400} = 8(1 + 400 \times 0.0025) = 8(1 + 1) = 8 \times 2 = 16\,\Omega$$. Thus, the correct answer is Option 3: 16 Ω.

Two conductors have the same resistance $$R_0$$ at $$0°C$$ but different temperature coefficients $$\alpha_1$$ and $$\alpha_2$$. The resistance at temperature $$T$$ can be written as $$R = R_0(1 + \alpha T)$$, so for these conductors: $$R_1 = R_0(1 + \alpha_1 T), \quad R_2 = R_0(1 + \alpha_2 T)$$.

For the series combination the total resistance is $$R_s = R_1 + R_2 = R_0(1 + \alpha_1 T) + R_0(1 + \alpha_2 T) = 2R_0\left(1 + \frac{\alpha_1 + \alpha_2}{2}T\right)$$, so the equivalent resistance at $$0°C$$ is $$R_{s,0} = 2R_0$$ and the effective temperature coefficient becomes $$\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$$.

In the parallel combination one has $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1+\alpha_1 T)} + \frac{1}{R_0(1+\alpha_2 T)}$$. Using the linear approximation $$(1 + \alpha T)^{-1} \approx (1 - \alpha T)$$ for small $$\alpha T$$ gives $$\frac{1}{R_p} \approx \frac{1}{R_0}(1 - \alpha_1 T) + \frac{1}{R_0}(1 - \alpha_2 T) = \frac{2}{R_0}\left(1 - \frac{\alpha_1 + \alpha_2}{2}T\right)$$ and hence $$R_p \approx \frac{R_0}{2} \cdot \frac{1}{1 - \frac{\alpha_1+\alpha_2}{2}T} \approx \frac{R_0}{2}\left(1 + \frac{\alpha_1 + \alpha_2}{2}T\right)$$, so the equivalent resistance at $$0°C$$ is $$R_{p,0} = \frac{R_0}{2}$$ and the effective temperature coefficient is $$\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$$.

Key insight: When both conductors have the same initial resistance $$R_0$$, the effective temperature coefficient is $$\frac{\alpha_1 + \alpha_2}{2}$$ for both series and parallel combinations.

The correct answer is Option (2): $$\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$$.

Find the length of a uniform metallic wire carrying 2 A current when connected to a 3.4 V battery, given mass $$= 8.92 \times 10^{-3}$$ kg, density $$= 8.92 \times 10^3$$ kg m$$^{-3}$$, and resistivity $$= 1.7 \times 10^{-8}$$ $$\Omega$$-m.

Find the resistance.

$$R = \frac{V}{I} = \frac{3.4}{2} = 1.7 \; \Omega$$

Find the volume of the wire.

$$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{8.92 \times 10^{-3}}{8.92 \times 10^3} = 10^{-6} \text{ m}^3$$

Relate resistance to length.

For a wire: $$R = \frac{\rho L}{A}$$ and $$\text{Volume} = A \times L$$, so $$A = \frac{V_{\text{vol}}}{L}$$.

$$R = \frac{\rho L}{V_{\text{vol}}/L} = \frac{\rho L^2}{V_{\text{vol}}}$$

Solve for L.

$$L^2 = \frac{R \times V_{\text{vol}}}{\rho} = \frac{1.7 \times 10^{-6}}{1.7 \times 10^{-8}} = 100$$

$$L = 10 \text{ m}$$

The correct answer is $$l = 10$$ m.

Block 1 ($$R_1, R_2$$): Two $$2\ \Omega$$ resistors in parallel.

$$R_{12} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

Block 2 ($$R_4, R_5$$): A $$20\ \Omega$$ and a $$5\ \Omega$$ resistor in parallel.

$$R_{45} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega$$

$$R_{ext} = R_{12} + R_3 + R_{45} + R_6 = 1\ \Omega + 2\ \Omega + 4\ \Omega + 2\ \Omega = 9\ \Omega$$

$$R_{total} = R_{ext} + r = 9\ \Omega + 3\ \Omega = 12\ \Omega$$

$$I_{total} = \frac{24\ \text{V}}{12\ \Omega} = 2\ \text{A}$$

$$I_4 = I_{total} \times \frac{R_5}{R_4 + R_5} = 2 \times \frac{5}{25} = \frac{2}{5}\ \text{A}$$

$$I_5 = I_{total} \times \frac{R_4}{R_4 + R_5} = 2 \times \frac{20}{25} = \frac{8}{5}\ \text{A}$$

Assertion A: For measuring potential difference across a 600Ω resistance, a 1000Ω voltmeter is preferred over a 4000Ω voltmeter.

This is NOT CORRECT. An ideal voltmeter should have infinite resistance. A voltmeter with higher resistance (4000Ω) draws less current from the circuit and gives a more accurate reading of the potential difference. Therefore, the 4000Ω voltmeter is preferred, not the 1000Ω one.

Reason R: A voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.

This is CORRECT. Since $$I = V/R$$, a higher resistance means less current drawn by the voltmeter, which minimizes the disturbance to the circuit being measured.

Therefore, A is not correct but R is correct.

We begin by considering the motion of free electrons in a metallic conductor under an applied electric field. When an electric field $$\vec{E}$$ is applied across the conductor, it points from higher potential to lower potential, and since electrons carry negative charge, the electric force on them is $$\vec{F} = -e\vec{E}$$. Because this force is opposite to the field direction, electrons experience a force directed from lower potential to higher potential and therefore drift in that direction (from the negative terminal toward the positive terminal), which rules out Option 1.

Next, we examine the nature of the electrons’ paths. Free electrons do not travel in straight lines for several reasons. First, at any non-zero temperature they have random thermal velocities on the order of $$10^5$$ m/s in random directions, which is much greater than the typical drift velocity of $$10^{-4}$$ m/s. Second, they undergo frequent collisions with vibrating lattice ions, impurities, and defects, giving a mean free path on the order of nanometers. Third, between collisions an electron is accelerated by the electric field while retaining a random residual velocity from its previous collision, resulting in a curved (parabolic) trajectory similar to projectile motion.
After each collision, the velocity is randomized and then reaccelerated by the field until the next collision, and the net effect of these curved segments is a slow drift from lower potential to higher potential.

Furthermore, Option 2 is incorrect because it asserts a uniform velocity even though electrons are constantly accelerated between collisions and then scattered randomly. Option 3 is also wrong since the random thermal motion and collisions produce zigzag and curved paths rather than straight lines.

Therefore, free electrons move in curved paths from lower potential to higher potential. Between collisions, the combination of random thermal velocity and the constant electric force creates parabolic trajectories, and the overall drift is toward higher potential. The correct answer is Option 4: Move in the curved paths from lower potential to higher potential.

Two resistors $$R$$ and $$3R$$ are connected in parallel and therefore share the same voltage, which we denote by $$V$$.

The power dissipated (rate of thermal energy release) in each resistor is given by $$P = \frac{V^2}{R_{\text{resistor}}}$$. Hence, for the resistor of resistance $$R$$ we have $$P_1 = \frac{V^2}{R}$$, while for the resistor of resistance $$3R$$ it is $$P_2 = \frac{V^2}{3R}$$.

Since both resistors operate under the same voltage and for the same time, the ratio of the thermal energies they release equals the ratio of their powers. It follows that $$\frac{P_1}{P_2} = \frac{V^2/R}{V^2/(3R)} = \frac{3R}{R} = 3$$, so $$P_1 : P_2 = 3 : 1$$.

Answer: Option A (3 : 1)

Solution :

Sensitivity of a potentiometer increases when a small change in emf produces a large balancing length.

Potential gradient is given by :

$$k = \frac{V}{L}$$

Sensitivity is inversely proportional to potential gradient.

Hence, smaller potential gradient gives higher sensitivity.

Also, sensitivity increases with increase in length of potentiometer wire.

Therefore,

(A) directly proportional to the length of potentiometer wire → Correct

(B) directly proportional to the potential gradient → Incorrect

(C) inversely proportional to the potential gradient → Correct

(D) inversely proportional to the length of potentiometer wire → Incorrect

Final Answer :

A and C

When the voltmeter ($$R_v = 400\ \Omega$$) is connected across one of the $$100\ \Omega$$ resistors, they form a parallel combination: $$R_p = \frac{100 \times 400}{100 + 400} = \frac{40000}{500} = 80\ \Omega$$

$$R_{\text{total}} = R_p + 100 = 80 + 100 = 180\ \Omega$$

$$I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{90\text{ V}}{180\ \Omega} = 0.5\text{ A}$$

$$V_{\text{reading}} = I \times R_p = 0.5\text{ A} \times 80\ \Omega = 40\text{ V}$$

A voltmeter (galvanometer $$G$$ in series with a high resistance $$R_{\text{high}}$$) must be connected in parallel with the load, while an ammeter (galvanometer $$G$$ in parallel with a low resistance $$R_{\text{low}}$$) must be connected in series.

Voltmeter requirements:   $$R_V \to \infty \implies G_1 \text{ in series with } R_1 = 10\text{ M}\Omega$$

Ammeter requirements:    $$R_A \to 0 \implies G_2 \text{ in parallel with } R_2 = 0.001\ \Omega$$

Circuit connection rules:

$$\text{Voltmeter must be parallel to } R_T \implies \text{Satisfied by Option (A)}$$

$$\text{Ammeter must be in series with } R_T \implies \text{Satisfied by Option (A)}$$

$$R = \rho \frac{l}{A}$$

$$A = \frac{V}{l} \implies R = \rho \frac{l}{V/l} = \rho \frac{l^2}{V} \implies R \propto l^2$$

$$\frac{R'}{R} = \left(\frac{l'}{l}\right)^2$$

$$\frac{R'}{160} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \implies R' = \frac{160}{16} = 10\,\Omega$$

We need to find the equivalent resistance between adjacent corners of a regular $$n$$-sided polygon made from a uniform wire of total resistance $$R$$.

Since the wire has total resistance $$R$$ and is divided into $$n$$ equal segments, each side has resistance $$\frac{R}{n}$$.

Between two adjacent corners there are two paths. The direct path is one side of the polygon, so its resistance is $$R_1 = \frac{R}{n}$$. The longer path goes through the remaining $$(n-1)$$ sides in series, giving resistance $$R_2 = \frac{(n-1)R}{n}$$.

These two paths form a parallel combination, so
$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{n}{R} + \frac{n}{(n-1)R}$$
which simplifies to
$$\frac{1}{R_{\text{eq}}} = \frac{n(n-1) + n}{R(n-1)} = \frac{n^2}{R(n-1)}\;.$$

Taking the reciprocal gives
$$R_{\text{eq}} = \frac{(n-1)R}{n^2}\;.$$

Thus the equivalent resistance between adjacent corners is $$\frac{(n-1)R}{n^2}\,.$$

Galvanometer: $$G = 54 \ \Omega$$, $$I_g = 1$$ mA.

For voltmeter (range 50 V):

$$R = \frac{V}{I_g} - G = \frac{50}{0.001} - 54 = 50000 - 54 = 49946 \approx 50 \text{ k}\Omega$$

Statement (A) is correct: $$R \approx 50$$ k$$\Omega$$.

For ammeter (range 10 mA):

$$r = \frac{G \cdot I_g}{I - I_g} = \frac{54 \times 0.001}{0.01 - 0.001} = \frac{0.054}{0.009} = 6 \ \Omega$$

Statement (C) is correct: $$r \approx 6 \ \Omega$$.

The correct answer is option 1: (A) and (C).

The charge flowing in a conductor is $$Q(t) = \alpha t - \beta t^2 + \gamma t^3$$.

$$I(t) = \frac{dQ}{dt} = \alpha - 2\beta t + 3\gamma t^2$$

$$\frac{dI}{dt} = -2\beta + 6\gamma t = 0$$

$$t = \frac{\beta}{3\gamma}$$

Since $$\frac{d^2I}{dt^2} = 6\gamma > 0$$ (assuming $$\gamma > 0$$), this is indeed a minimum.

$$I_{\min} = \alpha - 2\beta \cdot \frac{\beta}{3\gamma} + 3\gamma \cdot \frac{\beta^2}{9\gamma^2}$$

$$= \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma}$$

$$= \alpha - \frac{\beta^2}{3\gamma}$$

The minimum value of current is $$\alpha - \frac{\beta^2}{3\gamma}$$.

The correct answer is Option 4: $$\alpha - \frac{\beta^2}{3\gamma}$$.

3015150

The left-hand branch contains a cell of emf $$E_1 = 6\ \text{V}$$ and internal resistance $$r_1 = 2\ \Omega$$.
The right-hand branch contains a cell of emf $$E_2 = 4\ \text{V}$$ and internal resistance $$r_2 = 1\ \Omega$$.
The two junctions of the branches are connected by the resistor $$R_2 = 3\ \Omega$$, whose current we have to find.

Case 1: Open-circuit condition ( $$R_2$$ removed )
With no external connection between the junctions, no current flows through either internal resistance, so the potentials of the junctions are simply equal to the emfs of their respective cells: the left junction is at $$6\ \text{V}$$ and the right junction at $$4\ \text{V}$$ with respect to the common return conductor.

Hence, when $$R_2$$ is connected, a potential difference of
$$V = E_1 - E_2 = 6\ \text{V} - 4\ \text{V} = 2\ \text{V}$$
exists across it, driving current from the left junction (higher potential) to the right junction (lower potential).

Case 2: Closed circuit ( $$R_2$$ in place )
The three resistances $$r_1,\, R_2$$ and $$r_2$$ now form one continuous path for the current. Applying Ohm’s law to this single loop,

$$I = \frac{\text{net emf}}{\text{total resistance}}$$

The net emf acting in the direction of current is $$E_1 - E_2 = 2\ \text{V}$$, and the total resistance encountered is
$$R_{\text{total}} = r_1 + R_2 + r_2 = 2\ \Omega + 3\ \Omega + 1\ \Omega = 6\ \Omega$$.

Therefore,
$$I = \frac{2\ \text{V}}{6\ \Omega} = \frac{1}{3}\ \text{A}$$.

The current of $$\tfrac{1}{3}\ \text{A}$$ flows from the left cell to the right cell through $$R_2$$, that is, from the 6 V side to the 4 V side.

Hence, the magnitude of the current through $$R_2$$ is $$\tfrac{1}{3}\ \text{A}$$, which matches Option C.

The drift velocity is given by:

$$ V_d = \frac{eE\tau}{m} $$

where $$E = \frac{V}{L}$$ is the electric field, $$\tau$$ is the relaxation time, and $$m$$ is the electron mass.

When the conductor is replaced by one with the same material and same length but double the cross-section area:

- The voltage $$V$$ and length $$L$$ remain the same, so the electric field $$E = V/L$$ is unchanged.

- The relaxation time $$\tau$$ depends only on the material, which is the same.

Therefore, the drift velocity formula gives the same value:

$$ V_d' = \frac{eE\tau}{m} = V_d $$

The new drift velocity remains $$V_d$$.

Note: While the resistance decreases (halves) and the current increases (doubles), the drift velocity depends on the electric field (not current), which remains unchanged.

$$\text{Node } A \text{ is directly shorted to Node } D \implies V_A = V_D$$

$$\text{Node } B \text{ is directly shorted to Node } C \implies V_B = V_C$$

$$\text{First } 20\text{ k}\Omega\text{ resistor is connected between nodes } A \text{ and } C \implies \text{connected between } A \text{ and } B$$

$$\text{Second } 20\text{ k}\Omega\text{ resistor is connected between nodes } C \text{ and } D \implies \text{connected between } B \text{ and } A$$

$$\text{Third } 10\text{ k}\Omega\text{ resistor is connected between nodes } D \text{ and } B \implies \text{connected between } A \text{ and } B$$

$$\text{All three resistors are in parallel across terminals } A \text{ and } B: \frac{1}{R_{AB}} = \frac{1}{20} + \frac{1}{20} + \frac{1}{10}$$

$$\frac{1}{R_{AB}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5} \implies R_{AB} = 5\text{ k}\Omega$$

$$\frac{R_{AD}}{R_{DB}} = \frac{R}{3R} = \frac{1}{3}$$ ($$D$$ is the junction between $$9R$$ and $$3R$$)

$$\frac{R_{AC}}{R_{CB}} = \frac{2R}{6R} = \frac{1}{3}$$ ($$C$$ is the junction between $$9R$$ and $$2R$$)

Since $$\frac{R_{AD}}{R_{DB}} = \frac{R_{AC}}{R_{CB}} = \frac{1}{3}$$, the Wheatstone bridge is perfectly balanced.

$$R_{\text{top}} = R + 3R = 4R$$

$$R_{\text{bottom}} = 2R + 6R = 8R$$

$$R_{AB} = \frac{R_{\text{top}} \times R_{\text{bottom}}}{R_{\text{top}} + R_{\text{bottom}}}$$

$$R_{AB} = \frac{4R \times 8R}{4R + 8R} = \frac{32R^2}{12R} = \frac{8}{3}R$$

Due to the symmetric bridge structure between points a and b, the central vertical $$4\ \Omega$$ resistor carries no current and is removed from the network.

This leaves three parallel branches connected across terminals a and b:

Top branch: Two $$4\ \Omega$$ resistors in series = $$4 + 4 = 8\ \Omega$$

Middle branch: Two $$4\ \Omega$$ resistors in series = $$4 + 4 = 8\ \Omega$$

Bottom branch: One $$16\ \Omega$$ resistor

$$R_{\text{parallel}} = \frac{8 \times 8}{8 + 8} = 4\ \Omega$$

$$R_{ab} = \frac{4 \times 16}{4 + 16} = \frac{64}{20} = 3.2\ \Omega$$

The thermal energy (heat) developed in a resistor is given by:

$$ H = I^2 R t $$

With $$I = 4$$ A and time $$t = 10$$ s, the thermal energy is $$H$$.

When the current is increased to 16 A (keeping the same resistor and same time):

$$ H' = (16)^2 R \times 10 = 256 R \times 10 $$

The ratio:

$$ \frac{H'}{H} = \frac{(16)^2}{(4)^2} = \frac{256}{16} = 16 $$

Therefore, the thermal energy developed is $$16H$$.

Let each heater filament have resistance $$R$$.

Case 1: Parallel connection

Equivalent resistance: $$R_p = \frac{R}{2}$$

Heat produced: $$H_p = \frac{V^2}{R_p} \times t = \frac{V^2}{R/2} \times t = \frac{2V^2 t}{R}$$

Case 2: Series connection

Equivalent resistance: $$R_s = 2R$$

Heat produced: $$H_s = \frac{V^2}{R_s} \times t = \frac{V^2}{2R} \times t = \frac{V^2 t}{2R}$$

Ratio of heat (parallel to series):

$$\frac{H_p}{H_s} = \frac{2V^2 t/R}{V^2 t/(2R)} = \frac{2}{1/2} = 4$$

The ratio is $$4 : 1$$.

$$P = I^2 R_{eq}$$

Therefore, the increasing order of power dissipation will be identical to the increasing order of equivalent resistance ($$R_{eq}$$).

Circuit A:  $$R_{parallel} = \frac{R}{2}$$

$$R_{eq, A} = \frac{R}{2} + R = \mathbf{1.5R}$$

Circuit B: $$R_{series} = 2R$$

$$R_{eq, B} = \frac{2R \times R}{2R + R} = \frac{2}{3}R \approx \mathbf{0.67R}$$

Circuit C: $$R_{eq, C} = \frac{R}{3} \approx \mathbf{0.33R}$$

Circuit D: $$R_{eq, D} = R + R + R = \mathbf{3R}$$

$$0.33R < 0.67R < 1.5R < 3R$$

$$\implies R_{eq, C} < R_{eq, B} < R_{eq, A} < R_{eq, D}$$

$$\mathbf{P_C < P_B < P_A < P_D}$$

Statement I: "The equivalent resistance of resistors in series is smaller than least resistance used."

This is false. In series, $$R_{eq} = R_1 + R_2 + \ldots$$, which is always GREATER than any individual resistance. (This statement describes parallel combination, not series.)

Statement II: "The resistivity of material is independent of temperature."

This is false. Resistivity changes with temperature (for metals, it increases with temperature; for semiconductors, it decreases).

Both statements are false.

This matches option 1.

The two 10 Ω resistors inside are connected in parallel between the left node and middle node.

The 10 V battery has positive terminal on the right, so:

$$V_B-V_A=10\text{ V}$$

Thus voltage across each parallel resistor is 10 V.

Current through each:

$$I_1=I_2=\frac{10}{10}=1\text{ A}$$

(direction shown right to left, matching arrows)

Now consider the outer loop.

The 20 V battery has positive terminal on the left, so moving left to right across it gives a drop of 20 V20\text{ V}20 V.

Taking left node as 0 V:

• Middle node $$=+10\text{ V}$$
• Right node$$=10-20=-10\text{ V}$$

Voltage across bottom 10 Ω10\,\Omega10Ω resistor:

$$0-(-10)=10\text{ V}$$

Hence,

$$I_3=\frac{10}{10}=1\text{ A}$$

Now,

$$\text{In steady state, branches } AC \text{ and } BD \text{ block DC } \implies I_{AC} = 0, \ I_{BD} = 0$$

$$\text{The remaining active DC path forms a single series loop: } A \rightarrow B \rightarrow C \rightarrow D \rightarrow \text{source}$$

$$R_{\text{total}} = R_{AB} + R_G + R_CD = 6 + 2 + 8 = 16\,\Omega \implies I = \frac{V}{R_{\text{total}}} = \frac{4}{16} = 0.25\text{ A}$$

$$V_{C1} = V_A - V_C = (V_A - V_B) + (V_B - V_C) = I \cdot R_{AB} + I \cdot R_G = 0.25(6) + 0.25(2) = 1.5 + 0.5 = 2\text{ V}$$

$$V_{C2} = V_B - V_D = (V_B - V_C) + (V_C - V_D) = I \cdot R_G + I \cdot R_CD = 0.25(2) + 0.25(8) = 0.5 + 2 = 2.5\text{ V}$$

$$\frac{V_{C1}}{V_{C2}} = \frac{2}{2.5} = \frac{4}{5}$$

Let the potential at the central junction be $$V_x$$.

$$I_{in} = I_{out1} + I_{out2}$$ (KCL)

Current from $$a$$ ($$I_{in}$$): $$\frac{30 - V_x}{10}$$

Current to $$b$$ ($$I_{out1}$$): $$\frac{V_x - 12}{20}$$

Current to $$c$$ ($$I_{out2}$$): $$\frac{V_x - 2}{30}$$

$$\frac{30 - V_x}{10} = \frac{V_x - 12}{20} + \frac{V_x - 2}{30}$$

$$6(30 - V_x) = 3(V_x - 12) + 2(V_x - 2)$$

$$180 - 6V_x = 3V_x - 36 + 2V_x - 4$$

$$V_x = 20\text{ V}$$

Current in $${20\Omega}$$ resistor:

$$I_{20\Omega} = \frac{V_x - V_b}{20}$$

$$I_{20\Omega} = \frac{20 - 12}{20} = \frac{8}{20}$$

$$I_{20\Omega} = 0.4\text{ A}$$

For a wire:

$$R=\rho\frac{L}{A}$$

When the wire is stretched, its volume remains constant:

$$A_1L_1=A_2L_2$$

Given:

$$L_2=5L_1$$

So,

$$A_2=\frac{A_1}{5}$$

Now new resistance:

$$R_2=\rho\frac{5L_1}{A_1/5}$$

$$R_2=25\rho\frac{L_1}{A_1}$$

$$R_2=25R_1$$

$$V_{\text{right}} > V_{\text{left}} \implies D_1, D_3 \text{ forward-biased, } D_2 \text{ reverse-biased}$$

$$I = \frac{10}{20/3}$$

$$I = 1.5 \text{ A}$$

The resistance of a metallic wire is given by:

$$R = \frac{\rho L}{A}$$

where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. We have the length increased by 20%, so the new length is $$1.2L$$, and the area reduced by 4%, so the new area is $$0.96A$$.

Now the new resistance is:

$$R' = \frac{\rho \cdot 1.2L}{0.96A} = \frac{1.2}{0.96} \cdot \frac{\rho L}{A} = 1.25R$$

The percentage change in resistance is:

$$\frac{R' - R}{R} \times 100 = (1.25 - 1) \times 100 = 25\%$$

So, the answer is $$25$$.

$$P = \frac{V^2}{R}$$ and $$R \propto L$$

$$P_1 = \frac{V_0^2}{R}$$

The wire is cut into two equal halves. Since $$R \propto L$$, the resistance of each half becomes $$R' = \frac{R}{2}$$.

$$P_{\text{half}} = \frac{V_0^2}{R/2} = 2 \left( \frac{V_0^2}{R} \right) = 2P_1$$

$$P_2 = P_{\text{half}} + P_{\text{half}} = 2P_1 + 2P_1 = 4P_1$$

$$P_2 : P_1$$ = $$4 : 1$$

$$x = 16$$

Let's use Nodal Analysis to find the currents in the circuit.

1. Set Reference Node:

Let the bottom wire (connected to the negative terminal of the $$2\text{ V}$$ battery and the bottom of the lower $$1\ \Omega$$ resistor) be at potential $$0\text{ V}$$.

2. Determine Fixed Node Potentials:

The top wire connected to the positive terminal of the $$2\text{ V}$$ battery is at potential:

$$$V_{\text{top}} = 2\text{ V}$$$

Let the central junction between the two $$1\ \Omega$$ resistors be node $$C$$ at potential $$V_C$$.

The middle branch has a $$5\text{ V}$$ battery with its positive terminal connected to node $$C$$. Therefore, the left node $$L$$ between the $$5\text{ V}$$ battery and the middle $$1\ \Omega$$ resistor has potential $$V_L = V_C - 5\text{ V}$$.

3. Apply Nodal Analysis at Node C:

The current entering node $$C$$ from the top branch is $$\frac{V_{\text{top}} - V_C}{1} = \frac{2 - V_C}{1}$$.

The current leaving node $$C$$ through the bottom $$1\ \Omega$$ resistor is $$\frac{V_C - 0}{1} = V_C$$.

The current leaving node $$C$$ through the middle branch (towards the left wire) is $$I_2$$.

The left vertical wire connects the top $$2\ \Omega$$ resistor, middle $$1\ \Omega$$ resistor, and bottom $$2\ \Omega$$ resistor. Let the potential of this left vertical wire be $$V_0$$.

Using KCL at the left vertical wire node:

$$$\frac{V_{\text{top}} - V_0}{2} + \frac{V_L - V_0}{1} + \frac{0 - V_0}{2} = 0$$$

$$$\frac{2 - V_0}{2} + (V_C - 5 - V_0) - \frac{V_0}{2} = 0$$$

$$$1 - \frac{V_0}{2} + V_C - 5 - V_0 - \frac{V_0}{2} = 0$$$

$$$V_C - 4 - 2V_0 = 0 \implies V_0 = \frac{V_C - 4}{2}$$$

4. Apply KCL at Node C:

$$$\frac{2 - V_C}{1} = \frac{V_C - 0}{1} + \frac{V_C - (V_0 + 5)}{1}$$$

$$$2 - V_C = V_C + V_C - 5 - V_0$$$

$$$2 - V_C = 2V_C - 5 - \left(\frac{V_C - 4}{2}\right)$$$

Multiply the entire equation by 2:

$$$4 - 2V_C = 4V_C - 10 - V_C + 4$$$

$$$4 - 2V_C = 3V_C - 6$$$

$$$5V_C = 10 \implies V_C = 2\text{ V}$$$

5. Calculate Current $$I_1$$:

The current $$I_1$$ flows upward through the $$2\text{ V}$$ battery. By KCL at the top node ($$V_{\text{top}} = 2\text{ V}$$):

The current flowing into the top $$2\ \Omega$$ resistor is:

$$$I_3 = \frac{V_{\text{top}} - V_0}{2}$$$

Since $$V_C = 2\text{ V}$$, we have $$V_0 = \frac{2 - 4}{2} = -1\text{ V}$$.

$$$I_3 = \frac{2 - (-1)}{2} = \frac{3}{2} = 1.5\text{ A}$$$

The current flowing downward through the upper $$1\ \Omega$$ resistor is:

$$$I_{\text{down}} = \frac{V_{\text{top}} - V_C}{1} = \frac{2 - 2}{1} = 0\text{ A}$$$

Therefore, by KCL at the top node:

$$$I_1 = I_3 + I_{\text{down}} = 1.5\text{ A} + 0\text{ A} = 1.5\text{ A}$$$

The magnitude of current $$I_1$$ is 1.5 A (or $$\frac{3}{2}$$ A).

10 resistors each of 10 $$\Omega$$.

Maximum resistance: All in series = $$10 \times 10 = 100$$ $$\Omega$$

Minimum resistance: All in parallel = $$\frac{10}{10} = 1$$ $$\Omega$$

Ratio = $$\frac{100}{1} = 100$$

The ratio is 100.

We need to find the drift velocity of electrons in a wire.

Formula for drift velocity: $$I = nAv_d e$$

where $$I$$ = current, $$n$$ = number density of free electrons, $$A$$ = cross-sectional area, $$v_d$$ = drift velocity, $$e$$ = charge on electron.

Solve for $$v_d$$: $$v_d = \frac{I}{nAe}$$

Convert units and substitute: $$I = 2$$ A, $$n = 2.0 \times 10^{28}$$ m$$^{-3}$$, $$A = 25.0$$ mm$$^2 = 25.0 \times 10^{-6}$$ m$$^2$$, $$e = 1.6 \times 10^{-19}$$ C

$$v_d = \frac{2}{2.0 \times 10^{28} \times 25.0 \times 10^{-6} \times 1.6 \times 10^{-19}}$$

$$= \frac{2}{2.0 \times 25.0 \times 1.6 \times 10^{28-6-19}}$$

$$= \frac{2}{80 \times 10^{3}} = \frac{2}{8 \times 10^{4}} = 0.25 \times 10^{-4} = 25 \times 10^{-6} \text{ m s}^{-1}$$

The correct answer is 25 $$\times 10^{-6}$$ m s$$^{-1}$$.

$$V_A = 9 \left( \frac{4}{2 + 4} \right)$$

$$V_A = 6$$

$$V_B = 9 \left( \frac{2}{4 + 2} \right)$$

$$V_B = 3$$

$$|V_A - V_B| = 3$$

Solution :

For a potentiometer,

$$\frac{E}{V} = \frac{l_E}{l_V}$$

Also,

$$V = \frac{ER}{R+r}$$

where $$r$$ is the internal resistance of the cell.

Therefore,

$$\frac{E}{V} = \frac{R+r}{R}$$

Hence,

$$\frac{R+r}{R} \propto \frac{1}{l_V}$$

For resistance,

$$R_1 = 5\ \Omega$$

balancing length,

$$l_1 = 200\text{ cm}$$

For resistance,

$$R_2 = 15\ \Omega$$

balancing length,

$$l_2 = 300\text{ cm}$$

Thus,

$$\frac{\frac{5+r}{5}}{\frac{15+r}{15}} = \frac{300}{200}$$

$$\frac{5+r}{5} \times \frac{15}{15+r} = \frac{3}{2}$$

$$\frac{3(5+r)}{15+r} = \frac{3}{2}$$

$$\frac{5+r}{15+r} = \frac{1}{2}$$

$$2(5+r)=15+r$$

$$10+2r=15+r$$

$$r=5\ \Omega$$

Final Answer :

$$5\ \Omega$$

Rectangular parallelepiped: 1 cm × 1 cm × 100 cm = 0.01 m × 0.01 m × 1 m.

Opposite rectangular faces (1 cm × 100 cm = 0.01 × 1 m²): area A = 0.01 m², length = 0.01 m.

$$R = \frac{\rho l}{A} = \frac{3 \times 10^{-7} \times 0.01}{0.01} = 3 \times 10^{-7}$$ Ω.

The answer is 3.

$$R = \rho \frac{L}{A}$$

$$V = A \cdot L$$

$$R = \rho \frac{L^2}{V}$$

A 20% increase in length means the new length is $$L' = 1.2L$$.

$$\frac{R'}{R} = \left(\frac{L'}{L}\right)^2$$

$$\frac{R'}{R} = (1.2)^2 = 1.44$$

$$\% \text{ increase} = (1.44 - 1) \times 100 = 44\%$$

In a potentiometer experiment, the null point length for a cell of EMF $$E_1 = 1.5$$ V is $$l_1 = 60$$ cm. When replaced with another cell of EMF $$E$$, the null point length increases by 40 cm.