JEE Current & Resistance Questions

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.

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

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

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

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

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

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:

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

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

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

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

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

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)

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.

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:

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

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 $$1694°C$$.

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.

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:

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

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

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): $$3.75 \times 10^{-3} \, \Omega$$.

Treat it as a Wheatstone bridge. At balance:

$$\frac{AB}{BC}=\frac{AD}{DC}$$

Given:
AB = 0.8 mΩ
AD = 1 mΩ
DC = 3 mΩ

So,

Initially (at 25°C), BC = 3 mΩ
After cooling, it becomes 2.4 mΩ

step 2: temperature change

Cooling rate = 2 °C/s
Time = 10 s

$$\Delta T=20^{\circ}C$$

Final temperature = 25 − 20 = 5°C

step 3: use resistance-temperature relation

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

Here ΔT = −20°C:

$$2.4=3(1+\alpha(-20))$$

$$2.4=3(1-20\alpha)$$

$$0.8=1-20\alpha$$

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): $$96 \, \Omega$$.

By conservation of momentum: $$\vec{p}_1 + \vec{p}_2 = 0$$ (nucleus initially at rest).

So $$m_1v_1 = m_2v_2$$, and they move in opposite directions.

$$\frac{v_1}{v_2} = \frac{m_2}{m_1} = \frac{1}{2}$$.

The heavier nucleus (mass ratio 2) moves slower, and the lighter one (mass ratio 1) moves faster, in opposite directions with speeds in ratio 1:2.

The correct answer is Option 4: in opposite directions with speed in the ratio of 1:2 respectively.

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

Earth's angular momentum is conserved: $$I\omega = const$$. $$I = \frac{2}{5}MR^2$$.

$$I_1\omega_1 = I_2\omega_2 \Rightarrow R^2 \omega_1 = (3R/4)^2 \omega_2 \Rightarrow \omega_2 = \frac{16}{9}\omega_1$$.

$$T_2 = \frac{2\pi}{\omega_2} = \frac{9}{16} T_1 = \frac{9}{16} \times 24 = 13.5$$ hours = 13 hours 30 minutes.

The answer is 13.

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

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

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

$$T-27 = 1000 \implies T = 1027$$°C.

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

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.

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.

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

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.

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)

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 $$\frac{(n-1)R}{n^2}\,.$$

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

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.

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.

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:

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

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

$$R_2=25R_1$$

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 $$25$$.

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$$^{-1}$$.

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 $$L' = 1.2L$$.

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

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

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

We have number density $$n = 8 \times 10^{28}$$ m$$^{-3}$$, area $$A = 2 \times 10^{-6}$$ m$$^2$$, and current $$I = 3.2$$ A.

The relation between current and drift speed is:

$$I = neAv_d$$

Solving for drift speed:

$$v_d = \frac{I}{neA}$$

$$v_d = \frac{3.2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$$

$$v_d = \frac{3.2}{8 \times 1.6 \times 2 \times 10^{28-19-6}}$$

$$v_d = \frac{3.2}{25.6 \times 10^{3}}$$

$$v_d = \frac{3.2}{25600} = 1.25 \times 10^{-4} \text{ m/s} = 125 \times 10^{-6} \text{ m/s}$$

So, the answer is $$125$$.

The resistance of a uniform conductor is given by the formula $$R=\dfrac{\rho\,L}{A}$$, where $$\rho$$ is the resistivity, $$L$$ is the length and $$A$$ is the cross-sectional area perpendicular to the current.

For a hollow cylindrical conductor, the current flows through the annular (ring-shaped) cross-section.

Outer diameter $$=8\text{ mm}=8\times10^{-3}\text{ m}$$, so outer radius $$R_o=4\times10^{-3}\text{ m}$$.
Inner diameter $$=4\text{ mm}=4\times10^{-3}\text{ m}$$, so inner radius $$R_i=2\times10^{-3}\text{ m}$$.

Area of an annulus:
$$A=\pi\!\left(R_o^{\,2}-R_i^{\,2}\right)$$

Substituting the radii:
$$A=\pi\!\left[(4\times10^{-3})^{2}-(2\times10^{-3})^{2}\right]$$
$$=\pi\!\left(16\times10^{-6}-4\times10^{-6}\right)$$
$$=\pi\,(12\times10^{-6})$$
$$=12\pi\times10^{-6}\ \text{m}^2$$

Given length $$L=3.14\ \text{m}$$ and resistivity $$\rho=2.4\times10^{-8}\ \Omega\!\,\text{m}$$, the resistance is

$$R=\dfrac{\rho\,L}{A} =\dfrac{(2.4\times10^{-8})(3.14)}{12\pi\times10^{-6}}$$

Because $$\pi\approx3.14$$, the factor $$3.14/\pi$$ is approximately $$1$$, so

$$R\approx\dfrac{2.4\times10^{-8}}{12\times10^{-6}} =\dfrac{2.4}{12}\times10^{-2} =0.2\times10^{-2} =2\times10^{-3}\ \Omega$$

Thus $$R = n\times10^{-3}\ \Omega$$ with $$n=2$$.

Answer : $$n = 2$$

We need to identify the correct statements about the drift velocity of electrons in a conductor.

Key Formula for Drift Velocity:

The drift velocity is given by:

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

where $$e$$ is the electron charge, $$E$$ is the electric field, $$\tau$$ is the mean relaxation time, and $$m$$ is the electron mass.

Since the electric field $$E = \frac{V}{L}$$ (where $$V$$ is the potential difference and $$L$$ is the conductor length):

$$v_d = \frac{e\tau}{m} \cdot \frac{V}{L}$$

Also, from $$I = neAv_d$$, we get $$v_d = \frac{I}{neA}$$, and using Ohm's law $$I = \frac{V}{R} = \frac{VA}{\rho L}$$:

$$v_d = \frac{V}{neA} \cdot \frac{A}{\rho L} = \frac{V}{ne\rho L}$$

Statement A: "The drift velocity of electrons decreases with increase in temperature."

When temperature increases, lattice vibrations increase, causing more frequent collisions with electrons. This reduces the relaxation time $$\tau$$.

Since $$v_d = \frac{eE\tau}{m}$$, a decrease in $$\tau$$ leads to a decrease in $$v_d$$ (for a given electric field).

Also, since resistivity $$\rho$$ increases with temperature, from $$v_d = \frac{V}{ne\rho L}$$, $$v_d$$ decreases when $$\rho$$ increases (for a given $$V$$ and $$L$$).

Statement A is CORRECT.

Statement B: "The drift velocity is inversely proportional to the area of cross-section."

From $$v_d = \frac{e\tau}{m} \cdot \frac{V}{L}$$, the drift velocity depends on $$V$$ and $$L$$ but NOT on the cross-sectional area $$A$$.

Similarly, from $$v_d = \frac{V}{ne\rho L}$$, the area $$A$$ does not appear.

Note: While $$v_d = \frac{I}{neA}$$ has $$A$$ in the denominator, for a given $$V$$, $$I$$ itself is proportional to $$A$$ (since $$R = \frac{\rho L}{A}$$, so $$I = \frac{VA}{\rho L}$$). These effects cancel out, making $$v_d$$ independent of $$A$$.

Statement B is INCORRECT.

Statement C: "The drift velocity does not depend on the applied potential difference."

From $$v_d = \frac{e\tau}{m} \cdot \frac{V}{L}$$, the drift velocity is directly proportional to the applied potential difference $$V$$.

Statement C is INCORRECT.

Statement D: "The drift velocity of electron is inversely proportional to the length of the conductor."

From $$v_d = \frac{e\tau}{m} \cdot \frac{V}{L}$$, for a given potential difference $$V$$, the drift velocity is inversely proportional to the length $$L$$ of the conductor.

This is because a longer conductor has a weaker electric field $$E = V/L$$ for the same applied voltage, resulting in lower drift velocity.

Statement D is CORRECT.

Statement E: "The drift velocity increases with increase in temperature."

As explained in Statement A, drift velocity decreases (not increases) with temperature, because the relaxation time $$\tau$$ decreases due to increased lattice vibrations.

Statement E is INCORRECT.

The correct statements are A and D.

Answer: Option B: A and D only

A wire of resistance $$R_1$$ is drawn out so that its length increases by twice its original length. We need to find the ratio of new resistance to original resistance.

The length is increased by twice the original length, meaning the new length is:

$$L' = L + 2L = 3L$$

When a wire is drawn out, its volume remains constant:

$$A \cdot L = A' \cdot L'$$

$$A' = \frac{AL}{3L} = \frac{A}{3}$$

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

Original resistance: $$R_1 = \frac{\rho L}{A}$$

New resistance: $$R_2 = \frac{\rho \cdot 3L}{A/3} = \frac{9\rho L}{A} = 9R_1$$

$$\frac{R_2}{R_1} = \frac{9}{1}$$

Hence, the correct answer is Option A: 9 : 1.

Statement I: A uniform wire of resistance 80 $$\Omega$$ is cut into four equal parts. Each part has resistance $$\frac{80}{4} = 20 \ \Omega$$. When four resistances of 20 $$\Omega$$ each are connected in parallel, the equivalent resistance is $$R_{eq} = \frac{20}{4} = 5 \ \Omega$$. So Statement I is correct.

Statement II: Two resistances $$2R$$ and $$3R$$ are connected in parallel. In a parallel combination, both resistors have the same potential difference $$V$$ across them. The thermal energy (power) developed in each resistor is $$P = \frac{V^2}{R_{\text{resistor}}}$$.

So the power in $$3R$$ is $$P_1 = \frac{V^2}{3R}$$ and the power in $$2R$$ is $$P_2 = \frac{V^2}{2R}$$.

The ratio of thermal energy developed in $$3R$$ to that in $$2R$$ is $$\frac{P_1}{P_2} = \frac{V^2/3R}{V^2/2R} = \frac{2R}{3R} = \frac{2}{3}$$, i.e., $$2:3$$.

The statement claims the ratio is $$3:2$$, which is incorrect. So Statement II is incorrect.

Hence, the correct answer is Option C.

We have the Assertion that alloys such as constantan and manganin are used in making standard resistance coils, and the Reason that these alloys have a very small value of temperature coefficient of resistance.

The assertion is true because constantan and manganin are indeed the preferred materials for standard resistance coils used in precision instruments like Wheatstone bridges and potentiometers.

The reason is also true: these alloys have extremely low temperature coefficients of resistance (of the order of $$10^{-5}$$ per °C), which means their resistance remains nearly constant over a wide range of temperatures.

Now, the reason correctly explains the assertion, because the primary requirement for a standard resistance coil is that its resistance should not change appreciably with temperature. Since constantan and manganin satisfy this requirement owing to their very small temperature coefficient of resistance, R is indeed the correct explanation for A.

Hence, the correct answer is Option 1.

We have a wire of total length 1 m broken into two parts: X of length $$l$$ and Y of length $$(1 - l)$$, both having the same cross-sectional area $$A$$ and resistivity $$\rho$$.

The resistance of Y is $$R_Y = \frac{\rho(1 - l)}{A}$$.

Now, wire X (length $$l$$, area $$A$$) is stretched to form wire W of length $$2l$$. Since the volume is conserved during stretching, $$l \cdot A = 2l \cdot A_W$$, which gives the new cross-sectional area $$A_W = \frac{A}{2}$$.

The resistance of W is $$R_W = \frac{\rho \cdot 2l}{A/2} = \frac{4\rho l}{A}$$.

We are given that $$R_W = 2R_Y$$, so: $$\frac{4\rho l}{A} = 2 \cdot \frac{\rho(1 - l)}{A}$$

Simplifying: $$4l = 2(1 - l) = 2 - 2l$$, which gives $$6l = 2$$, so $$l = \frac{1}{3}$$.

Therefore the length of X is $$\frac{1}{3}$$ m and the length of Y is $$\frac{2}{3}$$ m. The ratio $$\frac{X}{Y} = \frac{1/3}{2/3} = \frac{1}{2}$$.

Hence, the correct answer is Option 2.

The wire $$AB$$ is of length $$L = 1\,$$m and has uniform resistivity $$\rho$$, but its radius varies linearly from
$$r_A = 0.2\text{ mm} = 0.0002\,$$m at $$A$$ to $$r_B = 1\text{ mm} = 0.001\,$$m at $$B$$.

Let the distance measured from end $$A$$ be $$x$$. Because the radius changes linearly,

$$r(x) = r_A + (r_B - r_A)\frac{x}{L} = 0.0002 + 0.0008\,x \quad (0 \le x \le 1)$$

The cross-sectional area at position $$x$$ is $$A(x) = \pi\,r(x)^2$$.

For an element of length $$dx$$ the resistance is

$$dR = \frac{\rho\,dx}{A(x)} = \frac{\rho\,dx}{\pi\,[\,0.0002 + 0.0008x\,]^2}$$

Integrating gives the resistance between any two points. The integral needed is

$$\int \frac{dx}{(a + bx)^2} = -\frac{1}{b\,(a + bx)} \qquad\text{where } a = 0.0002,\; b = 0.0008$$

Resistance of segment $$AC$$ (first 0.5 m)

$$R_{AC} = \frac{\rho}{\pi}\,\frac{1}{b} \Bigg[\,\frac{1}{a} - \frac{1}{a + 0.5\,b}\,\Bigg]$$

Substituting $$\frac{1}{a} = \frac{1}{0.0002} = 5000$$ $$\frac{1}{a + 0.5\,b} = \frac{1}{0.0002 + 0.0004} = \frac{1}{0.0006} = 1666.67$$

$$\Rightarrow R_{AC} \propto 5000 - 1666.67 = 3333.33$$

Resistance of segment $$CB$$ (last 0.5 m)

$$R_{CB} = \frac{\rho}{\pi}\,\frac{1}{b} \Bigg[\,\frac{1}{a + 0.5\,b} - \frac{1}{a + b}\,\Bigg]$$

Substituting $$\frac{1}{a + b} = \frac{1}{0.0002 + 0.0008} = \frac{1}{0.001} = 1000$$

$$\Rightarrow R_{CB} \propto 1666.67 - 1000 = 666.67$$

Ratio of the two wire resistances

$$\frac{R_{AC}}{R_{CB}} = \frac{3333.33}{666.67} = 5$$

The galvanometer is connected at the midpoint of the wire, forming a Wheatstone bridge with arms:

Left arm: $$R_1$$ in series with $$R_{AC}$$ Right arm: $$R_2$$ in series with $$R_{CB}$$

For zero deflection (bridge balance),

$$\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}} = 5$$

Given $$R_2 = 1\,\Omega$$,

$$R_1 = 5 \times 1 = 5\,\Omega$$

Therefore, the required value of $$X$$ is 5.

We are given a copper wire of length $$L = 1 \text{ m}$$, carrying current $$I = 1 \text{ A}$$, with cross-section area $$A = 2.0 \text{ mm}^2 = 2.0 \times 10^{-6} \text{ m}^2$$ and resistivity $$\rho = 1.7 \times 10^{-8} \text{ } \Omega \text{ m}$$. We need to find the force on a moving electron.

The resistance of the wire is:

$$R = \frac{\rho L}{A} = \frac{1.7 \times 10^{-8} \times 1}{2.0 \times 10^{-6}} = 8.5 \times 10^{-3} \text{ } \Omega$$

The voltage across the wire is:

$$V = IR = 1 \times 8.5 \times 10^{-3} = 8.5 \times 10^{-3} \text{ V}$$

The electric field inside the wire is:

$$E = \frac{V}{L} = \frac{8.5 \times 10^{-3}}{1} = 8.5 \times 10^{-3} \text{ V/m}$$

The force experienced by an electron in the electric field is:

$$F = eE = 1.6 \times 10^{-19} \times 8.5 \times 10^{-3}$$

$$F = 13.6 \times 10^{-22} = 136 \times 10^{-23} \text{ N}$$

The force experienced by the moving electron is $$136 \times 10^{-23} \text{ N}$$.

In a meter bridge, the balance condition is:

$$\dfrac{P}{Q} = \dfrac{l}{100 - l}$$

We start by finding the ratio $$P/Q$$ from the initial balance point where the balancing length is $$l_1 = 40 \text{ cm}$$. Substituting into the balance condition gives

$$\dfrac{P}{Q} = \dfrac{40}{100 - 40} = \dfrac{40}{60} = \dfrac{2}{3}$$

This implies $$P = \dfrac{2Q}{3}$$ ...(i)

Next, when an unknown resistance $$x$$ is connected in series with $$P$$, the new balance length becomes $$l_2 = 80 \text{ cm}$$. Applying the same condition yields

$$\dfrac{P + x}{Q} = \dfrac{80}{100 - 80} = \dfrac{80}{20} = 4$$

Hence, $$P + x = 4Q$$ ...(ii)

Substituting equation (i) into equation (ii) gives

$$\dfrac{2Q}{3} + x = 4Q$$

Therefore,

$$x = 4Q - \dfrac{2Q}{3} = \dfrac{12Q - 2Q}{3} = \dfrac{10Q}{3}$$

Since $$Q = 6 \text{ }\Omega$$ (standard value given in the circuit), it follows that

$$x = \dfrac{10 \times 6}{3} = 20 \text{ }\Omega$$

Therefore, the value of $$x$$ is 20 $$\Omega$$.

Since the current density in the cylindrical wire is uniform, $$J = 1.0 \times 10^6$$ A/m$$^2$$, and the radius of the wire is $$r = 4.0$$ mm $$= 4.0 \times 10^{-3}$$ m, we need to find the current through the outer portion between $$\frac{r}{2}$$ and $$r$$.

For a uniform current density, the current through an annular region between radii $$r_1$$ and $$r_2$$ is given by

$$I = J \times A = J \times \pi(r_2^2 - r_1^2)$$

Here $$r_1 = \frac{r}{2} = 2.0 \times 10^{-3}$$ m and $$r_2 = r = 4.0 \times 10^{-3}$$ m. Using these values, the area of the outer portion becomes

$$A = \pi(r^2 - (r/2)^2) = \pi\left(r^2 - \frac{r^2}{4}\right) = \pi \times \frac{3r^2}{4}$$

Substituting $$r = 4.0 \times 10^{-3}$$ m gives

$$A = \frac{3\pi}{4} \times (4.0 \times 10^{-3})^2 = \frac{3\pi}{4} \times 16 \times 10^{-6} = 12\pi \times 10^{-6} \text{ m}^2$$

Therefore, the current through this region is

$$I = J \times A = 1.0 \times 10^6 \times 12\pi \times 10^{-6} = 12\pi \text{ A}$$

Comparing this with $$x\pi$$ A yields $$x = 12$$.

The answer is $$\boxed{12}$$.

We have two bulbs connected in series: Bulb 1 rated at 220 V, 100 W and Bulb 2 rated at 220 V, 60 W. The total voltage across the combination is 220 V.

The resistance of each bulb is found from its rating. For Bulb 1, $$R_1 = \frac{V^2}{P_1} = \frac{220^2}{100} = \frac{48400}{100} = 484 \; \Omega$$. For Bulb 2, $$R_2 = \frac{V^2}{P_2} = \frac{220^2}{60} = \frac{48400}{60} = \frac{2420}{3} \; \Omega$$.

Since the bulbs are in series, the total resistance is $$R = R_1 + R_2 = 484 + \frac{2420}{3} = \frac{1452 + 2420}{3} = \frac{3872}{3} \; \Omega$$.

The current through the series combination is $$I = \frac{V}{R} = \frac{220}{\frac{3872}{3}} = \frac{660}{3872} \text{ A}$$.

The power consumed by Bulb 1 (the 100 W bulb) is $$P = I^2 R_1 = \left(\frac{660}{3872}\right)^2 \times 484$$.

Now, $$\frac{660}{3872} = \frac{165}{968}$$, so $$I^2 = \frac{165^2}{968^2} = \frac{27225}{937024}$$.

Therefore, $$P = \frac{27225 \times 484}{937024} = \frac{13176900}{937024} \approx 14.06 \text{ W}$$.

Hence, the power consumed by the 100 W bulb is about $$\textbf{14}$$ W.

In a meter bridge experiment, the null point is at $$30 \text{ cm}$$ from the left side at point $$D$$, and $$R = 5.6 \text{ k}\Omega$$. We need to find the unknown resistance $$S$$.

In a meter bridge (based on the Wheatstone bridge principle), at the null point (balance point), the ratio of the two resistances equals the ratio of the two lengths of the bridge wire:

$$\frac{S}{R} = \frac{l}{100 - l}$$

where $$l = 30 \text{ cm}$$ is the distance of the null point from the left end, $$S$$ is in the left gap, and $$R$$ is in the right gap.

$$\frac{S}{R} = \frac{30}{100 - 30} = \frac{30}{70} = \frac{3}{7}$$

Given $$R = 5.6 \text{ k}\Omega = 5600 \text{ }\Omega$$:

$$S = R \times \frac{3}{7} = 5600 \times \frac{3}{7}$$

$$S = \frac{5600 \times 3}{7} = \frac{16800}{7} = 2400 \text{ }\Omega$$

The value of the unknown resistance $$S$$ is $$2400 \text{ }\Omega$$.

We need to find the current through the outer portion of a cylindrical wire between radial distances R/2 and R, where R = 4 mm and the current density J = $$4 \times 10^6$$ A/m². Because the current density is uniform, the total current is given by $$I = J \times A$$, where A is the area of the annular region between R/2 and R.

To determine this area, note that $$A = \pi R^2 - \pi \left(\frac{R}{2}\right)^2 = \pi R^2 - \frac{\pi R^2}{4} = \frac{3\pi R^2}{4}$$.

Here, $$R = 4 \text{ mm} = 4 \times 10^{-3} \text{ m}$$. Substituting into the expression for the current gives

$$I = J \times \frac{3\pi R^2}{4} = 4 \times 10^6 \times \frac{3\pi \times (4 \times 10^{-3})^2}{4}$$

$$I = 4 \times 10^6 \times \frac{3\pi \times 16 \times 10^{-6}}{4}$$

$$I = 4 \times 10^6 \times 12\pi \times 10^{-6}$$

$$I = 48\pi \text{ A}$$

The current through the outer portion is 48$$\pi$$ A.

Thermal energy $$E_1 = 300$$ J is developed in time $$t_1 = 15$$ s with a current of $$I_1 = 2$$ A. Using Joule’s law of heating, $$E = I^2 R t$$, the resistance of the resistor is determined by $$R = \frac{E_1}{I_1^2 t_1} = \frac{300}{(2)^2 \times 15} = \frac{300}{60} = 5\ \Omega$$.

When the current is increased to $$I_2 = 3$$ A for $$t_2 = 10$$ s, the energy developed is $$E_2 = I_2^2 R t_2 = (3)^2 \times 5 \times 10 = 9 \times 50 = 450\ \text{J}$$. Thus, the energy developed in 10 s is 450 J.

The resistance of a wire is given by:

$$R_{\text{wire}} = \dfrac{\rho l}{A} = \dfrac{\rho l}{\pi (d/2)^2} = \dfrac{4\rho l}{\pi d^2}$$

Each wire has length $$l$$ and diameter $$d$$, so its resistance is

$$R_1 = \dfrac{4\rho l}{\pi d^2}$$

When eight such wires are connected in parallel, the equivalent resistance becomes

$$R = \dfrac{R_1}{8} = \dfrac{4\rho l}{8\pi d^2} = \dfrac{\rho l}{2\pi d^2}$$

Now consider a single wire of length $$2l$$ and diameter $$D$$ having the same resistance $$R$$. Its resistance is given by

$$R = \dfrac{4\rho (2l)}{\pi D^2} = \dfrac{8\rho l}{\pi D^2}$$

Equating the two expressions for $$R$$ yields

$$\dfrac{\rho l}{2\pi d^2} = \dfrac{8\rho l}{\pi D^2}$$

which simplifies to

$$\dfrac{1}{2d^2} = \dfrac{8}{D^2}$$

and hence

$$D^2 = 16d^2$$

$$D = 4d$$

Therefore, the diameter of the single wire is $$\boxed{4}d$$.

We need to determine the new balancing length in a meter bridge experiment when the radius of the wire $$AB$$ is doubled.

1. Understand the Working Principle of a Meter Bridge

A meter bridge operates on the principle of a balanced Wheatstone bridge. When the galvanometer shows null deflection (zero current), the ratio of the resistances in the two parallel arms is equal:

$$\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$$

Where:

• $$R_1$$ and $$R_2$$ are the known and unknown resistances in the gaps.
• $$R_{AC}$$ is the resistance of the balancing wire segment of length $$l_1$$.
• $$R_{CB}$$ is the resistance of the remaining wire segment of length $$l_2 = (100 - l_1)$$.

2. Analyze the Resistance of the Wire Segments

The resistance of a uniform wire of length $$l$$, resistivity $$\rho$$, and cross-sectional area $$A$$ is given by:

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

For the two segments of the same uniform wire:

• $$R_{AC} = \rho \frac{l_1}{A}$$
• $$R_{CB} = \rho \frac{(100 - l_1)}{A}$$

Substituting these into the balancing condition:

$$\frac{R_1}{R_2} = \frac{\rho \frac{l_1}{A}}{\rho \frac{(100 - l_1)}{A}} = \frac{l_1}{100 - l_1}$$

Notice that the resistivity ($$\rho$$) and the cross-sectional area ($$A$$) cancel out completely from the ratio. The balancing length depends only on the ratio of the resistances $$R_1$$ and $$R_2$$, and the total length of the wire.

3. Effect of Doubling the Radius

When the radius of the wire $$AB$$ is doubled, the cross-sectional area ($$A = \pi r^2$$) increases uniformly across its entire length.

Since the area changes identically for both segments ($$AC$$ and $$CB$$), it still cancels out in the balance equation. As long as the resistors $$R_1$$ and $$R_2$$ remain unchanged, the balancing length $$l_1$$ remains exactly the same.

Conclusion

Since the balancing length is independent of the thickness or radius of the uniform bridge wire, it will remain unchanged.

The balancing length will be 40 cm.

Given: The length of a cylindrical wire is increased to double its original length.

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

where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area.

When the wire is stretched, its volume remains constant:

$$A_1 L_1 = A_2 L_2$$

If $$L_2 = 2L_1$$, then:

$$A_2 = \frac{A_1 L_1}{2L_1} = \frac{A_1}{2}$$

$$R_1 = \frac{\rho L_1}{A_1}$$

$$R_2 = \frac{\rho L_2}{A_2} = \frac{\rho \times 2L_1}{A_1/2} = \frac{4\rho L_1}{A_1} = 4R_1$$

$$\% \text{ increase} = \frac{R_2 - R_1}{R_1} \times 100 = \frac{4R_1 - R_1}{R_1} \times 100 = 3 \times 100 = 300\%$$

Hence, the percentage increase in resistance is 300%.

Given: $$R_1 = 2 \, \Omega$$ at $$T_1 = 10°$$C, and $$R_2 = 3 \, \Omega$$ at $$T_2 = 30°$$C.

Write the resistance-temperature relation.

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

where $$R_0$$ is the resistance at 0°C and $$\alpha$$ is the temperature coefficient.

Set up equations for the two temperatures.

$$2 = R_0(1 + 10\alpha)$$ ... (i)

$$3 = R_0(1 + 30\alpha)$$ ... (ii)

Divide equation (ii) by equation (i).

$$\frac{3}{2} = \frac{1 + 30\alpha}{1 + 10\alpha}$$

$$3(1 + 10\alpha) = 2(1 + 30\alpha)$$

$$3 + 30\alpha = 2 + 60\alpha$$

$$1 = 30\alpha$$

$$\alpha = \frac{1}{30} = 0.033°\text{C}^{-1}$$

The correct answer is Option A.

We have two metallic wires of identical dimensions (same length $$l$$ and same cross-sectional area $$A$$) connected in series, with conductivities $$\sigma_1$$ and $$\sigma_2$$ respectively.

The resistance of each wire is given by $$R = \frac{l}{\sigma A}$$, since resistivity $$\rho = \frac{1}{\sigma}$$. So the resistances are $$R_1 = \frac{l}{\sigma_1 A}$$ and $$R_2 = \frac{l}{\sigma_2 A}$$.

In series, the total resistance is $$R_{total} = R_1 + R_2 = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A}\left(\frac{1}{\sigma_1} + \frac{1}{\sigma_2}\right) = \frac{l}{A} \cdot \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$.

The combination of two wires in series can be treated as a single wire of total length $$2l$$ and the same cross-sectional area $$A$$. If the effective conductivity is $$\sigma_{eff}$$, then $$R_{total} = \frac{2l}{\sigma_{eff} A}$$.

Equating the two expressions: $$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \cdot \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$.

Simplifying: $$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}$$, which gives $$\sigma_{eff} = \frac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$$.

Hence, the correct answer is Option B.

The resistance of a 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. When the wire is stretched, its volume remains constant, so $$LA = L'A'$$.

The original length is $$L = 1$$ m and the original resistance is $$R = 1 \, \Omega$$. After stretching, the new length is $$L' = 1.25L = 1.25$$ m. Since volume is conserved, $$A' = \frac{LA}{L'} = \frac{A}{1.25}$$.

The new resistance is $$R' = \frac{\rho L'}{A'} = \frac{\rho (1.25L)}{A/1.25} = \rho \frac{L}{A} \times (1.25)^2 = R \times 1.5625 = 1.5625 \, \Omega$$.

The percentage change in resistance is $$\frac{R' - R}{R} \times 100 = \frac{1.5625 - 1}{1} \times 100 = 56.25\%$$, which to the nearest integer is $$56\%$$.

The thermal energy developed in a resistor is given by $$H = I^2 R t$$, where $$I$$ is the current, $$R$$ is the resistance, and $$t$$ is the time.

From the first case, $$500 = (1.5)^2 \times R \times 20 = 2.25 \times 20R = 45R$$. Solving for resistance: $$R = \frac{500}{45} = \frac{100}{9}$$ $$\Omega$$.

When the current is increased to $$3$$ A, the energy developed in $$20$$ s is $$H_2 = (3)^2 \times \frac{100}{9} \times 20 = 9 \times \frac{100}{9} \times 20 = 100 \times 20 = 2000$$ J.

Alternatively, since $$H \propto I^2$$ (with $$R$$ and $$t$$ unchanged), $$\frac{H_2}{H_1} = \frac{(3)^2}{(1.5)^2} = \frac{9}{2.25} = 4$$, giving $$H_2 = 4 \times 500 = 2000$$ J.

We have a galvanometer of unknown resistance $$G$$ connected in parallel with a shunt resistance $$S = 5\;\Omega$$. Both elements share the same potential difference because they are in parallel. Let the total current entering the parallel combination be $$I$$. According to the statement, only 2 % of this total current flows through the galvanometer.

Hence the current through the galvanometer is $$I_g = 0.02\,I$$ and the current through the shunt is $$I_s = I - I_g = I - 0.02\,I = 0.98\,I$$.

For two resistors in parallel, the potential drops are equal, so we write the condition

$$I_g\,G = I_s\,S.$$

Substituting the expressions for $$I_g$$ and $$I_s$$ we get

$$0.02\,I \, G = 0.98\,I \, S.$$

We can cancel the common factor $$I$$ from both sides, giving

$$0.02\,G = 0.98\,S.$$

Now we isolate $$G$$ by dividing both sides by $$0.02$$:

$$G = \frac{0.98}{0.02}\,S.$$

Next we calculate the numerical ratio:

$$\frac{0.98}{0.02} = \frac{98}{2} = 49.$$

So we obtain

$$G = 49\,S.$$

Substituting the given value of the shunt resistance $$S = 5\;\Omega$$, we have

$$G = 49 \times 5\;\Omega = 245\;\Omega.$$

Hence, the correct answer is Option A.

We start by recalling the resistance formula for a uniform wire:

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

where $$\rho$$ is the resistivity, $$L$$ is the length and $$A$$ is the cross-sectional area.

The two wires (iron and copper-nickel alloy) have

diameter $$d = 2\ \text{mm}=0.2\ \text{cm}\;,$$

so the radius is

$$r=\frac{d}{2}=1\ \text{mm}=0.1\ \text{cm}.$$

The area of a circle is $$A=\pi r^{2},$$ hence

$$A=\pi(0.1\ \text{cm})^{2}=\pi(0.01)\ \text{cm}^{2}=0.01\pi\ \text{cm}^{2}.$$

Both wires are required to have the same unknown length $$L$$ and are connected in parallel. For two resistances $$R_{1}$$ and $$R_{2}$$ in parallel we use

$$\frac{1}{R_{\text{eq}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}.$$

The given resistivities are

$$\rho_{\text{Fe}} = 12\ \mu\Omega\text{ cm}=12\times10^{-6}\ \Omega\text{ cm},$$

$$\rho_{\text{Cu-Ni}} = 51\ \mu\Omega\text{ cm}=51\times10^{-6}\ \Omega\text{ cm}.$$

Therefore the individual resistances are

$$R_{\text{Fe}}=\rho_{\text{Fe}}\frac{L}{A},\qquad R_{\text{Cu-Ni}}=\rho_{\text{Cu-Ni}}\frac{L}{A}.$$

The equivalent resistance is given to be

$$R_{\text{eq}} = 3\ \Omega.$$

Substituting the expressions for the two resistances into the parallel-combination formula, we get

$$\frac{1}{3}=\frac{1}{\rho_{\text{Fe}}\dfrac{L}{A}}+\frac{1}{\rho_{\text{Cu-Ni}}\dfrac{L}{A}} =\frac{A}{L}\left(\frac{1}{\rho_{\text{Fe}}}+\frac{1}{\rho_{\text{Cu-Ni}}}\right).$$

Now we isolate $$L$$:

$$\frac{1}{3}=\frac{A}{L}\left(\frac{\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}}{\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}}\right)$$

$$\Rightarrow\;L=3A\frac{\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}}{\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}}.$$

We substitute the numerical values:

$$A = 0.01\pi\ \text{cm}^{2},\qquad \rho_{\text{Fe}} = 12\times10^{-6}\ \Omega\text{ cm},\qquad \rho_{\text{Cu-Ni}} = 51\times10^{-6}\ \Omega\text{ cm}.$$

First, add the resistivities:

$$\rho_{\text{Fe}}+\rho_{\text{Cu-Ni}}=(12+51)\times10^{-6}=63\times10^{-6}\ \Omega\text{ cm}.$$

Next, multiply them:

$$\rho_{\text{Fe}}\rho_{\text{Cu-Ni}}= (12\times10^{-6})(51\times10^{-6})=612\times10^{-12}\ \Omega^{2}\text{ cm}^{2}.$$

Insert everything into the length expression:

$$L = 3(0.01\pi)\frac{63\times10^{-6}}{612\times10^{-12}} =0.03\pi\cdot63\times10^{-6}\times\frac{1}{612\times10^{-12}}.$$

Simplifying the powers of ten:

$$\frac{10^{-6}}{10^{-12}} = 10^{6},$$

so

$$L = 0.03\pi\cdot63\cdot10^{6}\times\frac{1}{612}.$$

Compute the numerical factor:

$$0.03\times63 = 1.89,$$

$$\frac{1.89}{612}=0.0030882,$$

$$0.0030882\times\pi\approx0.009702.$$

Therefore

$$L\approx0.009702\times10^{6}\ \text{cm}=9702\ \text{cm}.$$

Finally convert centimetres to metres:

$$L=\frac{9702}{100}\ \text{m}\approx97.0\ \text{m}.$$

Hence, the correct answer is Option A.

The resistance of a 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. The original current is $$I = \frac{V}{R} = \frac{VA}{\rho l}$$.

When the length is doubled ($$l' = 2l$$) and the area of cross-section is halved ($$A' = \frac{A}{2}$$), the new resistance becomes $$R' = \frac{\rho \cdot 2l}{A/2} = \frac{4\rho l}{A}$$.

Assuming the same potential difference $$V$$ is applied, the new current is $$I' = \frac{V}{R'} = \frac{V}{\frac{4\rho l}{A}} = \frac{VA}{4\rho l} = \frac{1}{4} \cdot \frac{VA}{\rho l}$$.

The relation between current, number density of free electrons, cross-sectional area, and drift velocity is $$I = nAv_d e$$, where $$n$$ is the number of free electrons per cubic metre, $$A$$ is the cross-sectional area, $$v_d$$ is the drift velocity, and $$e = 1.6 \times 10^{-19}$$ C is the electron charge.

Given: $$I = 10$$ A, $$A = 5 \text{ mm}^2 = 5 \times 10^{-6} \text{ m}^2$$, $$v_d = 2 \times 10^{-3}$$ m/s. Solving for $$n$$: $$n = \frac{I}{Av_d e} = \frac{10}{5 \times 10^{-6} \times 2 \times 10^{-3} \times 1.6 \times 10^{-19}}$$.

Computing the denominator: $$5 \times 10^{-6} \times 2 \times 10^{-3} = 10^{-8}$$, and $$10^{-8} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-27}$$. Therefore $$n = \frac{10}{1.6 \times 10^{-27}} = 6.25 \times 10^{27} = 625 \times 10^{25}$$.

The correct answer is option 2: $$625 \times 10^{25}$$.

The current $$I$$ through the cross-section is related to the current density $$\vec{J}$$ by $$I = \vec{J} \cdot \vec{A} = JA\cos\theta$$, where $$\theta = 60°$$ is the angle between the current density direction and the area vector.

Solving for $$J$$: $$J = \frac{I}{A\cos\theta} = \frac{5}{0.04 \times \cos 60°} = \frac{5}{0.04 \times 0.5} = \frac{5}{0.02} = 250$$ A m$$^{-2}$$.

The magnitude of the electric field is related to the current density by $$E = \rho J$$, where $$\rho = 44 \times 10^{-8}$$ $$\Omega$$m:

$$E = 44 \times 10^{-8} \times 250 = 11000 \times 10^{-8} = 11 \times 10^{-5}$$ V m$$^{-1}$$.

We are told that the galvanometer shows a total deflection of 50 divisions when a potential difference of 50 mV is applied across its terminals.

Current sensitivity is given as 2 divisions per milliampere. By definition,

$$S_i=\frac{\text{deflection in divisions}}{\text{current in ampere}}.$$

Re-arranging, the current needed for any given deflection is

$$I=\frac{\text{deflection}}{S_i}.$$

For the full-scale deflection of 50 divisions we therefore have

$$I=\frac{50\ \text{div}}{2\ \text{div}\!/\!\text{mA}}.$$

Divisions cancel, leaving

$$I=25\ \text{mA}.$$

Now we already know that the corresponding voltage is 50 mV, i.e.

$$V=50\ \text{mV}=50\times10^{-3}\ \text{V}.$$

Ohm’s law states $$V=IR,$$ so the resistance of the galvanometer is

$$R=\frac{V}{I}=\frac{50\times10^{-3}\ \text{V}}{25\times10^{-3}\ \text{A}}.$$

Both the $$10^{-3}$$ factors cancel, giving

$$R=\frac{50}{25}=2\ \Omega.$$

Hence, the correct answer is Option B.

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Here is the formatted, step-by-step solution to find the equivalent resistance of the given circuit network.

Problem Statement

The equivalent resistance of the given circuit between terminals A and B is:

• (1) $$0\ \Omega$$
• (2) $$3\ \Omega$$
• (3) $$\frac{9}{2}\ \Omega$$
• (4) $$1\ \Omega$$
• Look closely at the $$5\ \Omega$$ resistor on the far left. Both of its ends are connected directly together by a continuous conducting wire (short-circuit path).
• Because both ends are connected to the same potential wire (the bottom wire line), the potential difference across it is zero ($$V = 0$$), so no current flows through it ($$i = 0$$).
• We can completely remove this $$5\ \Omega$$ resistor from our calculation.
• First Parallel Combination:
• Series Combination:
• Second Parallel Combination:

Step-by-Step Solution

To find the total equivalent resistance between terminals A and B, we analyze the circuit starting from the leftmost side and simplify it step-by-step moving towards the right.

Step 1: Analyze the Leftmost $$5\ \Omega$$ Resistor

Step 2: Parallel and Series Reductions

After removing the $$5\ \Omega$$ resistor, the first series $$2\ \Omega$$ resistor on the top branch is now connected in parallel across the middle vertical $$2\ \Omega$$ resistor.

$$R_{p1} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

This $$1\ \Omega$$ equivalent resistance is now connected directly in series with the second top-branch $$2\ \Omega$$ resistor.

$$R_{\text{series}} = 1\ \Omega + 2\ \Omega = 3\ \Omega$$

On the far right, near terminals A and B, we have two vertical $$3\ \Omega$$ resistors connected in parallel with each other.

$$R_{p2} = \frac{3 \times 3}{3 + 3} = \frac{9}{6} = \frac{3}{2}\ \Omega$$

Step 3: Final Equivalent Resistance ($$R_{\text{eq}}$$)

Now, the remaining circuit is simply our simplified series branch ($$3\ \Omega$$) connected in parallel across the terminal pair's combined resistance ($$\frac{3}{2}\ \Omega$$):

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

$$R_{\text{eq}} = \frac{\frac{9}{2}}{\frac{9}{2}} = 1\ \Omega$$

Correct Answer

The correct option is (4) $$1\ \Omega$$.

We have two resistors whose measured values are

$$R_1 = 4 \;\Omega \pm 0.8 \;\Omega , \qquad R_2 = 4 \;\Omega \pm 0.4 \;\Omega.$$

When resistors are connected in parallel, the equivalent resistance $$R_p$$ is obtained from the relation

$$\frac{1}{R_p} \;=\; \frac{1}{R_1} + \frac{1}{R_2}.$$

Instead of working with reciprocals, it is convenient to use the algebraically equivalent form

$$R_p \;=\; \frac{R_1 R_2}{R_1 + R_2}.$$

First we calculate the central (nominal) value by substituting the central values $$R_1 = 4 \;\Omega$$ and $$R_2 = 4 \;\Omega$$:

$$R_p = \frac{4 \times 4}{4 + 4} = \frac{16}{8} = 2 \;\Omega.$$

Now we must estimate the maximum possible error in $$R_p$$ that arises from the individual errors in $$R_1$$ and $$R_2$$. For a function of several variables, the (maximum) absolute error is obtained by adding the absolute values of the first‐order differentials:

$$\Delta R_p \;=\; \left| \frac{\partial R_p}{\partial R_1} \right|\! \Delta R_1 \;+\; \left| \frac{\partial R_p}{\partial R_2} \right|\! \Delta R_2.$$

So we need the partial derivatives of $$R_p = \dfrac{R_1 R_2}{R_1 + R_2}$$ with respect to each resistor.

Starting with $$R_1$$:

$$\frac{\partial R_p}{\partial R_1} = \frac{(R_1 + R_2)\,R_2 - R_1 R_2}{(R_1 + R_2)^2} = \frac{R_2^2}{(R_1 + R_2)^2}.$$

Similarly, for $$R_2$$ we have

$$\frac{\partial R_p}{\partial R_2} = \frac{R_1^2}{(R_1 + R_2)^2}.$$

Substituting the nominal values $$R_1 = R_2 = 4 \;\Omega$$ gives

$$R_1 + R_2 = 8 \;\Omega \;\;\Longrightarrow\;\; (R_1 + R_2)^2 = 64,$$

$$\frac{\partial R_p}{\partial R_1} = \frac{4^2}{64} = \frac{16}{64} = 0.25,$$

$$\frac{\partial R_p}{\partial R_2} = \frac{4^2}{64} = 0.25.$$

Now insert the absolute errors $$\Delta R_1 = 0.8 \;\Omega$$ and $$\Delta R_2 = 0.4 \;\Omega$$:

$$\Delta R_p = 0.25 \times 0.8 \;+\; 0.25 \times 0.4 = 0.20 + 0.10 = 0.30 \;\Omega.$$

Therefore, the equivalent resistance with its maximum possible error is

$$R_p = 2 \;\Omega \pm 0.3 \;\Omega.$$

Hence, the correct answer is Option D.

We are given the current as a function of time: $$i = \alpha_0 t + \beta t^2$$, where $$\alpha_0 = 20$$ A s$$^{-1}$$ and $$\beta = 8$$ A s$$^{-2}$$.

The charge flowing through a section of wire in time $$t$$ is obtained by integrating the current: $$q = \int_0^{t} i \, dt$$.

Substituting, $$q = \int_0^{15} (\alpha_0 t + \beta t^2) \, dt = \left[\frac{\alpha_0 t^2}{2} + \frac{\beta t^3}{3}\right]_0^{15}$$.

Now we compute each term. The first term is $$\frac{\alpha_0 \times 15^2}{2} = \frac{20 \times 225}{2} = \frac{4500}{2} = 2250$$.

The second term is $$\frac{\beta \times 15^3}{3} = \frac{8 \times 3375}{3} = \frac{27000}{3} = 9000$$.

So the total charge is $$q = 2250 + 9000 = 11250$$ C.

Hence, the correct answer is Option B.

We are given that the resistance of the conductor is $$R_1 = 16\;\Omega$$ at the initial temperature $$T_1 = 15^{\circ}\text{C}$$ and becomes $$R_2 = 20\;\Omega$$ at the higher temperature $$T_2 = 100^{\circ}\text{C}$$.

For a metallic conductor, the variation of resistance with temperature is described by the linear relation

$$R = R_0\left[1 + \alpha\,(T - T_0)\right],$$

where

$$R_0$$ = resistance at the reference temperature $$T_0,$$

$$R$$ = resistance at temperature $$T,$$

$$\alpha$$ = temperature coefficient of resistance (per °C).

Applying this formula to our data, we write

$$R_2 = R_1\left[1 + \alpha\,(T_2 - T_1)\right].$$

Substituting the known values, we get

$$20 = 16\left[1 + \alpha\,(100 - 15)\right].$$

First evaluate the temperature difference:

$$100 - 15 = 85^{\circ}\text{C}.$$

So the equation becomes

$$20 = 16\left[1 + 85\,\alpha\right].$$

Now divide both sides by 16 to isolate the bracketed term:

$$\frac{20}{16} = 1 + 85\,\alpha.$$

Simplify the left-hand fraction:

$$\frac{20}{16} = 1.25.$$

Thus we have

$$1.25 = 1 + 85\,\alpha.$$

Subtract 1 from both sides:

$$1.25 - 1 = 85\,\alpha.$$

$$0.25 = 85\,\alpha.$$

Finally, divide by 85 to solve for $$\alpha$$:

$$\alpha = \frac{0.25}{85}.$$

Perform the division step by step:

$$\alpha = \frac{25}{8500} = \frac{1}{340} \;\text{per}\;^{\circ}\text{C}.$$

Numerically,

$$\alpha \approx 0.00294\;^{\circ}\text{C}^{-1} \approx 0.003\;^{\circ}\text{C}^{-1}.$$

This matches the third option given in the list.

Hence, the correct answer is Option C.

We need to determine how physical properties like drift velocity, electric field, and electron current change as we move from point $$P$$ to point $$Q$$ along a non-uniform conductor.

1. Analyze the Geometry of the Conductor

From the problem details :

• The conductor has a tapering cross-section.
• The radius at $$P$$ is $$r_1$$, and the radius at $$Q$$ is $$r_2$$, given that $$r_2 < r_1$$.
• As you move from $$P$$ to $$Q$$, the radius decreases, which means the cross-sectional area ($$A = \pi r^2$$) decreases.

2. Analyze individual Physical Quantities

• Electron Current ($$I$$):
  The conductor is connected in a single series loop with a battery. Due to the conservation of charge, the steady-state electric current must remain constant at every cross-section along the conductor.

  $$\text{Current } I = \text{constant}$$

  Therefore, the statement "Electron current decreases" is incorrect.


• Drift Velocity ($$v_d$$):
  The relationship between electric current and drift velocity is given by the formula:

  $$I = n e A v_d \implies v_d = \frac{I}{n e A}$$

  Where $$n$$ is the free electron density and $$e$$ is the elementary charge (both constant for a given material). Since current $$I$$ is constant and the area $$A$$ decreases as we approach $$Q$$, the drift velocity must increase:

  $$v_d \propto \frac{1}{A}$$



• Electric Field ($$E$$):
  According to the microscopic form of Ohm's Law, current density ($$J$$) is related to the electric field by $$J = \sigma E$$, where $$\sigma$$ is the electrical conductivity. Since $$J = \frac{I}{A}$$:

  $$E = \frac{I}{\sigma A} \implies E \propto \frac{1}{A}$$

  As area $$A$$ decreases from $$P$$ to $$Q$$, the electric field strength must increase. Therefore, the statement "Electric field decreases" is incorrect.

Conclusion

As one moves from $$P$$ to $$Q$$, the narrowing area causes the drift velocity of the electrons to increase. 

We start with the basic electrical power-energy relations. The instantaneous electrical power converted to heat in a resistor is given by the formula $$P = I^{2} R$$, where $$P$$ is power in watts, $$I$$ is current in amperes and $$R$$ is resistance in ohms. The thermal energy (heat) produced in a time interval $$t$$ is the product of power and time, i.e. $$E = P\,t$$.

From the data provided for the first situation: energy dissipated $$E_{1} = 192\ \text{J}$$ in time $$t_{1} = 1\ \text{s}$$ with current $$I_{1} = 4\ \text{A}$$.

Using $$E = P\,t$$, we obtain the initial power:

$$P_{1} = \frac{E_{1}}{t_{1}} = \frac{192\ \text{J}}{1\ \text{s}} = 192\ \text{W}.$$

Now we employ $$P = I^{2} R$$ to determine the resistance of the resistor. Substituting $$P_{1} = 192\ \text{W}$$ and $$I_{1} = 4\ \text{A}$$:

$$192 = (4)^{2} R.$$

Since $$(4)^{2} = 16$$, we get

$$192 = 16\,R.$$

Solving for $$R$$:

$$R = \frac{192}{16} = 12\ \Omega.$$

Next, the current is doubled, so the new current is

$$I_{2} = 2\,I_{1} = 2 \times 4\ \text{A} = 8\ \text{A}.$$

With the same resistor, the new power $$P_{2}$$ is again obtained from $$P = I^{2} R$$:

$$P_{2} = I_{2}^{2} R = (8)^{2} \times 12.$$

Calculating $$(8)^{2} = 64$$, we have

$$P_{2} = 64 \times 12 = 768\ \text{W}.$$

This power acts for a new time interval of $$t_{2} = 5\ \text{s}$$. The thermal energy produced in this interval is therefore

$$E_{2} = P_{2}\,t_{2} = 768\ \text{W} \times 5\ \text{s}.$$

Multiplying, we obtain

$$E_{2} = 768 \times 5 = 3840\ \text{J}.$$

So, the answer is $$3840\ \text{J}$$.

We begin by recalling the formula for the electric power developed across a resistor. For a resistor of resistance $$R$$ connected to a potential difference $$V$$, the power dissipated is given by the Joule-Lenz law:

$$P=\dfrac{V^{2}}{R}.$$

Now the original heating wire has a resistance of $$36\;\Omega$$ and is connected directly across a potential difference of $$240\;{\rm V}$$. Substituting these values in the power formula we get

$$P_{1}= \dfrac{(240)^{2}}{36}.$$

Next we cut the wire into two equal halves. Because the wire is uniform, its resistance is proportional to its length. Halving the length therefore halves the resistance. Thus the resistance of each half becomes

$$R_{\text{half}} = \dfrac{36\;\Omega}{2}=18\;\Omega.$$

Each half-wire is now connected independently across the same potential difference of $$240\;{\rm V}$$. Applying the power formula to one half, we obtain

$$P_{\text{half}} = \dfrac{(240)^{2}}{18}.$$

There are two such identical halves, so the total power dissipated in the second arrangement is the sum of the powers in the two halves:

$$P_{2} = 2\,P_{\text{half}} = 2\left(\dfrac{(240)^{2}}{18}\right)=\dfrac{(240)^{2}}{9}.$$

We are asked for the ratio of the power in the first case to the total power in the second case. Writing the ratio explicitly,

$$\dfrac{P_{1}}{P_{2}} = \dfrac{\dfrac{(240)^{2}}{36}}{\dfrac{(240)^{2}}{9}}.$$

The common factor $$(240)^{2}$$ cancels out, leaving

$$\dfrac{P_{1}}{P_{2}} = \dfrac{1/36}{1/9}= \dfrac{9}{36}= \dfrac{1}{4}.$$

Hence the ratio $$P_{1}:P_{2}$$ is $$1:4$$, so the value of $$x$$ is $$4$$.

Hence, the correct answer is Option 4.

We first recall the basic power-voltage-resistance relation for a purely resistive load: $$P=\dfrac{V^{2}}{R}\,.$$

The bulb is rated $$200\ \text{W}$$ at $$100\ \text{V}$$. Substituting these rated values in the above formula gives us its resistance:

$$R_{\text{bulb}}=\dfrac{V^{2}}{P}=\dfrac{(100\ \text{V})^{2}}{200\ \text{W}}=\dfrac{10000}{200}=50\ \Omega.$$

Thus the filament of the bulb behaves like a $$50\ \Omega$$ resistor when it is glowing at its rated conditions.

Now the circuit will be connected to a supply of $$200\ \text{V}$$, but we still want the bulb itself to have only its rated voltage of $$100\ \text{V}$$ across it so that it continues to deliver $$200\ \text{W}$$ of power. To find the current that must flow through the bulb under rated conditions, we use the power formula $$P=VI$$ (power equals voltage times current):

$$I=\dfrac{P}{V}=\dfrac{200\ \text{W}}{100\ \text{V}}=2\ \text{A}.$$

Therefore a current of $$2\ \text{A}$$ must flow through the bulb as well as through any series element we add.

Because the source voltage is $$200\ \text{V}$$ and the bulb needs only $$100\ \text{V}$$, the remaining voltage that must be dropped across the series resistor is

$$V_{\text{series}}=V_{\text{supply}}-V_{\text{bulb}}=200\ \text{V}-100\ \text{V}=100\ \text{V}.$$

Using Ohm’s law $$V=IR$$ for the series resistor, and noting that the same current $$2\ \text{A}$$ flows through it, we get

$$R_{\text{series}}=\dfrac{V_{\text{series}}}{I}=\dfrac{100\ \text{V}}{2\ \text{A}}=50\ \Omega.$$

So, the answer is $$50\ \Omega$$.

The radius of the cylindrical wire is $$r = 0.5$$ mm $$= 0.5 \times 10^{-3}$$ m, the conductivity is $$\sigma = 5 \times 10^7$$ S m$$^{-1}$$, and the applied electric field is $$E = 10$$ mV m$$^{-1} = 10 \times 10^{-3} = 10^{-2}$$ V m$$^{-1}$$.

The current density is given by $$J = \sigma E = 5 \times 10^7 \times 10^{-2} = 5 \times 10^5$$ A m$$^{-2}$$.

The cross-sectional area of the wire is $$A = \pi r^2 = \pi \times (0.5 \times 10^{-3})^2 = \pi \times 0.25 \times 10^{-6} = 0.25\pi \times 10^{-6}$$ m$$^2$$.

The current through the wire is $$I = J \times A = 5 \times 10^5 \times 0.25\pi \times 10^{-6} = 1.25 \times 10^{-1} \times \pi = 0.125\pi$$ A $$= 125\pi$$ mA.

Comparing with the given expression $$I = x^3 \pi$$ mA, we get $$x^3 = 125$$, which gives $$x = 5$$.

Therefore, the value of $$x$$ is $$5$$.

We start by noting that the wire originally forms a square, and along each of its four sides the resistance is given as $$3\;\Omega$$. Resistance of a conductor is directly proportional to its length (for the same material and cross-section). Therefore the total resistance around the complete square is obtained by simply adding the resistances of the four equal sides:

$$R_{\text{square}} = 3\;\Omega + 3\;\Omega + 3\;\Omega + 3\;\Omega = 12\;\Omega.$$

Now the square is reshaped into a perfect circle. The material, length and cross-section of the wire all remain unchanged, so its total resistance also remains unchanged. Hence the resistance of the whole circumference of the new circle is still

$$R_{\text{circle}} = 12\;\Omega.$$

We are asked to find the resistance between two diametrically opposite points on this circle. Choosing those opposite points divides the circle into two equal semicircular arcs. Because the wire is uniform, each semicircle owns exactly half the total length and therefore half the total resistance.

So, the resistance of each semicircle is

$$R_{\text{semi}} = \dfrac{R_{\text{circle}}}{2} = \dfrac{12\;\Omega}{2} = 6\;\Omega.$$

Between the chosen diametrically opposite points, current can travel through either the upper semicircle or the lower semicircle; these two paths connect the same pair of nodes, so they are effectively in parallel.

For two resistors $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance is found using the parallel-combination formula

$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}.$$

Substituting $$R_1 = R_2 = 6\;\Omega$$ gives

$$\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{6\;\Omega} + \dfrac{1}{6\;\Omega} = \dfrac{2}{6\;\Omega} = \dfrac{1}{3\;\Omega}.$$

Taking the reciprocal yields the equivalent resistance:

$$R_{\text{eq}} = 3\;\Omega.$$

So, the answer is $$3\;\Omega$$.

Two wires of the same length $$l$$ and the same thickness (cross-sectional area $$A$$) have specific resistivities $$\rho_1 = 6\,\Omega\,\text{cm}$$ and $$\rho_2 = 3\,\Omega\,\text{cm}$$. Their resistances are $$R_1 = \frac{\rho_1 l}{A}$$ and $$R_2 = \frac{\rho_2 l}{A}$$.

When connected in parallel, the equivalent resistance is $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{A}{\rho_1 l} + \frac{A}{\rho_2 l} = \frac{A}{l}\left(\frac{1}{6} + \frac{1}{3}\right) = \frac{A}{l} \cdot \frac{1}{2}$$, so $$R_{eq} = \frac{2l}{A}$$.

The parallel combination of two identical-geometry wires is equivalent to a single conductor of length $$l$$ and cross-section $$2A$$. The effective resistivity satisfies $$R_{eq} = \frac{\rho \cdot l}{2A}$$, giving $$\frac{2l}{A} = \frac{\rho l}{2A}$$, hence $$\rho = 4\,\Omega\,\text{cm}$$.

The value of $$\rho$$ is $$4$$.

The energy dissipated by a resistor is given by $$E = I^2 R t$$.

Given: $$E = 10$$ mJ $$= 10 \times 10^{-3}$$ J, $$I = 2$$ mA $$= 2 \times 10^{-3}$$ A, and $$t = 1$$ s.

Substituting: $$10 \times 10^{-3} = (2 \times 10^{-3})^2 \times R \times 1$$.

$$10 \times 10^{-3} = 4 \times 10^{-6} \times R$$

$$R = \frac{10 \times 10^{-3}}{4 \times 10^{-6}} = \frac{10^{-2}}{4 \times 10^{-6}} = 2500 \; \Omega$$.

The resistance is $$\boxed{2500}$$ $$\Omega$$.

We begin by noting that the electric fuse has to carry the total current drawn by all the appliances that may be switched on together. The current is obtained from power with the well-known relation:

$$P = VI$$

where $$P$$ is the electric power in watts (W), $$V$$ is the supply voltage in volts (V) and $$I$$ is the current in amperes (A). Rearranging the formula, we obtain

$$I = \dfrac{P}{V}$$

To use this expression, we first calculate the total power consumption of the building by adding the power ratings of every device present.

We have 15 bulbs rated at 45 W each. So their combined power is

$$P_1 = 15 \times 45\ \text{W} = 675\ \text{W}$$

Now, 15 bulbs are rated at 100 W each. Hence

$$P_2 = 15 \times 100\ \text{W} = 1500\ \text{W}$$

Next, 15 small fans are rated at 10 W each, giving

$$P_3 = 15 \times 10\ \text{W} = 150\ \text{W}$$

Finally, there are 2 heaters, each of 1 kW. Remembering that $$1\ \text{kW}=1000\ \text{W}$$, we write

$$P_4 = 2 \times 1\ \text{kW} = 2 \times 1000\ \text{W} = 2000\ \text{W}$$

Adding all these powers gives the total power demand:

$$\begin{aligned} P_{\text{total}} &= P_1 + P_2 + P_3 + P_4 \\ &= 675\ \text{W} + 1500\ \text{W} + 150\ \text{W} + 2000\ \text{W} \\ &= 4325\ \text{W} \end{aligned}$$

The supply voltage of the mains is given as 220 V. Substituting $$P_{\text{total}}$$ and $$V=220\ \text{V}$$ in the rearranged power formula, we get the total current:

$$\begin{aligned} I_{\text{total}} &= \dfrac{P_{\text{total}}}{V} \\ &= \dfrac{4325\ \text{W}}{220\ \text{V}} \\ &= 19.659\ \text{A} \end{aligned}$$

Numerically, this equals approximately $$19.7\ \text{A}$$. A fuse is chosen so that its rating is just above the maximum expected current; if it were below, the fuse would blow during normal operation. The next available standard rating higher than $$19.7\ \text{A}$$ is $$20\ \text{A}$$.

Hence, the correct answer is Option D.

We model the torch battery as an ideal emf source $$\mathcal E$$ in series with its internal resistance, which is the resistance of the electrolyte that fills the space between the inner solid rod (radius $$a$$) and the outer cylindrical shell (inner radius $$b$$). Our first task is therefore to calculate this internal resistance, which we shall denote by $$r$$.

To find $$r$$, we remember the differential formula for the resistance of a conductor:

$$dR=\rho\,\dfrac{dl}{A},$$

where $$\rho$$ is the resistivity, $$dl$$ is the infinitesimal length of the current path, and $$A$$ is the cross-sectional area perpendicular to the current.

In the present arrangement the current flows radially outward from the inner rod to the outer shell. Choose an imaginary cylindrical surface of radius $$r$$ (with $$a<r<b$$) and thickness $$dr$$. The current crosses this surface normally, so:

• The path length for the current element is the radial thickness $$dl=dr$$.
• The area normal to the current is the curved area of the cylinder, $$A = 2\pi r\,l$$, where $$l$$ is the length of the battery.

Substituting these into the differential resistance formula, we obtain

$$dR=\rho\,\dfrac{dr}{2\pi r\,l}.$$

To get the total internal resistance, integrate from the inner radius $$a$$ to the outer radius $$b$$:

$$r = \int_{a}^{b}dR = \int_{a}^{b}\rho\,\dfrac{dr}{2\pi r\,l} = \dfrac{\rho}{2\pi l}\int_{a}^{b}\dfrac{dr}{r} = \dfrac{\rho}{2\pi l}\,\Bigl[\ln r\Bigr]_{a}^{b} = \dfrac{\rho}{2\pi l}\,\ln\!\left(\dfrac{b}{a}\right).$$

Now the circuit consists of the emf $$\mathcal E$$, the internal resistance $$r$$ and the external resistance $$R$$ in series. The Joule heating (power) in the external resistor is

$$P=\dfrac{\mathcal E^{2}R}{\left(R+r\right)^{2}}.$$

The maximum power transfer theorem states that, for a fixed internal resistance $$r$$, the power delivered to $$R$$ is maximum when

$$R=r.$$

Therefore, using the explicit form of $$r$$ we just calculated, the condition for maximum Joule heating is

$$R=\dfrac{\rho}{2\pi l}\,\ln\!\left(\dfrac{b}{a}\right).$$

This expression matches Option B.

Hence, the correct answer is Option B.

We begin by recalling the definition of resistivity. For any homogeneous conductor we have the relation

$$R=\rho\,\frac{\ell}{A},$$

where $$R$$ is the resistance of the specimen, $$\ell$$ its length, $$A$$ its cross-sectional area and $$\rho$$ the material‐dependent constant called resistivity. A smaller value of $$\rho$$ therefore corresponds to a better electrical conductor, while a larger value corresponds to a poorer conductor.

The standard tabulated resistivities at room temperature (≈ 20 °C) for the four materials mentioned in the question are:

$$\rho_C \;( \text{copper}) \approx 1.72\times10^{-8}\;\Omega\text{m},$$

$$\rho_A \;( \text{aluminium}) \approx 2.8\times10^{-8}\;\Omega\text{m},$$

$$\rho_T \;( \text{tungsten}) \approx 5.6\times10^{-8}\;\Omega\text{m},$$

$$\rho_M \;( \text{mercury}) \approx 9.5\times10^{-7}\;\Omega\text{m}.$$

Now we arrange these values in descending order (largest first):

$$\rho_M \;>\;\rho_T \;>\;\rho_A \;>\;\rho_C.$$

Next we compare this true order with each option supplied in the problem statement.

A. $$\rho_C > \rho_A > \rho_T$$ contradicts the numerical data because $$\rho_C$$ is the smallest, not the largest.

C. $$\rho_A > \rho_T > \rho_C$$ is also wrong because $$\rho_T$$ is greater than $$\rho_A$$, not the other way round.

D. $$\rho_A > \rho_M > \rho_C$$ is incorrect because $$\rho_M$$ is the largest of all; it cannot be less than $$\rho_A$$.

B. $$\rho_M > \rho_A > \rho_C$$ does not mention tungsten, but it is completely consistent with the accurate ordering of the three quantities that do appear, namely $$\rho_M > \rho_A$$ and $$\rho_A > \rho_C$$. Because the other three options are plainly false, Option B is the only statement that matches the established numerical hierarchy.

Hence, the correct answer is Option B.

We are told that the galvanometer coil has an internal resistance $$G = 100\ \Omega$$ and that it produces full-scale deflection when a current of $$I_g = 1\ \text{mA} = 1\times10^{-3}\ \text{A}$$ flows through it.

To convert this galvanometer into a voltmeter that reads full scale at a potential difference of $$V = 10\ \text{V}$$, we must connect an additional resistance $$R_s$$ in series with the galvanometer. In this series arrangement, the same current $$I_g$$ will still correspond to full-scale deflection, but now the total voltage across the series combination must be $$V$$.

Ohm’s law states that for a resistor, $$V = I R.$$ Applying it to the entire series combination, we have $$V = I_g\,(G + R_s).$$

We substitute the known values: $$10\ \text{V} = (1\times10^{-3}\ \text{A})\,(100\ \Omega + R_s).$$

Now we divide both sides by $$1\times10^{-3}\ \text{A}$$ to isolate the bracket: $$\frac{10}{1\times10^{-3}} = 100\ \Omega + R_s.$$

The left side simplifies because $$\frac{10}{1\times10^{-3}} = 10 \times 10^{3} = 10\,000.$$ So we get $$10\,000\ \Omega = 100\ \Omega + R_s.$$

Next, we subtract $$100\ \Omega$$ from both sides to solve for $$R_s$$: $$R_s = 10\,000\ \Omega - 100\ \Omega.$$

Simplifying the subtraction, $$R_s = 9\,900\ \Omega.$$

Expressing this resistance in kilohms: $$R_s = 9.9\ \text{k}\Omega.$$

Hence, the correct answer is Option D.

We are told that the line is supplying a useful power of $$P_{\text{out}} = 1\,\text{kW} = 1000\,\text{W}$$ to a load whose terminal voltage is $$V = 220\,\text{V}$$.

First, we find the current flowing through the line. We recall the electric‐power formula

$$P = VI,$$

where $$P$$ is the power delivered to the load, $$V$$ is the load voltage, and $$I$$ is the current. Solving this formula for $$I$$, we have

$$I = \frac{P}{V}.$$

Substituting $$P = 1000\,\text{W}$$ and $$V = 220\,\text{V}$$ gives

$$I = \frac{1000}{220}\,\text{A} \approx 4.545\,\text{A}.$$

Now, the transmission line itself has a total resistance of $$R = 2\,\Omega$$. Whenever current flows through a resistance, there is power lost as heat. The formula for this loss (Joule heating) is

$$P_{\text{loss}} = I^{2}R.$$

Substituting $$I \approx 4.545\,\text{A}$$ and $$R = 2\,\Omega$$, we obtain

$$\begin{aligned} P_{\text{loss}} &= (4.545\,\text{A})^{2}\times 2\,\Omega \\ &= 20.66 \times 2 \\ &\approx 41.32\,\text{W}. \end{aligned}$$

The total power that must be supplied at the sending end of the line is therefore the sum of the useful power delivered and the power lost:

$$P_{\text{in}} = P_{\text{out}} + P_{\text{loss}} = 1000\,\text{W} + 41.32\,\text{W} \approx 1041.32\,\text{W}.$$

Finally, the efficiency $$\eta$$ of the transmission line is defined as

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times 100\%.$$

Substituting the numbers we have just found,

$$\eta = \frac{1000}{1041.32}\times 100\% \approx 0.9603 \times 100\% \approx 96\%.$$

So the efficiency of this transmission line is about $$96\%$$.

Hence, the correct answer is Option D.

In a meter bridge, the condition for a balanced Wheatstone bridge is given by the relation

$$\dfrac{R}{S} \;=\; \dfrac{l}{100 - l},$$

where $$R$$ is the resistance in the resistance box arm, $$S$$ is the standard resistance, $$l$$ is the balancing length measured from one end, and the total length of the bridge wire is taken as 100 cm.

We are first told that the balancing length is $$l = 25\text{ cm}$$. Substituting this value into the above relation, we obtain

$$\dfrac{R}{S} \;=\; \dfrac{25}{100 - 25} \;=\; \dfrac{25}{75} \;=\; \dfrac{1}{3}.$$

Hence we conclude that

$$\dfrac{R}{S} = \dfrac13 \quad\Longrightarrow\quad R = \dfrac{S}{3}.$$

Now the wire of resistance $$R$$ is replaced by another wire made of the same material but having half the length and half the diameter. To find the new resistance $$R'$$ of this wire, we recall the resistance formula

$$R = \rho\,\dfrac{L}{A},$$

where $$\rho$$ is the resistivity of the material (unchanged because the material is the same), $$L$$ is the length, and $$A$$ is the cross-sectional area. For a cylindrical wire, the area is

$$A = \pi\Bigl(\dfrac{d}{2}\Bigr)^2 = \dfrac{\pi d^{2}}{4},$$

with $$d$$ being the diameter.

Let the original wire have length $$L$$ and diameter $$d$$, so its area is $$A = \dfrac{\pi d^{2}}{4}$$ and its resistance is

$$R = \rho\,\dfrac{L}{A}.$$

The new wire has

• length $$L' = \dfrac{L}{2},$$
• diameter $$d' = \dfrac{d}{2}.$$

The new area is therefore

$$A' = \dfrac{\pi (d')^{2}}{4} = \dfrac{\pi}{4}\Bigl(\dfrac{d}{2}\Bigr)^{2} = \dfrac{\pi d^{2}}{16} = \dfrac{A}{4}.$$

Using the resistance formula again, the new resistance $$R'$$ is

$$R' = \rho\,\dfrac{L'}{A'} = \rho\,\dfrac{\dfrac{L}{2}}{\dfrac{A}{4}} = \rho\,\dfrac{L}{2}\cdot\dfrac{4}{A} = 2\,\rho\,\dfrac{L}{A} = 2R.$$

Thus we have the simple relation

$$R' = 2R.$$

Because the standard resistance $$S$$ remains unchanged, the new balance condition for the meter bridge becomes

$$\dfrac{R'}{S} = \dfrac{l'}{100 - l'},$$

where $$l'$$ is the new balancing length to be found. Substituting $$R' = 2R$$ and the earlier ratio $$\dfrac{R}{S} = \dfrac13$$, we get

$$\dfrac{R'}{S} = \dfrac{2R}{S} = 2\left(\dfrac{R}{S}\right) = 2\left(\dfrac13\right) = \dfrac23.$$

Hence we must satisfy

$$\dfrac{l'}{100 - l'} = \dfrac23.$$

Cross-multiplying, we write

$$3\,l' = 2\,(100 - l').$$

Expanding the right-hand side gives

$$3l' = 200 - 2l'.$$

Next we collect the $$l'$$ terms on one side:

$$3l' + 2l' = 200,$$

which simplifies to

$$5l' = 200.$$

Finally, dividing by 5 yields

$$l' = \dfrac{200}{5} = 40\text{ cm}.$$

So, the answer is $$40\text{ cm}$$.

We have a current $$I = 5\ \text{A}$$ passing through a copper wire whose resistivity is given as $$\rho = 1.7 \times 10^{-8}\ \Omega\,\text{m}$$. The radius of the circular cross-section is $$r = 5\ \text{mm}$$, which in SI units is

$$r = 5\ \text{mm} = 5 \times 10^{-3}\ \text{m}.$$

The drift velocity of the charge carriers is provided as $$v_d = 1.1 \times 10^{-3}\ \text{m s}^{-1}$$. We wish to find the mobility $$\mu$$ of these carriers.

First, we recall the relation between mobility, drift velocity and electric field:

$$\mu = \frac{v_d}{E}.$$

Therefore, we must determine the electric field $$E$$ inside the conductor. Ohm’s microscopic form gives us

$$E = \rho J,$$

where $$J$$ is the current density. Current density itself is defined by

$$J = \frac{I}{A},$$

with $$A$$ being the cross-sectional area of the wire. Because the cross-section is a circle, its area is

$$A = \pi r^2.$$

Substituting the numerical value of the radius, we get

$$A = \pi \left(5 \times 10^{-3}\ \text{m}\right)^2 = \pi \times 25 \times 10^{-6}\ \text{m}^2 = 25\pi \times 10^{-6}\ \text{m}^2.$$ Numerically,

$$A = 25 \times 3.1416 \times 10^{-6}\ \text{m}^2 \approx 78.54 \times 10^{-6}\ \text{m}^2 = 7.85 \times 10^{-5}\ \text{m}^2.$$

Now we find the current density:

$$J = \frac{I}{A} = \frac{5\ \text{A}}{7.85 \times 10^{-5}\ \text{m}^2} \approx 6.37 \times 10^{4}\ \text{A m}^{-2}.$$

Next, we calculate the electric field using $$E = \rho J$$:

$$E = (1.7 \times 10^{-8}\ \Omega\,\text{m}) \left(6.37 \times 10^{4}\ \text{A m}^{-2}\right) = 1.7 \times 6.37 \times 10^{-8 + 4}\ \text{V m}^{-1} = 10.829 \times 10^{-4}\ \text{V m}^{-1} \approx 1.08 \times 10^{-3}\ \text{V m}^{-1}.$$

Finally, we obtain the mobility:

$$\mu = \frac{v_d}{E} = \frac{1.1 \times 10^{-3}\ \text{m s}^{-1}} {1.08 \times 10^{-3}\ \text{V m}^{-1}} \approx 1.02\ \text{m}^2 \text{V}^{-1} \text{s}^{-1}.$$

Rounding to one significant figure, $$\mu \approx 1.0\ \text{m}^2 \text{V}^{-1} \text{s}^{-1}$$, which matches the third option in the list.

Hence, the correct answer is Option C.

First, recall the fundamental relation between current and charge. The electric current $$I$$ through any conductor (or here, through the leads of the capacitor) is defined as the time‐rate of change of charge $$q$$ flowing past a given point. Mathematically, the definition is

$$I \;=\;\dfrac{dq}{dt}$$

We are given a graph of charge on the capacitor plate as a function of time. The question asks for the value of the current at the particular instant $$t = 4\;\text{s}$$. From the definition above, this current is nothing but the slope of the $$q$$-versus-$$t$$ curve at that instant.

Now, observe the portion of the graph that covers the interval from $$t = 3\;\text{s}$$ to $$t = 5\;\text{s}$$. In this stretch, the plotted charge remains perfectly horizontal, indicating that the value of charge does not change with time. Expressed algebraically, during this interval we have

$$q(t) = \text{constant}$$

Since the function is constant, its derivative with respect to time is zero:

$$\dfrac{dq}{dt} = 0$$

Substituting this derivative back into the current definition, we obtain

$$I = \dfrac{dq}{dt} = 0$$

Hence, at $$t = 4\;\text{s}$$—which lies squarely in the mentioned flat region—the current through the circuit is zero.

Hence, the correct answer is Option C.

We know that the resistance of a uniform wire is governed by the relation $$R=\rho\frac{L}{A},$$ where $$\rho$$ is the resistivity of the material, $$L$$ is the length of the wire and $$A$$ is its cross-sectional area.

The original wire has resistance $$R_0 = 3\;\Omega$$ and let its original length and cross-sectional area be $$L_0$$ and $$A_0$$ respectively. So for the first wire we may write

$$R_0 = \rho\frac{L_0}{A_0}=3\;\Omega.$$

Now the wire is stretched so that its new length becomes double, that is

$$L_1 = 2L_0.$$

While stretching, the volume of the wire remains constant. Volume $$V$$ is given by $$V=L\times A$$, hence

$$L_0A_0 = L_1A_1 \quad\Longrightarrow\quad A_1=\frac{L_0}{L_1}A_0=\frac{L_0}{2L_0}A_0=\frac{A_0}{2}.$$

So the new cross-sectional area is one-half of the original. Substituting $$L_1$$ and $$A_1$$ in the resistance formula we obtain the resistance of the elongated straight wire:

$$R_1=\rho\frac{L_1}{A_1}=\rho\frac{2L_0}{A_0/2}=\rho\frac{2L_0}{A_0}\times2=4\,\rho\frac{L_0}{A_0}=4R_0=4\times3\;\Omega=12\;\Omega.$$

This wire of resistance $$12\;\Omega$$ is now bent into a complete circle. Because the material remains uniform, resistance is directly proportional to the length of any arc. The total circumference of the circle corresponds to $$12\;\Omega$$.

Two points on the circle are chosen such that the angle subtended at the centre is $$60^\circ$$. The circle has $$360^\circ$$ in all, so the fraction of the circumference between the two points along the shorter arc is

$$\frac{60^\circ}{360^\circ}=\frac16.$$

Therefore, the resistance of the shorter arc is

$$R_{\text{short}} = 12\;\Omega \times \frac16 = 2\;\Omega.$$

The longer arc spans the remaining angle $$360^\circ-60^\circ = 300^\circ$$, i.e. a fraction

$$\frac{300^\circ}{360^\circ}=\frac56,$$

so its resistance is

$$R_{\text{long}} = 12\;\Omega \times \frac56 = 10\;\Omega.$$

Between the chosen two points these two arcs provide two parallel conducting paths. For two resistances $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance is given by the formula

$$R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2}.$$

Putting $$R_1 = 2\;\Omega$$ and $$R_2 = 10\;\Omega$$, we get

$$R_{\text{eq}} = \frac{2\;\Omega \times 10\;\Omega}{2\;\Omega + 10\;\Omega} = \frac{20}{12}\;\Omega = \frac{5}{3}\;\Omega.$$

Hence, the correct answer is Option A.

Let the resistance of each resistor be $$R$$ ohms. The battery maintains a constant potential difference, say $$V$$ volts.

First, we use the relation between power, voltage and resistance for a resistor:

$$P \;=\;\dfrac{V^2}{R_{\text{eq}}}$$

where $$P$$ is the electric power and $$R_{\text{eq}}$$ is the equivalent resistance connected across the battery.

We have the two resistances in series initially, so their equivalent resistance is

$$R_{\text{series}} \;=\; R + R \;=\; 2R.$$

The power consumed in this series arrangement is given to be 60 W. Hence,

$$P_{\text{series}} \;=\; 60 \;=\;\dfrac{V^2}{R_{\text{series}}} \;=\;\dfrac{V^2}{2R}.$$

Multiplying both sides by $$2R$$, we get

$$V^2 \;=\; 60 \times 2R \;=\; 120R.$$

Now the same two resistances are connected in parallel. The equivalent resistance of two equal resistors in parallel is

$$R_{\text{parallel}} \;=\;\dfrac{R}{2}.$$

The power consumed in the parallel arrangement is therefore

$$P_{\text{parallel}} \;=\;\dfrac{V^2}{R_{\text{parallel}}} \;=\;\dfrac{V^2}{\dfrac{R}{2}} \;=\; V^2 \times \dfrac{2}{R} \;=\; \dfrac{2V^2}{R}.$$

We substitute the value of $$V^2$$ obtained earlier, $$V^2 = 120R$$:

$$P_{\text{parallel}} \;=\;\dfrac{2 \times 120R}{R} \;=\; 240 \text{ W}.$$

Hence, the correct answer is Option B.

We recall the resistor colour-code convention. For the significant digits we have:
Black → 0, Brown → 1, Red → 2, Orange → 3, Yellow → 4, Green → 5, Blue → 6, Violet → 7, Grey → 8, White → 9.

In an ordinary three-band code the first two coloured rings give the first two digits of the resistance, while the third ring gives the multiplier, i.e. a factor of $$10^{\text{(third-band digit)}}$$.

The given resistance is $$200\;\Omega$$. Let us work out which colours must appear.

We want to write $$200$$ in the form $$(\text{first digit}\times 10 + \text{second digit}) \times 10^{\,\text{multiplier digit}}.$$

Observing $$200 = 20 \times 10^{1}$$ we identify

 • First digit $$=2$$, so the first band is Red.
 • Second digit $$=0$$, so the second band is Black.
 • Multiplier digit $$=1$$, so the third band is Brown (because Brown represents 1 and hence the factor $$10^{1}$$).

Thus the original colour sequence is Red - Black - Brown, giving the required $$200\;\Omega$$.

Now the question says that every Red band is replaced by Green. In our code only the first band is Red, so only the first significant digit changes.

Green represents the digit $$5$$. Therefore, after the replacement we have

 • First digit $$=5$$ (Green).
 • Second digit $$=0$$ (Black).
 • Multiplier digit remains $$=1$$ (Brown).

So the new resistance is calculated as follows:

We have $$\bigl(5 \times 10 + 0\bigr) \times 10^{1}$$.

First we evaluate the bracket:
$$5 \times 10 + 0 = 50.$$

Next we apply the multiplier $$10^{1}=10$$:
$$50 \times 10 = 500.$$

Hence the new resistance is $$500\;\Omega$$.

Hence, the correct answer is Option D.

We have a uniform metallic wire whose total resistance is given as $$18\,\Omega$$. Because a metallic wire is uniform, its resistance is directly proportional to its length, as stated by the basic relation $$R=\rho\frac{L}{A},$$ where $$\rho$$ is the resistivity, $$L$$ is the length, and $$A$$ is the cross-sectional area. Since $$\rho$$ and $$A$$ remain the same for the whole wire, halving or thirding the length simply halves or thirds the resistance.

Now the wire is bent to form an equilateral triangle. An equilateral triangle has three sides of equal length, so the original length of the wire is divided equally into three parts. Therefore each side uses exactly one-third of the original length. Because resistance is proportional to length, each side acquires one-third of the original resistance:

$$R_{\text{side}}=\frac{1}{3}\times 18\,\Omega = 6\,\Omega.$$

Label the three vertices as $$A,\,B,\,C$$. Each of the three sides $$AB,\,BC,\,CA$$ now has a resistance of $$6\,\Omega$$.

The question asks for the resistance measured between any two vertices; let us choose vertices $$A$$ and $$B$$. There are two distinct electrical paths connecting these vertices:

1. The direct side $$AB$$ itself, whose resistance is $$6\,\Omega$$.

2. The indirect path that goes from $$A$$ to $$C$$ and then from $$C$$ to $$B$$, passing through two sides in series. The resistance of this longer path is

$$R_{AC}+R_{CB}=6\,\Omega+6\,\Omega=12\,\Omega.$$

Thus, between vertices $$A$$ and $$B$$ we effectively have two resistances—one of $$6\,\Omega$$ and the other of $$12\,\Omega$$—connected in parallel. For two resistors $$R_1$$ and $$R_2$$ in parallel, the equivalent resistance $$R_{\text{eq}}$$ is given by the formula

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

Substituting $$R_1=6\,\Omega$$ and $$R_2=12\,\Omega$$ we get

$$\frac{1}{R_{\text{eq}}}=\frac{1}{6}+\frac{1}{12} =\frac{2}{12}+\frac{1}{12} =\frac{3}{12} =\frac{1}{4}.$$

Taking the reciprocal on both sides,

$$R_{\text{eq}}=4\,\Omega.$$

This equivalent resistance is the same no matter which pair of vertices is chosen, because all three sides are identical.

Hence, the correct answer is Option D.

The total resistance of the wire is R. Since it is bent into a square ABCD, each side has a resistance of R/4.

Point E is the midpoint of side $CD$. This divides the side CD into two segments, DE and EC, each having a resistance of:

$$R_{DE} = R_{EC} = \frac{1}{2} \left( \frac{R}{4} \right) = \frac{R}{8}$$

To find the effective resistance between E and C, we identify two parallel paths:

1. Path 1 (Direct): Only the segment EC.$$R_1 = \frac{R}{8}$$
2. Path 2 (The rest of the square): Segments $$ED \to DA \to AB \to BC$$
3. .$$R_2 = \frac{R}{8} + \frac{R}{4} + \frac{R}{4} + \frac{R}{4} = \frac{R + 2R + 2R + 2R}{8} = \frac{7R}{8}$$

The effective resistance $R_{eq}$ is calculated using the parallel formula:

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

$$\frac{1}{R_{eq}} = \frac{1}{R/8} + \frac{1}{7R/8}$$

$$\frac{1}{R_{eq}} = \frac{8}{R} + \frac{8}{7R}$$

Taking the common denominator $7R$:

$$\frac{1}{R_{eq}} = \frac{56 + 8}{7R} = \frac{64}{7R}$$

Solving for $$R_{eq}$$:

$$\boxed{R_{eq} = \frac{7}{64}R}$$

We have a current $$I = 1.5\ \text{A}$$ flowing through a copper wire of circular cross-sectional area $$A = 5\ \text{mm}^2$$. First we convert the area into SI units:

$$1\ \text{mm}^2 = 10^{-6}\ \text{m}^2 \; \Longrightarrow \; A = 5 \times 10^{-6}\ \text{m}^2.$$

The electron (number) density in copper is given as $$n = 9 \times 10^{28}\ \text{m}^{-3}.$$ The charge of a single electron is $$e = 1.6 \times 10^{-19}\ \text{C}.$$

The basic relation connecting the drift speed $$v_d$$ with these quantities is

$$I = n\,e\,A\,v_d.$$

We want $$v_d,$$ so we rearrange the formula:

$$v_d = \frac{I}{n\,e\,A}.$$

Substituting the numerical values step by step, we first evaluate the product $$n\,e$$:

$$n\,e = \left(9 \times 10^{28}\right)\left(1.6 \times 10^{-19}\right) = 14.4 \times 10^{9} = 1.44 \times 10^{10}\ \text{C m}^{-3}.$$

Now we include the area $$A = 5 \times 10^{-6}\ \text{m}^2$$:

$$n\,e\,A = \left(1.44 \times 10^{10}\right)\left(5 \times 10^{-6}\right) = 7.2 \times 10^{4}\ \text{C m}^{-2}.$$

Dividing the current by this product gives the drift speed:

$$v_d = \frac{I}{n\,e\,A} = \frac{1.5}{7.2 \times 10^{4}} = 2.0833 \times 10^{-5}\ \text{m s}^{-1}.$$

To express this speed in millimetres per second, we recall that $$1\ \text{m} = 1000\ \text{mm}$$:

$$v_d = 2.0833 \times 10^{-5}\ \text{m s}^{-1} \times 1000\ \frac{\text{mm}}{\text{m}} = 2.0833 \times 10^{-2}\ \text{mm s}^{-1}.$$

Numerically, $$2.0833 \times 10^{-2}\ \text{mm s}^{-1} \approx 0.02\ \text{mm s}^{-1}.$$

Hence, the correct answer is Option D.

Step-by-Step Solution

1. Identify the Equation of a Straight Line

From the statement, the graph plotted is a straight line. The standard equation for a downward-sloping straight line with a positive vertical intercept is given by:

$$y = -mx + c$$

Where:

• $$y$$ represents the variable on the vertical axis.
• $$x$$ represents the variable on the horizontal axis.
• $$m$$ is the positive slope magnitude.
• $$c$$ is the positive vertical intercept.
• The vertical axis variable ($$y$$) is $$\ln(R)$$.
• The horizontal axis variable ($$x$$) is $$\frac{1}{T^2}$$.
• $$R_0 = e^c$$
• $$m = T_0^2$$

2. Match Axis Variables to the Graph

By looking at the provided mathematical steps:

Substituting these coordinates into the straight-line equation:

$$\ln(R) = -m\left(\frac{1}{T^2}\right) + c$$

3. Convert logarithmic form to Exponential form

To find the direct relationship for resistance $$R$$ as a function of temperature $$T$$, take the exponential base $$e$$ on both sides of the equation:

$$e^{\ln(R)} = e^{-\frac{m}{T^2} + c}$$

Using exponential properties ($$e^{a+b} = e^a \cdot e^b$$):

$$R = e^c \cdot e^{-\frac{m}{T^2}}$$

Let the constant term $$e^c$$ be defined as the baseline resistance $$R_0$$, and the slope constant $$m$$ be defined as a characteristic temperature squared parameter $$T_0^2$$:

Substituting these constants back gives the function:

$$R(T) = R_0 e^{-\frac{T_0^2}{T^2}}$$

Final Answer

The variation of resistance $$R$$ with temperature $$T$$ that satisfies the linear graph is given by the exponential relationship:

$$R = R_0 e^{-\frac{T_0^2}{T^2}}$$

For every electric bulb the stated rating, for example “(25 W, 220 V)”, tells us that when a potential difference of 220 V is applied across its terminals it consumes 25 W of power. Using the basic electrical relation $$P=\dfrac{V^{2}}{R}\,,$$ we can obtain the resistance of each filament at its working temperature.

For the 25 W bulb we have

$$R_{25}=\dfrac{V^{2}}{P}=\dfrac{(220)^{2}}{25} =\dfrac{48400}{25}=1936\;\Omega.$$

For the 100 W bulb we write

$$R_{100}=\dfrac{V^{2}}{P}=\dfrac{(220)^{2}}{100} =\dfrac{48400}{100}=484\;\Omega.$$

The two bulbs are now connected in series across the same 220 V supply, so their resistances simply add. The total series resistance is therefore

$$R_{\text{series}}=R_{25}+R_{100}=1936+484=2420\;\Omega.$$

The same current flows through both resistances in series. Ohm’s law gives this current as

$$I=\dfrac{V_{\text{source}}}{R_{\text{series}}} =\dfrac{220}{2420}\;{\rm A} =\dfrac{1}{11}\;{\rm A}\approx 0.090909\;{\rm A}.$$

Now, to find the actual power dissipated in each bulb under this new condition, we employ the formula $$P=I^{2}R.$$ For the 25 W bulb we calculate

$$P_{1}=I^{2}R_{25} =\left(\dfrac{1}{11}\right)^{2}\times 1936 =\dfrac{1}{121}\times 1936 =\dfrac{1936}{121} =16\;{\rm W}.$$

Next, for the 100 W bulb we write

$$P_{2}=I^{2}R_{100} =\left(\dfrac{1}{11}\right)^{2}\times 484 =\dfrac{1}{121}\times 484 =\dfrac{484}{121} =4\;{\rm W}.$$

Thus under series connection the 25 W bulb actually consumes $$P_{1}=16\;{\rm W}$$ while the 100 W bulb consumes only $$P_{2}=4\;{\rm W}.$$

Comparing with the given choices, we see that this corresponds to Option D.

Hence, the correct answer is Option D.

First we interpret the colour bands of the carbon resistor. The standard colour-digit table is:

$$\text{Black}=0,\; \text{Brown}=1,\; \text{Red}=2,\; \text{Orange}=3,\; \text{Yellow}=4,\; \text{Green}=5,\; \text{Blue}=6,\; \text{Violet}=7,\; \text{Grey}=8,\; \text{White}=9.$$

In the given sequence green, black, red, silver we have:

Green gives the first digit $$a = 5,$$

Black gives the second digit $$b = 0,$$

Red gives the multiplier power $$c = 2,$$ because red corresponds to the multiplier $$10^2$$,

Silver gives the tolerance of $$\pm 10\%.$$(We shall use this tolerance a little later.)

The standard formula for the resistance of a 4-band resistor is

$$R = (10a + b)\times 10^{\,c}\;\Omega.$$

Substituting the digits just obtained we get

$$R = (10\times 5 + 0)\times 10^{\,2}\;\Omega.$$

Simplifying step by step, first evaluate the bracket:

$$10\times 5 + 0 = 50.$$

Now multiply by the power of ten supplied by the third band:

$$R = 50 \times 10^{\,2}\;\Omega.$$

Since $$10^{\,2}=100,$$ we have

$$R = 50 \times 100\;\Omega = 5000\;\Omega.$$

So the nominal resistance is

$$R_{\text{nom}} = 5\;\text{k}\Omega.$$

Because the tolerance is silver (±10 %), the actual resistance can be as low as

$$R_{\min}= R_{\text{nom}}\left(1-0.10\right).$$

Substituting the numerical value,

$$R_{\min}= 5\,000\;\Omega \times 0.90 = 4\,500\;\Omega.$$

To ensure the resistor never dissipates more than its rated power of 2 W, we use the power relation

$$P = I^{2}R.$$

For the maximum current we must employ the minimum resistance (because $$I^{2}R$$ must stay at 2 W). Rearranging the formula for current gives

$$I_{\max} = \sqrt{\dfrac{P}{R_{\min}}}.$$

Now substitute $$P = 2\;\text{W}$$ and $$R_{\min}=4\,500\;\Omega$$:

$$I_{\max} = \sqrt{\dfrac{2}{4\,500}}\;\text{A}.$$

Compute the fraction first:

$$\dfrac{2}{4\,500}=0.000444\overline{4}.$$

Taking the square root,

$$I_{\max}= \sqrt{0.000444\overline{4}}\;\text{A}\approx 0.0211\;\text{A}.$$

Convert amperes to milliamperes using $$1\;\text{A}=1000\;\text{mA}$$:

$$0.0211\;\text{A}\times 1000 = 21.1\;\text{mA}.$$

The nearest value among the given options is 20 mA.

Hence, the correct answer is Option C.

We have to find the resistivity $$\rho$$ of the conductor. The basic relation connecting the microscopic quantities with conductivity is first stated.

Formula (Drude model): the electrical conductivity $$\sigma$$ of a metal is

$$ \sigma \;=\; \frac{n e^{2}\,\tau}{m_e}, $$

where

$$n = 8.5 \times 10^{28}\;{\rm m^{-3}},\quad e = 1.6 \times 10^{-19}\;{\rm C},\quad \tau = 25\;{\rm fs}=25 \times 10^{-15}\;{\rm s},\quad m_e = 9.1 \times 10^{-31}\;{\rm kg}. $$

The required resistivity is simply the reciprocal of conductivity:

$$ \rho = \frac{1}{\sigma}. $$

Now we start substituting the numbers step by step.

First, square the electronic charge:

$$ e^{2} = \left(1.6 \times 10^{-19}\right)^{2} = 2.56 \times 10^{-38}\;{\rm C^{2}}. $$

Multiply this by the electron number density $$n$$:

$$ n e^{2} = \left(8.5 \times 10^{28}\right) \left(2.56 \times 10^{-38}\right). $$

We separate the coefficients and the powers of ten:

Coefficient → $$8.5 \times 2.56 = 21.76,$$

Power of ten → $$10^{28} \times 10^{-38} = 10^{-10}.$$

So

$$ n e^{2} = 21.76 \times 10^{-10} = 2.176 \times 10^{-9}. $$

Next, multiply by the mean free time $$\tau$$:

$$ n e^{2} \tau = \left(2.176 \times 10^{-9}\right) \left(25 \times 10^{-15}\right). $$

Again treat the numerical and exponential parts separately:

Coefficient → $$2.176 \times 25 = 54.4,$$

Power of ten → $$10^{-9} \times 10^{-15} = 10^{-24}.$$

Therefore

$$ n e^{2} \tau = 54.4 \times 10^{-24} = 5.44 \times 10^{-23}. $$

Now divide by the electron mass $$m_e$$ to obtain conductivity:

$$ \sigma = \frac{n e^{2} \tau}{m_e} = \frac{5.44 \times 10^{-23}}{9.1 \times 10^{-31}}. $$

To divide, divide the coefficients and subtract the exponents:

Coefficient → $$\dfrac{5.44}{9.1} \approx 0.5989 \approx 0.6,$$

Power of ten → $$10^{-23} \div 10^{-31} = 10^{8}.$$

Thus

$$ \sigma \approx 0.6 \times 10^{8} = 6 \times 10^{7}\;{\rm S\,m^{-1}}. $$

Finally, resistivity is the reciprocal:

$$ \rho = \frac{1}{\sigma} = \frac{1}{6 \times 10^{7}} \approx 1.67 \times 10^{-8}\;{\rm \Omega\,m}. $$

Since $$1.67 \times 10^{-8}\;{\rm \Omega\,m}$$ lies closest to the option written as $$10^{-8}\;{\rm \Omega\,m},$$ we match it to that choice.

Hence, the correct answer is Option D.

We recall the basic relation between conductivity and carrier parameters in a semiconductor. For one kind of charge carrier (only electrons in an N-type material) the conductivity is given by the formula $$\sigma = n\,e\,\mu$$ where

$$n = 10^{19}\ \text{m}^{-3}$$ is the electron concentration, $$e = 1.6 \times 10^{-19}\ \text{C}$$ is the electronic charge, and $$\mu = 1.6\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}$$ is the mobility of the electrons.

Substituting these values, we obtain

$$\sigma = (10^{19}\ \text{m}^{-3})(1.6 \times 10^{-19}\ \text{C})(1.6\ \text{m}^2\ \text{V}^{-1}\ \text{s}^{-1}).$$

First we multiply the numerical parts of the first two factors:

$$10^{19} \times 1.6 \times 10^{-19} = 1.6.$$

Now we include the mobility factor:

$$\sigma = 1.6 \times 1.6 = 2.56\ \text{S m}^{-1}.$$

Resistivity $$\rho$$ is the reciprocal of conductivity, so

$$\rho = \frac{1}{\sigma} = \frac{1}{2.56}\ \Omega\,\text{m}.$$

Carrying out the division gives

$$\rho \approx 0.390625\ \Omega\,\text{m} \approx 0.4\ \Omega\,\text{m}.$$

Hence, the correct answer is Option D.

We are given two concentric conducting spheres. Their radii are $$a$$ for the inner sphere and $$b$$ for the outer sphere, where, as stated, $$b > a$$. The space between them is completely filled with a material of resistivity $$\rho$$. Our task is to calculate the electrical resistance offered by this spherical shell of material when current flows radially from one sphere to the other.

To find resistance, we first recall the basic definition that connects resistivity with resistance for a small (differential) element:

$$dR=\rho\,\dfrac{dl}{A}$$

Here, $$dR$$ is the infinitesimal resistance of a small slab of material, $$dl$$ is its length along the direction of current, and $$A$$ is the cross-sectional area perpendicular to that current. We will now apply this formula to a thin spherical shell lying between the radii $$r$$ and $$r+dr$$.

For a radial current in a sphere, the current lines are directed along the radius, so:

• The infinitesimal length is simply $$dl = dr$$.
• The cross-sectional area through which current passes at radius $$r$$ is the entire spherical surface at that radius, whose area equals $$4\pi r^2$$.

Substituting these into the formula, we obtain the elemental resistance:

$$dR=\rho\,\dfrac{dr}{4\pi r^2}$$

Now we must sum (integrate) all these differential resistances from the inner radius $$a$$ up to the outer radius $$b$$, because each thin shell contributes in series along the radial path:

$$R=\displaystyle\int_a^{b}dR=\int_a^{b}\rho\,\dfrac{dr}{4\pi r^2}$$

The constants $$\rho$$ and $$4\pi$$ can be pulled outside the integral:

$$R=\dfrac{\rho}{4\pi}\,\int_a^{b}r^{-2}\,dr$$

We now integrate $$r^{-2}$$. The standard integral formula is:

$$\displaystyle\int r^{-2}\,dr = -\dfrac{1}{r}+C$$

Applying the limits from $$a$$ to $$b$$, we have:

$$\int_a^{b}r^{-2}\,dr = \left[-\dfrac{1}{r}\right]_{a}^{b} = \left(-\dfrac{1}{b}\right) - \left(-\dfrac{1}{a}\right) = \dfrac{1}{a}-\dfrac{1}{b}$$

Substituting this result back into our expression for $$R$$ yields:

$$R=\dfrac{\rho}{4\pi}\,\left(\dfrac{1}{a}-\dfrac{1}{b}\right)$$

This formula matches exactly with Option C in the given list.

Hence, the correct answer is Option C.

For a Wheatstone bridge to be balanced, we apply the well-known balance condition

$$\dfrac{R_1}{R_2}=\dfrac{R_3}{R_4}.$$

We are told that the carbon resistor used as $$R_1$$ has the colour code (orange, red, brown). Before proceeding, let us translate this colour code into an actual resistance value.

The standard 3-band colour code for a carbon resistor is:

• First band → first significant digit
• Second band → second significant digit
• Third band → multiplier $$10^n$$

Using the colour-digit table (black = 0, brown = 1, red = 2, orange = 3, yellow = 4, green = 5, blue = 6, violet = 7, grey = 8, white = 9), we have

$$R_1 :\; \text{orange (3)},\; \text{red (2)},\; \text{brown }(10^1).$$

So

$$R_1 = (3\,2)\times10^{1}\,\Omega = 32 \times 10 \,\Omega = 320\,\Omega.$$

The problem statement gives us

$$R_2 = 80\,\Omega,\qquad R_4 = 40\,\Omega.$$

Now we substitute these values into the balance condition:

$$\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}.$$

Rearranging to find the unknown $$R_3$$, we get

$$R_3 = R_1 \times \dfrac{R_4}{R_2}.$$

Substituting the numerical values,

$$R_3 = 320\,\Omega \times \dfrac{40\,\Omega}{80\,\Omega}.$$

Simplify the fraction inside:

$$\dfrac{40}{80} = \dfrac12 = 0.5.$$

Hence

$$R_3 = 320\,\Omega \times 0.5 = 160\,\Omega.$$

The next task is to convert $$160\,\Omega$$ into its colour code. Again using the 3-band scheme:

• First significant digit = 1 → brown
• Second significant digit = 6 → blue
• Multiplier required: $$10^{1}$$ (because $$16 \times 10^{1} = 160$$) → brown

Therefore, the colour sequence for $$R_3$$ is

brown, blue, brown.

Looking at the given options, this corresponds to Option B (which is numbered 2 in the list).

Hence, the correct answer is Option 2.

We have a uniform current $$I$$ flowing through a copper rod of cross-sectional area $$A$$. By definition, the current density is the current per unit area, so

$$J \;=\;\frac{I}{A}.$$

For charge carriers moving with an average drift velocity $$v_d$$, the current density can also be written in terms of the volume charge density. The volume charge density $$\rho$$ represents the amount of charge (not merely the number of carriers) present per unit volume. Each tiny volume element therefore carries charge $$\rho$$, and as this charge drifts with speed $$v_d$$, the amount of charge crossing unit area per unit time is

$$J \;=\;\rho\,v_d.$$

Equating the two expressions for $$J$$, we obtain

$$\frac{I}{A}\;=\;\rho\,v_d.$$

Now we solve this equation for the drift velocity $$v_d$$. Dividing both sides by $$\rho$$ gives

$$v_d \;=\;\frac{I}{\rho\,A}.$$

The time $$t$$ required for the charge to travel a distance $$d$$ is simply the distance divided by the drift velocity, because speed equals distance over time. Hence

$$t \;=\;\frac{d}{v_d}.$$

Substituting the expression we just found for $$v_d$$, we get

$$t \;=\;\frac{d}{\dfrac{I}{\rho\,A}} \;=\;d\;\frac{\rho\,A}{I}.$$

Simplifying the numerator and denominator, the final expression for the time is

$$t \;=\;\frac{\rho\,d\,A}{I}.$$

Observe that the temperature $$T$$ does not explicitly enter this relationship, so it does not appear in the final formula.

Hence, the correct answer is Option C.

We are told that the resistance of the heating element at room temperature is $$R_0 = 100\ \Omega.$$

When this element is connected across a supply of $$V = 220\ \text{V},$$ a steady current of $$I = 2\ \text{A}$$ is observed. Ohm’s law, stated as $$V = IR,$$ allows us to find the resistance of the element at its elevated operating temperature.

Substituting the given values, we have $$R = \dfrac{V}{I} = \dfrac{220\ \text{V}}{2\ \text{A}} = 110\ \Omega.$$

This resistance $$R = 110\ \Omega$$ corresponds to a temperature that is $$\Delta T = 500^\circ\text{C}$$ higher than room temperature.

Now we recall the formula that relates resistance to temperature for a conductor:

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

where $$\alpha$$ is the temperature coefficient of resistance, $$R_0$$ is the resistance at the reference temperature (room temperature here), and $$\Delta T$$ is the rise in temperature.

Substituting the known values, we get

$$110\ \Omega = 100\ \Omega \left(1 + \alpha \times 500^\circ\text{C}\right).$$

First divide both sides by $$100\ \Omega$$ to isolate the bracketed term:

$$\dfrac{110}{100} = 1 + 500\,\alpha.$$

Simplifying the left side gives

$$1.10 = 1 + 500\,\alpha.$$

Now subtract 1 from both sides:

$$1.10 - 1 = 500\,\alpha.$$

So we have

$$0.10 = 500\,\alpha.$$

Finally, divide both sides by 500 to solve for $$\alpha$$:

$$\alpha = \dfrac{0.10}{500} = 0.00020.$$

Expressing this in scientific notation,

$$\alpha = 2 \times 10^{-4}\ ^\circ\text{C}^{-1}.$$

Hence, the correct answer is Option B.

First, recall that in a potentiometer, the potential drop along a uniform wire is proportional to its length. If the potential gradient (potential per unit length) is $$k$$, then

$$\text{potential drop} = k \times \text{length}.$$

We have a cell of emf $$E$$ and unknown internal resistance $$r$$. When this cell is connected to the potentiometer without any external load, the galvanometer shows null deflection at a length of 52 cm. Hence

$$E = k \times 52.$$

Now the cell is “shunted”, that is, a resistor of $$5 \,\Omega$$ is connected directly across its terminals. Under null conditions the potentiometer draws no current, so the only current leaving the cell is the current through this 5 $$\Omega$$ shunt. Let the terminal voltage across the cell in this loaded condition be $$V$$. Balance is now obtained at 40 cm, giving

$$V = k \times 40.$$

Dividing the two balance-point equations eliminates $$k$$:

$$\frac{E}{V} = \frac{k \times 52}{k \times 40} = \frac{52}{40} = \frac{13}{10}.$$

Next, relate $$V$$ to $$E$$, $$r$$ and the external resistance. With the 5 $$\Omega$$ resistor connected across the cell, the circuit inside the cell consists of the internal resistor $$r$$ in series with the emf $$E$$, while outside we merely have the 5 $$\Omega$$ resistance. The current through the cell is therefore

$$I = \frac{E}{r + 5}.$$

The terminal voltage $$V$$ is the drop across the external 5 $$\Omega$$ resistor:

$$V = I \times 5 = \frac{E}{r + 5}\times 5 = E\,\frac{5}{\,r + 5\,}.$$

Taking the ratio $$E/V$$ from this expression, we obtain the theoretical relation

$$\frac{E}{V} = \frac{r + 5}{5}.$$

But from the potentiometer readings we already have $$E/V = 13/10$$. Equate the two results:

$$\frac{r + 5}{5} = \frac{13}{10}.$$

Cross-multiply to solve for $$r$$:

$$10\,(r + 5) = 5 \times 13,$$

$$10r + 50 = 65,$$

$$10r = 65 - 50 = 15,$$

$$r = \frac{15}{10} = 1.5 \,\Omega.$$

Hence, the correct answer is Option C.

We recall the basic relation between the resistance $$R$$ of a uniform wire, its length $$l$$, cross-sectional area $$A$$ and the material’s resistivity $$\rho$$. The formula is

$$R=\dfrac{\rho\,l}{A}.$$

The wire is melted and redrawn, so the material remains the same; therefore $$\rho$$ is unchanged and the total volume of the wire is conserved. We first write the volume of the original wire:

$$V_{\text{old}}=\pi r^{2}l.$$

After recasting, the new radius becomes $$\dfrac{r}{2}$$ and we denote the new length by $$l_2$$. The new volume must equal the old volume, so

$$V_{\text{new}}=\pi\left(\dfrac{r}{2}\right)^{2}l_2=\pi r^{2}l.$$

We simplify the left side:

$$\pi\left(\dfrac{r}{2}\right)^{2}l_2=\pi\left(\dfrac{r^{2}}{4}\right)l_2=\dfrac{\pi r^{2}l_2}{4}.$$

Equating the two volumes we have

$$\dfrac{\pi r^{2}l_2}{4}=\pi r^{2}l.$$

Dividing by $$\pi r^{2}$$ gives

$$\dfrac{l_2}{4}=l,$$

so

$$l_2=4l.$$

Now we calculate the resistance of the new wire. Its cross-sectional area is

$$A_2=\pi\left(\dfrac{r}{2}\right)^{2}=\dfrac{\pi r^{2}}{4}.$$

Applying the resistance formula to the new wire, we write

$$R_2=\dfrac{\rho\,l_2}{A_2}.$$

Substituting $$l_2=4l$$ and $$A_2=\dfrac{\pi r^{2}}{4}$$, we get

$$R_2=\dfrac{\rho\,(4l)}{\dfrac{\pi r^{2}}{4}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal, so

$$R_2= \rho\,(4l)\left(\dfrac{4}{\pi r^{2}}\right)=16\,\dfrac{\rho\,l}{\pi r^{2}}.$$

But the original resistance is

$$R_1=\dfrac{\rho\,l}{\pi r^{2}}=100\;\Omega.$$

Therefore

$$R_2 = 16R_1 = 16 \times 100\;\Omega = 1600\;\Omega.$$

Hence, the correct answer is Option A.

We are given a moving-coil galvanometer with an internal resistance $$R_g = 100\;\Omega$$. It shows full-scale deflection when the current through it is $$I_g = 1\ \text{mA} = 0.001\ \text{A}$$.

We wish to use this galvanometer as an ammeter that can measure a much larger full-scale current $$I = 10\ \text{A}$$. To achieve this, a very small resistance, called a shunt resistance, is connected in parallel with the galvanometer. The shunt diverts most of the current so that only the safe current $$I_g$$ flows through the delicate galvanometer coil.

First, we recall (and state) the standard relation for a shunt resistance $$S$$ connected in parallel with a galvanometer:

$$ S \;=\; \dfrac{I_g\,R_g}{I - I_g}. $$

This formula arises from the condition that the potential difference across the galvanometer and the shunt must be equal, because they are in parallel. The derivation is:

The current through the shunt is the remainder $$I - I_g$$. The potential drop across the galvanometer is

$$ V_g \;=\; I_g\,R_g. $$

Since the same potential drop appears across the shunt, Ohm’s law for the shunt gives

$$ V_g \;=\; (I - I_g)\,S. $$

Equating the two expressions for the same voltage, we have

$$ I_g\,R_g \;=\; (I - I_g)\,S. $$

Now we solve for $$S$$ by dividing both sides by $$(I - I_g):$$

$$ S \;=\; \dfrac{I_g\,R_g}{I - I_g}. $$

With the formula stated and derived, we now substitute the numerical values.

First, calculate the numerator $$I_g\,R_g$$:

$$ I_g\,R_g \;=\; (0.001\ \text{A})\,(100\ \Omega) \;=\; 0.1\ \text{V}. $$

Next, compute the denominator $$I - I_g$$:

$$ I - I_g \;=\; 10\ \text{A} - 0.001\ \text{A} \;=\; 9.999\ \text{A}. $$

Now form the fraction:

$$ S \;=\; \dfrac{0.1\ \text{V}}{9.999\ \text{A}}. $$

Divide the numbers:

$$ S \;=\; 0.010001\ \Omega. $$

The result rounds off neatly to

$$ S \;\approx\; 0.01\ \Omega. $$

This is exactly the value of resistance that must be connected in parallel with the galvanometer to convert it into an ammeter capable of reading up to $$10\ \text{A}$$ at full scale.

Hence, the correct answer is Option C.

In a galvanometer, the deflection ($$\theta$$) is directly proportional to the current ($$I$$) passing through it $$I = k\theta$$

$$I = \frac{V}{R + G}$$,$$V$$ is the battery voltage ($$2\ \text{V}$$). $$R$$ is the resistance from the Resistance Box (R.B.). $$G$$ is the internal resistance of the galvanometer.

Case 1: $$R_1 = 2400\ \Omega$$ resulting in a deflection $$\theta_1 = 40$$ divisions.

$$I_1 = \frac{2}{2400 + G} = 40k \quad \text{--- (Eq. 1)}$$

Case 2: $$R_2 = 4900\ \Omega$$ resulting in a deflection $$\theta_2 = 20$$ divisions.

$$I_2 = \frac{2}{4900 + G} = 20k \quad \text{--- (Eq. 2)}$$

Divide (Eq. 1) by (Eq. 2), $$\frac{I_1}{I_2} = \frac{4900 + G}{2400 + G} = \frac{40k}{20k}$$ $$\implies \frac{4900 + G}{2400 + G} = 2$$

$$4900 + G = 4800 + 2G$$ $$\implies G = 100\ \Omega$$

$$20k = \frac{2}{4900 + 100}$$ $$\implies 20k = \frac{2}{5000}$$

$$k = \frac{1}{20 \times 2500} = \frac{1}{50000}\ \text{A/division}$$ $$\implies k = 20\ \mu\text{A/division}$$

Option A: Current sensitivity is $$20\ \mu\text{A/division}$$. (Correct)

Option B: $$G = 100\ \Omega$$, not $$200\ \Omega$$. (Incorrect)

Option C: For $$\theta = 10$$ div, $$10(20 \times 10^{-6}) = \frac{2}{R + 100} \implies R = 9900\ \Omega$$. (Incorrect)

Option D: Full scale current ($$50$$ divisions) $$= 50 \times 20\ \mu\text{A} = 1\ \text{mA}$$. (Incorrect)

We are told that the resistance of the toaster varies with temperature according to

$$R(T)=R_0\left[1+\alpha\,(T-T_0)\right]$$

and that at the reference temperature $$T_0=300\ \text{K}$$ the resistance is $$R_0=100\ \Omega$$. We also know that at $$T=500\ \text{K}$$ the resistance is $$120\ \Omega$$. We first determine the temperature-coefficient $$\alpha$$.

Substituting $$T=500\ \text{K}$$ in the given relation, we have

$$120 = 100\left[1+\alpha\,(500-300)\right].$$

Hence

$$\frac{120}{100}=1+\alpha(200)\quad\Longrightarrow\quad1.2=1+200\alpha,$$

so

$$200\alpha = 0.2\quad\Longrightarrow\quad\alpha = 0.001\ \text{K}^{-1}.$$

Therefore the resistance as an explicit function of temperature is

$$R(T)=100\left[1+0.001\,(T-300)\right].$$

Expanding the bracket,

$$R(T)=100\bigl[1+0.001T-0.3\bigr]=100[0.7+0.001T]=70+0.1T.$$

We check: at $$T=300\ \text{K}$$, $$R=70+0.1(300)=100\ \Omega$$; at $$T=500\ \text{K}$$, $$R=70+0.1(500)=120\ \Omega$$, as required.

The toaster is connected to a fixed voltage source $$V=200\ \text{V}$$. The electrical power at any instant is given by Ohm’s law-power formula:

$$P(t)=\frac{V^2}{R(t)}.$$

The temperature is raised uniformly from 300 K to 500 K in 30 s; hence the temperature rises at the constant rate

$$\frac{\Delta T}{\Delta t} = \frac{500-300}{30}= \frac{200}{30}= \frac{20}{3}\ \text{K s}^{-1}.$$

If $$t$$ is the time (in seconds) measured from the start, the instantaneous temperature is

$$T(t)=300+\frac{20}{3}t.$$

Substituting this in our expression for resistance,

$$R(t)=70+0.1T(t)=70+0.1\left[300+\frac{20}{3}t\right].$$

Simplifying step by step,

$$R(t)=70+30+\frac{0.1\times20}{3}t=100+\frac{2}{3}t.$$

Thus the power as a function of time becomes

$$P(t)=\frac{200^2}{100+\frac{2}{3}t}=\frac{40000}{100+\frac{2}{3}t}.$$

The total electrical work done (energy supplied) while the temperature goes from 300 K to 500 K is the time integral of the power:

$$W=\int_{0}^{30}P(t)\,dt=\int_{0}^{30}\frac{40000}{100+\frac{2}{3}t}\,dt.$$

To evaluate the integral, we set

$$u = 100+\frac{2}{3}t \quad\Longrightarrow\quad du = \frac{2}{3}dt \quad\Longrightarrow\quad dt=\frac{3}{2}\,du.$$

When $$t=0$$, $$u=100$$; when $$t=30\ \text{s}$$, $$u=100+\frac{2}{3}\times30=100+20=120.$$

Hence

$$W = \int_{u=100}^{120}\frac{40000 \times \frac{3}{2}}{u}\,du = 40000\left(\frac{3}{2}\right)\int_{100}^{120}\frac{1}{u}\,du.$$

Multiplying the constants first,

$$40000\left(\frac{3}{2}\right)=60000.$$

Therefore

$$W = 60000\bigl[\ln u\bigr]_{100}^{120}=60000\ln\left(\frac{120}{100}\right).$$

Since $$\dfrac{120}{100}=\dfrac{6}{5},$$ we finally have

$$W = 60000\,\ln\left(\frac{6}{5}\right)\ \text{joules}.$$

Comparing with the given options, this matches Option A.

Hence, the correct answer is Option A.

We have a battery of e.m.f. $$E = 5\text{ V}$$ connected to a load resistance $$R = 50\;\Omega$$.

If no ammeter is present, the only resistance in the circuit is this 50 $$\Omega$$ resistor, so the current is obtained from Ohm’s law $$I_0 = \dfrac{E}{R}$$.

Substituting the given numbers,

$$I_0 = \dfrac{5\text{ V}}{50\;\Omega} = 0.10\ \text{A}$$

So the true current is $$0.10\ \text{A} = 100\ \text{mA}$$.

We now examine each proposed ammeter construction. In every case the ammeter is inserted in series with the 50 $$\Omega$$ resistor, so the total circuit resistance becomes

$$R_{\text{total}} = R + R_{\text{ammeter}}.$$

The requirement is that the current with the ammeter, $$I,$$ must differ from $$I_0$$ by less than 1 %, i.e.

$$\left|\dfrac{I - I_0}{I_0}\right| < 0.01.$$

The galvanometer resistance is $$R_g = 100\;\Omega.$$ We consider the four possibilities one by one.

Option A: $$r_S = 0.5\;\Omega$$ in series with galvanometer

Series combination: $$R_{\text{ammeter}} = R_g + r_S = 100 + 0.5 = 100.5\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 100.5 = 150.5\;\Omega.$$

Current: $$I = \dfrac{E}{R_{\text{total}}} = \dfrac{5}{150.5}\ \text{A} \approx 0.0332\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0332}{0.10}\times100\% \approx 66.8\%.$$ This is far larger than 1 %, so Option A is rejected.

Option B: $$r_S = 1\;\Omega$$ in series with galvanometer

Series combination: $$R_{\text{ammeter}} = 100 + 1 = 101\;\Omega.$$

Total resistance: $$50 + 101 = 151\;\Omega.$$

Current: $$I = \dfrac{5}{151}\ \text{A} \approx 0.0331\ \text{A}.$$

Error again exceeds 66 %, so Option B fails the requirement.

Option C: $$r_S = 1\;\Omega$$ in parallel with galvanometer

For two resistances in parallel the equivalent resistance is given by

$$R_{\text{eq}} = \dfrac{R_g\,r_S}{R_g + r_S}.$$

Substituting $$R_g = 100\;\Omega$$ and $$r_S = 1\;\Omega,$$

$$R_{\text{ammeter}} = R_{\text{eq}} = \dfrac{100 \times 1}{100 + 1}\;\Omega = \dfrac{100}{101}\;\Omega \approx 0.990\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 0.990 = 50.990\;\Omega.$$

Current: $$I = \dfrac{5}{50.990}\ \text{A} \approx 0.0981\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0981}{0.10}\times100\% \approx 1.9\%,$$ which is >1 %, so Option C is not acceptable.

Option D: $$r_S = 0.5\;\Omega$$ in parallel with galvanometer

Again using the parallel-resistance formula,

$$R_{\text{ammeter}} = \dfrac{R_g\,r_S}{R_g + r_S} = \dfrac{100 \times 0.5}{100 + 0.5}\;\Omega = \dfrac{50}{100.5}\;\Omega \approx 0.4975\;\Omega.$$

Total resistance: $$R_{\text{total}} = 50 + 0.4975 = 50.4975\;\Omega.$$

Current: $$I = \dfrac{5}{50.4975}\ \text{A} \approx 0.0990\ \text{A}.$$

Percentage error: $$\dfrac{0.10 - 0.0990}{0.10}\times100\% \approx 0.99\%.$$

This error is less than 1 %, fulfilling the stated condition.

Hence, the correct answer is Option D.

$$I = nAev_d$$

$$E = \frac{V}{l}$$. For a fixed length, $$E \propto V$$

$$v_d \propto \sqrt{E}$$

Substituting the proportionalities $$I \propto v_d$$ and $$E \propto V$$ into the above condition:

$$I \propto \sqrt{V}$$

$$I^2 \propto V \implies \mathbf{V = kI^2}$$

First, let us work out the magnitude of the electric field inside the wire. We know that electric field and potential difference are related by the formula $$E=\dfrac{V}{L},$$ where $$V$$ is the applied potential difference and $$L$$ is the length of the conductor. Substituting the given numbers, we have $$E=\dfrac{5\;\text{V}}{0.1\;\text{m}}=50\;\text{V m}^{-1}.$$

Next, we express the current-density $$J$$ in terms of the drift speed of electrons. The microscopic relation is stated as $$J = n e v_d,$$ where $$n$$ is the electron number density, $$e$$ is the magnitude of electronic charge, and $$v_d$$ is the drift speed. Using the given data $$n = 8\times 10^{28}\;\text{m}^{-3},\; e = 1.6\times 10^{-19}\;\text{C},\; v_d = 2.5\times 10^{-4}\;\text{m s}^{-1},$$ we obtain

$$\begin{aligned} J &= (8\times10^{28})\,(1.6\times10^{-19})\,(2.5\times10^{-4}) \\[4pt] &= 8 \times 1.6 \times 2.5 \times 10^{28-19-4} \\[4pt] &= 32 \times 10^{5} \\[4pt] &= 3.2 \times 10^{6}\;\text{A m}^{-2}. \end{aligned}$$

Resistivity $$\rho$$ is defined through Ohm’s microscopic form $$\rho = \dfrac{E}{J},$$ because $$\mathbf{E} = \rho \mathbf{J}.$$ Substituting the electric field and current density we have just found,

$$\begin{aligned} \rho &= \dfrac{50\;\text{V m}^{-1}}{3.2 \times 10^{6}\;\text{A m}^{-2}} \\[6pt] &= \left(\dfrac{50}{3.2}\right) \times 10^{-6}\;\Omega\;\text{m} \\[6pt] &= 15.625 \times 10^{-6}\;\Omega\;\text{m} \\[6pt] &= 1.5625 \times 10^{-5}\;\Omega\;\text{m}. \end{aligned}$$

The numerical value $$1.5625 \times 10^{-5}\;\Omega\;\text{m}$$ matches most closely with option A.

Hence, the correct answer is Option A.

Each of the four bulbs ($$B_1, B_2, B_3, B_4$$) is rated at $$100$$ W and connected to a $$220$$ V main. Since they are in parallel across the supply, the current $$I$$ through each individual bulb is $$I = \frac{P}{V} = \frac{100\text{ W}}{220\text{ V}} = \frac{5}{11}\text{ A} \approx 0.454\text{ A}$$

The ammeter measures the combined current required to power all subsequent bulbs in the parallel network: $$B_2$$, $$B_3$$, and $$B_4$$.

$$I_{ammeter} = I_{B2} + I_{B3} + I_{B4}$$

$$I_{ammeter} = 3 \times \left( \frac{5}{11}\text{ A} \right) = \frac{15}{11}\text{ A}$$

$$I_{ammeter} \approx 1.35\text{ A}$$

We have to find the current that can flow when all appliances in the building are switched on together, because the main fuse must be able to carry at least this much current without melting.

First, let us list every appliance with its power rating and then calculate the total power.

There are 15 bulbs each of 40 W. Their total power is

$$P_1 = 15 \times 40\ \text{W} = 600\ \text{W}.$$

Next, there are 5 bulbs each of 100 W. Their total power is

$$P_2 = 5 \times 100\ \text{W} = 500\ \text{W}.$$

We also have 5 fans, each rated at 80 W. Their total power becomes

$$P_3 = 5 \times 80\ \text{W} = 400\ \text{W}.$$

Finally, there is one heater rated at 1 kW, that is

$$P_4 = 1\ \text{kW} = 1000\ \text{W}.$$

Now we add all these individual powers to obtain the total power consumption when every appliance is on:

$$P_{\text{total}} = P_1 + P_2 + P_3 + P_4 = 600\ \text{W} + 500\ \text{W} + 400\ \text{W} + 1000\ \text{W} = 2500\ \text{W}.$$

The electric mains supply a voltage of 220 V. We use the electric power formula, which states

$$P = VI,$$

where $$P$$ is the power, $$V$$ the voltage and $$I$$ the current.

Rearranging for current, we get

$$I = \frac{P}{V}.$$

Substituting the total power and the mains voltage, we find

$$I_{\text{total}} = \frac{2500\ \text{W}}{220\ \text{V}} = \frac{2500}{220}\ \text{A}.$$

Carrying out the division step by step,

$$\frac{2500}{220} = \frac{25}{2.2} = 11.36\ \text{A} \;(\text{approximately}).$$

So, about 11.36 A of current will flow when everything is working simultaneously. The fuse must have a rating just higher than this value so that it does not blow during normal operation but will still protect the circuit if the current exceeds this level.

Among the given standard fuse ratings in the options — 8 A, 10 A, 12 A and 14 A — the smallest rating that safely exceeds 11.36 A is 12 A.

Hence, the correct answer is Option C.

In Circuit (d), the high resistance $$R$$ is placed in series with the main circuit branch. The low resistance shunt $$S$$ (along with key $$K_2$$) is connected in parallel directly across the galvanometer $$G$$. This is the correct configuration.

$$I_g = \frac{V}{R + G}$$

$$\frac{I_g}{2} = I \cdot \left(\frac{S}{G+S}\right) = \left(\frac{V}{R + \frac{GS}{G+S}}\right) \cdot \frac{S}{G+S} = \frac{V \cdot S}{R(G+S) + GS}$$

$$\frac{1}{2}\left(\frac{V}{R+G}\right) = \frac{V \cdot S}{R \cdot G + R \cdot S + G \cdot S}$$

$$R \cdot G + R \cdot S + G \cdot S = 2 \cdot S(R + G)$$

$$R \cdot G - G \cdot S = R \cdot S \implies G(R - S) = R \cdot S$$

$$G = \frac{R \cdot S}{R - S}$$

In a meter bridge experiment, we have a wire of length 100 cm. The null point occurs when the ratio of the resistances equals the ratio of the lengths of the wire segments. Initially, resistance X is balanced against resistance Y, and the null point is at 40 cm from the end where X is connected. Since X is less than Y, we use the formula:

$$\frac{X}{Y} = \frac{l_1}{100 - l_1}$$

Here, $$ l_1 = 40 $$ cm. Substituting the values:

$$\frac{X}{Y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$$

So, we have $$ \frac{X}{Y} = \frac{2}{3} $$, which means $$ X = \frac{2}{3} Y $$.

Now, we need to balance a resistance of 3X against Y. Let the new null point be at $$ l $$ cm from the same end where X was connected. The balance condition becomes:

$$\frac{3X}{Y} = \frac{l}{100 - l}$$

Substitute $$ \frac{X}{Y} = \frac{2}{3} $$ into the equation:

$$\frac{3X}{Y} = 3 \times \frac{X}{Y} = 3 \times \frac{2}{3} = 2$$

So, the equation simplifies to:

$$2 = \frac{l}{100 - l}$$

To solve for $$ l $$, multiply both sides by $$ 100 - l $$:

$$2 \times (100 - l) = l$$

Which gives:

$$200 - 2l = l$$

Now, bring all terms involving $$ l $$ to one side:

$$200 = l + 2l$$

$$200 = 3l$$

Dividing both sides by 3:

$$l = \frac{200}{3} \approx 66.67 \text{ cm}$$

This value is approximately 67 cm. Comparing with the options, 67 cm corresponds to option C.

Hence, the correct answer is Option C.

$$R_{PQ} = r$$

$$R_{QR} = \frac{r \times r}{r + r} = \frac{r}{2}$$

$$R_{PR} = \frac{r}{3}$$

Between P and Q: $$r$$ in parallel with $$(\frac{r}{2} + \frac{r}{3} = \frac{5r}{6})$$.

$$R_{net(PQ)} = \frac{r \times \frac{5r}{6}}{r + \frac{5r}{6}} = \frac{5}{11}r \approx 0.45r$$

Between Q and R: $$\frac{r}{2}$$ in parallel with $$(r + \frac{r}{3} = \frac{4r}{3})$$.

$$R_{net(QR)} = \frac{\frac{r}{2} \times \frac{4r}{3}}{\frac{r}{2} + \frac{4r}{3}} = \frac{\frac{4r^2}{6}}{\frac{11r}{6}} = \frac{4}{11}r \approx 0.36r$$

Between P and R: $$\frac{r}{3}$$ in parallel with $$(r + \frac{r}{2} = \frac{3r}{2})$$.

$$R_{net(PR)} = \frac{\frac{r}{3} \times \frac{3r}{2}}{\frac{r}{3} + \frac{3r}{2}} = \frac{\frac{r^2}{2}}{\frac{11r}{6}} = \frac{3}{11}r \approx 0.27r$$

We have a mains (supply) voltage of $$120\ \text{V}$$ feeding the room through lead-in wires whose total resistance is given as $$R_L = 6\ \Omega$$. Any current that flows must first pass through these leads, so whatever appliances are inside the room receive a slightly smaller voltage than the full 120 V because part of the supply voltage is dropped across $$R_L$$.

A 60 W incandescent bulb is already glowing. Because every manufacturer rates a lamp at the voltage that should appear directly across it, we can find its resistance from the power formula

$$P = \frac{V^2}{R}\; ,$$ so $$R = \frac{V^2}{P}.$$

Substituting the rating figures of the bulb,

$$R_B = \frac{(120\ \text{V})^2}{60\ \text{W}} = \frac{14400}{60} = 240\ \Omega.$$

First we calculate the situation before the heater is switched on. The circuit is simply the lead resistance $$R_L$$ in series with the bulb resistance $$R_B$$, and the current through the series combination is

$$I_1 = \frac{V_{\text{supply}}}{R_L + R_B} = \frac{120}{6 + 240} = \frac{120}{246} \approx 0.4878\ \text{A}.$$

The voltage that actually appears across the bulb is then

$$V_{B1} = I_1 \, R_B = 0.4878 \times 240 \approx 117.1\ \text{V}.$$

Thus the initial drop across the lead wires is $$120 - 117.1 \approx 2.9\ \text{V}$$, but we keep the exact figure for later comparison.

Now a 240 W heater is added in parallel with the bulb. We first find the heater’s resistance from the same power formula:

$$R_H = \frac{(120\ \text{V})^2}{240\ \text{W}} = \frac{14400}{240} = 60\ \Omega.$$

The bulb and heater form a parallel combination. The equivalent resistance of two resistors in parallel is given by

$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_B} + \frac{1}{R_H},$$

so

$$R_{\text{eq}} = \frac{R_B\,R_H}{R_B + R_H} = \frac{240 \times 60}{240 + 60} = \frac{14400}{300} = 48\ \Omega.$$

Now the supply sees the series combination of the lead resistance $$R_L = 6\ \Omega$$ and the equivalent load resistance $$R_{\text{eq}} = 48\ \Omega$$. Hence the total current drawn from the mains after the heater is switched on is

$$I_2 = \frac{V_{\text{supply}}}{R_L + R_{\text{eq}}} = \frac{120}{6 + 48} = \frac{120}{54} \approx 2.222\ \text{A}.$$

The voltage that now reaches the parallel junction (and therefore appears across both the bulb and the heater) is

$$V_{B2} = I_2 \, R_{\text{eq}} = 2.222 \times 48 \approx 106.7\ \text{V}.$$

Finally, the decrease in the bulb’s voltage is

$$\Delta V = V_{B1} - V_{B2} \approx 117.1\ \text{V} - 106.7\ \text{V} \approx 10.4\ \text{V}.$$

Hence, the correct answer is Option B.

The letter 'A' is made of uniform wire with a resistance of 1.0 Ω per cm. The sides (legs) are each 20 cm long, and the cross piece is 10 cm long. The apex angle is 60°. We need to find the resistance between the ends of the legs, labeled as points A and B.

First, visualize the structure. The apex angle at the top is 60°, and the legs CA and CB are both 20 cm. The cross piece is horizontal and connects the midpoints of the legs. Let D be the midpoint of CA and E be the midpoint of CB. Since the legs are 20 cm, CD = DA = CE = EB = 10 cm. The cross piece DE is 10 cm long. Given the apex angle of 60° and CD = CE = 10 cm, triangle CDE is equilateral, so DE = 10 cm, which matches.

The circuit has the following segments with resistances (since resistance per cm is 1 Ω):

• AD (from A to D): 10 cm → 10 Ω
• DC (from D to C): 10 cm → 10 Ω
• CE (from C to E): 10 cm → 10 Ω
• EB (from E to B): 10 cm → 10 Ω
• DE (from D to E): 10 cm → 10 Ω

The circuit between A and B can be represented as:

• A is connected to D by 10 Ω.
• D is connected to E by 10 Ω (DE) and to C by 10 Ω (DC).
• E is connected to B by 10 Ω and to C by 10 Ω (CE).

To find the equivalent resistance between A and B, note that between D and E, there are two parallel paths:

1. Direct path DE: 10 Ω
2. Path via C: DC + CE = 10 Ω + 10 Ω = 20 Ω

The equivalent resistance between D and E is the parallel combination:

$$ \frac{1}{R_{DE}} = \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} $$

$$ R_{DE} = \frac{20}{3} \Omega \approx 6.6667 \Omega $$

Now, the total path from A to B is: A to D (10 Ω), then D to E (20/3 Ω), then E to B (10 Ω). These are in series, so:

$$ R_{AB} = R_{AD} + R_{DE} + R_{EB} = 10 + \frac{20}{3} + 10 = 20 + \frac{20}{3} = \frac{60}{3} + \frac{20}{3} = \frac{80}{3} \Omega \approx 26.6667 \Omega $$

Comparing with the options:

• A. 50.0 Ω
• B. 10 Ω
• C. 36.7 Ω
• D. 26.7 Ω

The value 26.6667 Ω is closest to 26.7 Ω.

Hence, the correct answer is Option D.

Initial state ($$R_3 = 0$$):

In this case, the galvanometer is in parallel with $$R_2$$. The total resistance of the circuit is $$R_{total} = R_1 + \frac{G R_2}{G + R_2}$$. Since $$R_1 \gg R_2$$, the current from the battery is $$I \approx \frac{V}{R_1}$$. 

The current through the galvanometer ($$I_g$$) corresponding to deflection $$d$$ is $$I_g = I \left( \frac{R_2}{G + R_2} \right) = \frac{V}{R_1} \left( \frac{R_2}{G + R_2} \right) \quad \dots (1)$$

Final state ($$R_3 = 107\ \Omega$$):

The new current through the galvanometer ($$I_g'$$) corresponding to half deflection $$d/2$$ is $$I_g' = I \left( \frac{R_2}{G + R_3 + R_2} \right) = \frac{V}{R_1} \left( \frac{R_2}{G + 107 + R_2} \right) \quad \dots (2)$$

Since $$I_g' = \frac{1}{2} I_g$$, $$\frac{R_2}{G + 107 + R_2} = \frac{1}{2} \left( \frac{R_2}{G + R_2} \right)$$

$$2(G + R_2) = G + 107 + R_2$$

$$G + R_2 = 107$$

$$G + 30 = 107 \implies G = 77\ \Omega$$

The given relation is an experimental (empirical) relation for metals

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

where $$R_0$$ is the resistance at some reference temperature and $$\alpha$$ is called the (average) temperature coefficient of resistance between the two temperatures considered. Using this single-step linear relation does not require that the temperature change be infinitesimally small; it works quite accurately over a fairly wide range of temperatures for ordinary metals.

Evaluating Statement-1

The wire changes from
$$R_0 = 100 \,\Omega \quad\text{at}\quad T_0 = 27^{\circ}\text{C}$$
to
$$R = 150 \,\Omega \quad\text{at}\quad T = 227^{\circ}\text{C}$$

Hence $$\Delta T = 227 - 27 = 200^{\circ}\text{C}$$. Putting these numbers in the empirical formula:

$$150 = 100\bigl(1 + \alpha \times 200\bigr)$$

$$\Rightarrow\; 1 + 200\alpha = 1.5$$

$$\Rightarrow\; 200\alpha = 0.5 \;\;\Longrightarrow\;\; \alpha = 2.5\times10^{-3}/^{\circ}\text{C}$$

The value quoted in the statement is exactly this, so Statement-1 is true.

Evaluating Statement-2

Statement-2 claims that the linear formula is valid only when the temperature rise is small and consequently $$\Delta R \ll R_0$$. In practice, for common metals the resistance varies almost linearly with temperature over several hundreds of degrees Celsius. The same linear law is routinely used, for example, from room temperature up to the red-hot range in platinum resistance thermometers. Therefore the restriction “only when $$\Delta T$$ is small and $$\Delta R \ll R_0$$” is not correct; the formula remains a good empirical fit even when the change is large (as in the present problem where $$\Delta R = 0.5 R_0$$).

Hence Statement-2 is false.

Combining the results:

Statement-1 is true and Statement-2 is false.

Option A which is: Statement-1 is true, Statement-2 is false.

When a point current source injects a steady current $$I$$ into a large block of uniform resistivity $$\rho$$, the current spreads radially inside the conductor. Close to the entry point $$A$$ the current lines form a hemisphere of radius $$r$$ (only the half-space inside the block is available). The surface area of this hemisphere is $$2\pi r^{2}$$.

Case 1: Current $$I$$ entering at $$A$$
Current density $$j(r)$$ at distance $$r$$ from $$A$$ is obtained from the definition $$j = I/\text{area}$$:

$$j(r)=\frac{I}{2\pi r^{2}}$$

Ohm’s law for a conductor gives the magnitude of the electric field:
$$E(r)=\rho\,j(r)=\frac{\rho I}{2\pi r^{2}}$$

The field is radial, so the potential difference between two points at radii $$r_1$$ and $$r_2$$ from $$A$$ is the line integral of $$E$$:

$$V(r_2)-V(r_1)=-\int_{r_1}^{r_2}E(r)\,dr =-\int_{r_1}^{r_2}\frac{\rho I}{2\pi r^{2}}\,dr =\frac{\rho I}{2\pi}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

Choosing the reference potential at a very large distance (effectively $$r\rightarrow\infty$$, where the term $$1/r$$ vanishes) the absolute potential at a point a distance $$r$$ from $$A$$ is

$$V_A(r)=\frac{\rho I}{2\pi r}$$

In the block the points $$B$$ and $$C$$ lie on the line joining them to $$A$$ at distances
$$AB = a,\quad AC = a+b$$.
Hence

$$V_A(B)=\frac{\rho I}{2\pi a},\qquad V_A(C)=\frac{\rho I}{2\pi(a+b)}$$

Case 2: Current $$I$$ leaving at $$D$$
Exactly the same reasoning applies near $$D$$, but now the current is directed out of the conductor. The electric field near $$D$$ therefore points towards $$D$$, making its contribution to the scalar potential negative. In the given geometry points $$B$$ and $$C$$ are equidistant from $$D$$ (the figure shows them located symmetrically with respect to $$D$$). Consequently, the potentials produced by the exit current at $$B$$ and $$C$$ are equal and cancel in the difference $$\Delta V$$:

$$V_D(B)=V_D(C)\;\Longrightarrow\; V_D(B)-V_D(C)=0$$

Total potential difference $$\Delta V_{BC}$$
Superposing the two cases:

$$\Delta V_{BC}= \big[V_A(B)+V_D(B)\big]-\big[V_A(C)+V_D(C)\big] = V_A(B)-V_A(C)$$