A wire of resistance R$R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

Related Formula

For a wire of total resistance R$R$ divided into N$N$ equal segments, the resistance of each segment r$r$ is:

For a balanced Wheatstone bridge with resistors of resistance r$r$, the central arm can be neglected because no current flows through it.

Core Logic

The triangular pyramid has 6$6$ segments of equal length. Since the total resistance of the wire is R$R$:

Let the four vertices of the pyramid be A$A$, B$B$, C$C$, and D$D$. Terminals are at A$A$ and B$B$.

The segments are:

• AB$AB$ (direct path between terminals, resistance r$r$)
• AC$AC$, BC$BC$, AD$AD$, BD$BD$ (forming a closed quadrilateral network between A$A$ and B$B$ with bridge arm CD$CD$)
• CD$CD$ (bridge arm connecting the midpoints, resistance r$r$)

Step 1: Simplify the Network

By symmetry, the potentials at C$C$ and D$D$ are equal when a voltage is applied across A$A$ and B$B$. Thus, the bridge is balanced, and no current flows through the segment CD$CD$.

We can remove segment CD$CD$ from the calculations:

• The path A → C → B$A \to C \to B$ consists of two resistors in series: r + r = 2r$r + r = 2r$.
• The path A → D → B$A \to D \to B$ also consists of two resistors in series: r + r = 2r$r + r = 2r$.
• The direct path A → B$A \to B$ has a single resistor r$r$.

These three parallel branches are connected between A$A$ and B$B$.

Step 2: Calculate Equivalent Resistance

The equivalent resistance RAB$R_{AB}$ is:

Substitute r = (R)/(6)$r = \frac{R}{6}$:

Comparing with RAB = (R)/(n)$R_{AB} = \frac{R}{n}$, we find n = 12$n = 12$.

Pattern Recognition

Sees: Resistor network formed by a 3D pyramid (6 identical edges, 4 nodes). Shortcut: A 6-resistor regular tetrahedron has an equivalent resistance of r/2$r/2$ across any two vertices. Since the total wire resistance is R$R$ and it's cut into 6 pieces, r = R/6 Req = R/12$r = R/6 \implies R_{\text{eq}} = R/12$.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

where, V$V$ = Thermoelectric voltage (Seebeck voltage) S$S$ = Seebeck coefficient Δ T$\Delta T$ = Temperature difference

Core Logic

To maximize the efficiency of a thermoelectric device harvesting heat energy, two conditions must be fulfilled:

• Low thermal conductivity: This ensures that the temperature gradient (Δ T$\Delta T$) across the material is maintained and heat does not rapidly flow from the hot side to the cold side.
• High electrical conductivity: This minimizes internal Joule heating losses (I²R$I^2R$) when electrical current is drawn from the material.

Therefore, the material should possess low thermal conductivity and high electrical conductivity.

Pattern Recognition

thermoelectric figure of merit is given by Z = (S² σ)/(κ)$Z = \frac{S^2 \sigma}{\kappa}$, where σ$\sigma$ is electrical conductivity and κ$\kappa$ is thermal conductivity. To maximize Z$Z$, we inherently need high σ$\sigma$ and low κ$\kappa$.

Chapter Mix

Class 12 Physics: Current Electricity Class 11 Physics: Thermal Properties of Matter

Core Logic

Label the circuit nodes carefully to see how the elements are connected across terminal zones:

Tracing parallel loops reveals all inner subsets share identical path boundaries.

Step 1: Final Expression

The overall simplified system network value equals r9$\frac{\mathrm{r}}{9}$, matching option (3).

Pattern Recognition

Short-circuit tracking lines allow collapsing complex meshes into basic parallel branches. If all nodes connect symmetrically, apply simple division: R/N$\mathrm{R}/\mathrm{N}$.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

Resistance of a uniform wire:

Equivalent resistance of two identical resistors in parallel:

Core Logic

First, find the total resistance Rtotal$R_{\text{total}}$ of the straight wire:

• Length of wire, L = 25~m$L = 25\mathrm{~m}$
• Area of cross-section, A = 5~mm² = 5 × 10⁻⁶~m²$A = 5\mathrm{~mm}^2 = 5 \times 10^{-6}\mathrm{~m}^2$
• Resistivity, ρ = 2 × 10⁻⁶Ω~m$\rho = 2 \times 10^{-6}\Omega\mathrm{~m}$

Step 1: Splitting into Semicircles

When the wire is bent into a complete circle, measuring the resistance between two diametrically opposite points splits the wire into two parallel halves of equal length.

Each semicircle has a resistance of:

These two halves are connected in parallel between the diametric terminals:

Step 2: Conclusion & Discrepancy

The actual mathematically rigorous answer is 2.5Ω$2.5\Omega$. Since 2.5Ω$2.5\Omega$ is not present in the given options, the question is marked as a Bonus question by standard key evaluation guidelines. If forced to choose a theoretical option due to printing mistakes, some keys may relate it to 10Ω / 4 = 2.5Ω$10\Omega / 4 = 2.5\Omega$, but scientifically it stands as a bonus.

Pattern Recognition

Shortcut: A wire of total resistance R$R$ bent into a circle has an effective resistance across its diameter equal to Req = R/4$R_{\text{eq}} = R/4$. Memorize this ratio! Here R = 10Ω$R = 10\Omega$, so Req = 10/4 = 2.5Ω$R_{\text{eq}} = 10/4 = 2.5\Omega$.

Chapter Mix

Class 12 Physics: Current Electricity