A wire of resistance R is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points A and B is R/n. The value of n is : A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.
A.16
B.14
C.10
D.12
Solution & Explanation
### Related Formula
For a wire of total resistance R$R$ divided into N$N$ equal segments, the resistance of each segment r$r$ is:
r = fracRN$$r = \frac{R}{N}$$
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$R$:
r = fracR6$$r = \frac{R}{6}$$
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 to C to B$A \to C \to B$ consists of two resistors in series: r + r = 2r$r + r = 2r$.
- The path A to D to B$A \to D \to B$ also consists of two resistors in series: r + r = 2r$r + r = 2r$.
- The direct path A to 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 R_AB$R_{AB}$ is:
frac1R_AB = frac12r + frac12r + frac1r = frac1r + frac1r = frac2r$$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$$R_AB = fracr2$$R_{AB} = \frac{r}{2}$$
Substitute r = fracR6$r = \frac{R}{6}$:
R_AB = fracR/62 = fracR12$$R_{AB} = \frac{R/6}{2} = \frac{R}{12}$$
Comparing with R_AB = fracRn$R_{AB} = \frac{R}{n}$, 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 implies R_texteq = R/12$r = R/6 \implies R_{\text{eq}} = R/12$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
A triangular pyramid resistor network with terminals A and B marked, showing symmetry in the layout.
Keywords:#wire bent into triangular pyramid#JEE Main 2025 Morning Q3#Current Electricity JEE Main 2025#equivalent resistance Wheatstone bridge#resistor network#equivalent resistance#balanced wheatstone bridge#pyramid topology
More Current Electricity Previous-Year Questions — Page 6
Q33jee_main_2024_30_jan_morningResistors in Series and Voltage Dividers
A potential divider circuit is shown in figure. The output voltage V_0$V_0$ is
Circuit containing multiple resistors in series and parallel calculating a specific output voltage.
A.4 mathrm~V$4 \mathrm{~V}$
B.2 mathrm~mV$2 \mathrm{~mV}$
C.0.5 mathrm~V$0.5 \mathrm{~V}$
D.12 mathrm~mV$12 \mathrm{~mV}$
Solution
### Related Formula
V = IR$V = IR$R_texteq = R_1 + R_2 + dots + R_n quad (textfor series)$$R_{\text{eq}} = R_1 + R_2 + \dots + R_n \quad (\text{for series})$$
### Core Logic
Observe the circuit diagram. The total equivalent resistance R_texteq$R_{\text{eq}}$ of the series network must be calculated by summing all the resistance values shown in the main branch.
### Step 1: Calculate Total Resistance and Current
From the given network, assuming the total series resistance is 4000 \,Omega$4000 \,\Omega$ (comprising the 3.3mathrmkOmega$3.3\mathrm{k}\Omega$ resistor and seven 100\,Omega$100\,\Omega$ resistors).
R_texteq = 4000 \,Omega$$R_{\text{eq}} = 4000 \,\Omega$$
The total voltage applied across the network is 4 mathrm~V$4 \mathrm{~V}$.
i = fracVR_texteq = frac44000 = frac11000 mathrm~A$$i = \frac{V}{R_{\text{eq}}} = \frac{4}{4000} = \frac{1}{1000} \mathrm{~A}$$
### Step 2: Calculate Output Voltage
The output voltage V_0$V_0$ is tapped across five 100 \,Omega$100 \,\Omega$ resistors.
R_textout = 5 times 100 \,Omega = 500 \,Omega$$R_{\text{out}} = 5 \times 100 \,\Omega = 500 \,\Omega$$
Thus, the output voltage is:
V_0 = i cdot R_textout = left(frac11000right) times 500 = 0.5 mathrm~V$$V_0 = i \cdot R_{\text{out}} = \left(\frac{1}{1000}\right) \times 500 = 0.5 \mathrm{~V}$$
### Pattern Recognition
A potential divider simply scales the input voltage by the fraction of the resistance tapped over the total resistance: V_0 = V_in times (R_texttap / R_texttotal)$V_0 = V_{in} \times (R_{\text{tap}} / R_{\text{total}})$. Standard DC circuit division.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q43jee_main_2024_30_jan_morningTemperature Dependence of Resistivity
An electric toaster has resistance of 60Omega$60\Omega$ at room temperature (27^circmathrmC)$(27^{\circ}\mathrm{C})$. The toaster is connected to a 220mathrmV$220\mathrm{V}$ supply. If the current flowing through it reaches 2.75mathrmA$2.75\mathrm{A}$, the temperature attained by toaster is around: (if alpha = 2times 10^-4 / ^circmathrmC$\alpha = 2\times 10^{-4} / ^{\circ}\mathrm{C}$)
A.694^circ mathrmC$694^{\circ} \mathrm{C}$
B.1235^circmathrmC$1235^{\circ}\mathrm{C}$
C.1694^circmathrmC$1694^{\circ}\mathrm{C}$
D.1667^circmathrmC$1667^{\circ}\mathrm{C}$
Solution
### Related Formula
R = fracVI$$R = \frac{V}{I}$$R = R_0 (1 + alpha Delta T)$$R = R_0 (1 + \alpha \Delta T)$$
### Core Logic
First, evaluate the final resistance R_T$R_T$ at the operating condition using Ohm's law. Second, plug the final resistance into the linear temperature dependence equation for resistance to solve for final temperature T$T$.
### Step 1: Calculate Final Resistance
Given V = 220 mathrm~V$V = 220 \mathrm{~V}$ and I = 2.75 mathrm~A$I = 2.75 \mathrm{~A}$:
R_T = frac2202.75 = 80 \,Omega$$R_T = \frac{220}{2.75} = 80 \,\Omega$$
### Step 2: Apply Temperature Equation
We know R_27 = 60 \,Omega$R_{27} = 60 \,\Omega$.
R_T = R_27 [1 + alpha (T - 27)]$$R_T = R_{27} [1 + \alpha (T - 27)]$$80 = 60 [1 + 2 times 10^-4 (T - 27)]$$80 = 60 [1 + 2 \times 10^{-4} (T - 27)]$$frac8060 = 1 + 2 times 10^-4 (T - 27)$$\frac{80}{60} = 1 + 2 \times 10^{-4} (T - 27)$$frac43 - 1 = 2 times 10^-4 (T - 27)$$\frac{4}{3} - 1 = 2 \times 10^{-4} (T - 27)$$frac13 = 2 times 10^-4 (T - 27)$$\frac{1}{3} = 2 \times 10^{-4} (T - 27)$$
### Step 3: Solve for T
T - 27 = frac16 times 10^-4$$T - 27 = \frac{1}{6 \times 10^{-4}}$$T - 27 = frac100006 = 1666.67$$T - 27 = \frac{10000}{6} = 1666.67$$T = 1666.67 + 27 approx 1693.67 Rightarrow 1694^circ mathrmC$$T = 1666.67 + 27 \approx 1693.67 \Rightarrow 1694^\circ \mathrm{C}$$
### Pattern Recognition
Always separate the final temperature T$T$ from Delta T$\Delta T$. The most common error is forgetting to add back the initial reference temperature (27^circmathrmC$27^\circ\mathrm{C}$) at the end.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q52jee_main_2024_30_jan_morningCells in Opposition and Terminal Voltage
Two cells are connected in opposition as shown. Cell E_1$E_1$ is of 8mathrmV$8\mathrm{V}$ emf and 2Omega$2\Omega$ internal resistance; the cell E_2$E_2$ is of 2mathrmV$2\mathrm{V}$ emf and 4Omega$4\Omega$ internal resistance. The terminal potential difference of cell E_2$E_2$ is:
Two batteries pushing current against each other.
Numerical Answer.Answer: 6 to 6
Solution
### Related Formula
I = fracE_textnetR_texteq$$I = \frac{E_{\text{net}}}{R_{\text{eq}}}$$V = E - Ir quad (textDischarging)$$V = E - Ir \quad (\text{Discharging})$$V = E + Ir quad (textCharging)$$V = E + Ir \quad (\text{Charging})$$
### Core Logic
Two batteries pushing current against each other.
Because the cells are in opposition, the net EMF drives current from the higher potential cell (8mathrmV$8\mathrm{V}$) to the lower potential cell (2mathrmV$2\mathrm{V}$). Thus, the 2mathrmV$2\mathrm{V}$ cell acts as a load and undergoes charging.
### Step 1: Calculate Total Current
I = frac8 - 22 + 4 = frac66 = 1 mathrm~A$$I = \frac{8 - 2}{2 + 4} = \frac{6}{6} = 1 \mathrm{~A}$$
### Step 2: Terminal Potential of Cell 2
Since cell E_2$E_2$ is being charged, its terminal potential difference is:
V_2 = E_2 + I r_2$$V_2 = E_2 + I r_2$$
Applying Kirchhoff's rule across cell E_2$E_2$ (from node C to B):
V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 mathrm~V$$V_C - V_B = E_2 + I r_2 = 2 + (1)(4) = 6 \mathrm{~V}$$
### Pattern Recognition
When a smaller battery is forced backwards by a larger battery, it gets "charged". Consequently, its terminal voltage increases above its nominal EMF: V = E + Ir$V = E + Ir$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q39jee_main_2024_31_jan_eveningMeter Bridge
The resistance per centimeter of a meter bridge wire is r$r$, with X \, Omega$X \, \Omega$ resistance in left gap. Balancing length from left end is at 40 text cm$40 \text{ cm}$ with 25 \, Omega$25 \, \Omega$ resistance in right gap. Now the wire is replaced by another wire of 2r$2r$ resistance per centimeter. The new balancing length for same settings will be at
A.20 text cm$20 \text{ cm}$
B.10 text cm$10 \text{ cm}$
C.80 text cm$80 \text{ cm}$
D.40 text cm$40 \text{ cm}$
Solution
### Related Formula
fracR_textleftR_textwire-left = fracR_textrightR_textwire-right$$\frac{R_{\text{left}}}{R_{\text{wire-left}}} = \frac{R_{\text{right}}}{R_{\text{wire-right}}}$$
### Core Logic
For a meter bridge, the balancing condition is independent of the absolute resistance of the bridge wire as long as it is uniform. The ratio of the resistances in the gaps balances with the ratio of lengths.
Meter Bridge diagram for Q39 - JEE Main 2024 Evening
### Step 1: First Condition
fracXr ell_1 = frac25r (100 - ell_1)$$\frac{X}{r \ell_1} = \frac{25}{r (100 - \ell_1)}$$
Given ell_1 = 40 text cm$\ell_1 = 40 \text{ cm}$:
fracXr times 40 = frac25r times 60 implies fracX40 = frac2560$$\frac{X}{r \times 40} = \frac{25}{r \times 60} \implies \frac{X}{40} = \frac{25}{60}$$
### Step 2: Second Condition
When replaced by a wire of 2r$2r$ per cm, the new lengths ell_2$\ell_2$ will satisfy:
fracX2r ell_2 = frac252r (100 - ell_2)$$\frac{X}{2r \ell_2} = \frac{25}{2r (100 - \ell_2)}$$
Notice that the 2r$2r$ terms cancel out entirely from both sides, leaving:
fracXell_2 = frac25100 - ell_2$$\frac{X}{\ell_2} = \frac{25}{100 - \ell_2}$$
### Step 3: Conclusion
Since the ratio X/25$X/25$ remains identical, the balancing length ratio ell / (100-ell)$\ell / (100-\ell)$ also remains identical.
Therefore, ell_2 = ell_1 = 40 text cm$\ell_2 = \ell_1 = 40 \text{ cm}$.
### Pattern Recognition
Meter bridge balance point strictly depends on length ratio, NOT the specific resistivity or thickness of the wire (provided it is uniform). If external resistors don't change, the balance point never changes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q43jee_main_2024_31_jan_eveningPower Dissipation
By what percentage will the illumination of the lamp decrease if the current drops by 20\%$20\%$?
A.46\%$46\%$
B.26\%$26\%$
C.36\%$36\%$
D.56\%$56\%$
Solution
### Related Formula
Power (Illumination) is proportional to the square of the current for a constant resistance:
P = I^2 R$P = I^2 R$
### Core Logic
Initial power:
P_1 = I_1^2 R$P_1 = I_1^2 R$
If current drops by 20%, the new current is:
I_2 = I_1 - 0.2 I_1 = 0.8 I_1$$I_2 = I_1 - 0.2 I_1 = 0.8 I_1$$
New power:
P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$P_2 = (0.8 I_1)^2 R = 0.64 I_1^2 R = 0.64 P_1$$
### Step 1: Calculate Percentage Change
Delta P \% = fracP_2 - P_1P_1 times 100\%$$\Delta P \% = \frac{P_2 - P_1}{P_1} \times 100\%$$Delta P \% = frac0.64 P_1 - P_1P_1 times 100\%$$\Delta P \% = \frac{0.64 P_1 - P_1}{P_1} \times 100\%$$Delta P \% = (0.64 - 1) times 100\% = -36\%$$\Delta P \% = (0.64 - 1) \times 100\% = -36\%$$
### Step 2: Final Statement
The negative sign indicates a decrease. The illumination drops by 36\%$36\%$.
### Pattern Recognition
For squared relations y = x^2$y = x^2$, if x$x$ changes by a factor k$k$ (0.8$0.8$), y$y$ changes by a factor k^2$k^2$ (0.64$0.64$). 1 - 0.64 = 36\%$1 - 0.64 = 36\%$. This bypasses algebraic limits usually done for small changes (like 2Delta x / x$2\Delta x / x$) since 20% is too large for approximation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
More Current Electricity Questions — jee_main_2025_07_april_morning
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