A wire of length 25~m$25\mathrm{~m}$ and cross-sectional area 5~mm²$5\mathrm{~mm}^2$ having resistivity of 2× 10⁻⁶Ω~m$2\times 10^{-6}\Omega\mathrm{~m}$ is bent into a complete circle. The resistance between diametrically opposite points will be:
A.12.5Ω$12.5\Omega$
B.50Ω$50\Omega$
C.100Ω$100\Omega$
D.25Ω$25\Omega$
Solution & Explanation
Related Formula
Resistance of a uniform wire:
R = (ρ L)/(A)$$R = \frac{\rho L}{A}$$
Equivalent resistance of two identical resistors in parallel:
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.
Circular loop splitting resistance across diameter for Q5
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$R_{\text{eq}} = 10/4 = 2.5\Omega$.
Keywords:#wire bent into circle resistance#JEE Main 2025 Morning Q5#diametrically opposite resistance#Current electricity parallel loop
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The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9V with internal resistance of 1 Ω$1 \Omega$ is connected across these points is ____ J.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
The circuit is a balanced Wheatstone bridge between A and B, because the ratio of adjacent arms is (1)/(2) = (2)/(4)$\frac{1}{2} = \frac{2}{4}$. Thus, the middle 1 Ω$1\,\Omega$ resistor (if there is one connecting the middle nodes) is ineffective.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
The equivalent resistance across AB (RAB$R_{AB}$) involves the upper branch (1+2 = 3 Ω$1+2 = 3\,\Omega$) and the lower branch (2+4 = 6 Ω$2+4 = 6\,\Omega$) in parallel.
A bridge network with resistors 1 ohm, 2 ohms, 2 ohms, and 4 ohms forming the arms.
Total resistance in the circuit = RAB + rinternal = 2 + 1 = 3 Ω$R_{AB} + r_{\text{internal}} = 2 + 1 = 3\,\Omega$.
Total current from the battery i = 9V3 Ω = 3A$i = \frac{9\text{V}}{3\,\Omega} = 3\text{A}$.
The heat generated specifically between points A and B in 1 minute (60 seconds) is:
Check for balanced Wheatstone bridge first. Then use H = i² R t$H = i^2 R t$ strictly with the equivalent resistance of just the section AB$AB$ to find heat specifically generated across AB$AB$.
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The total length of potentiometer wire AB is 50 cm$50 \text{ cm}$ in the arrangement as shown in figure. If P is the point where the galvanometer shows zero reading then the length AP is ________ cm.
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When the galvanometer shows zero reading, it implies that the bridge is balanced. The resistance of the wire segments is proportional to their lengths.
Let the length of segment AP be lAP$l_{AP}$ and PB be lPB$l_{PB}$.
Null deflection over a uniform wire directly translates resistance ratios into length ratios. Standard Wheatstone bridge application wrapped in potentiometer geometry.
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Q35jee_main_2026_21_jan_eveningMaximum Power Transfer Theorem
P = I² R = ((E)/(R+r))² R$$P = I^2 R = \left(\frac{E}{R+r}\right)^2 R$$
Core Logic
To maximize power across the load resistance R$R$, we apply the Maximum Power Transfer Theorem. The power transferred to the load is maximum when the load resistance equals the internal resistance of the battery.
Power is maximum when the external resistance is strictly equal to the internal resistance (R=r$R=r$).
Pattern Recognition
Direct theoretical application of the Maximum Power Transfer Theorem.
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Two known resistance of R Ω$R\,\Omega$ and 2R Ω$2R\,\Omega$ and one unknown resistance X Ω$X\,\Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X Ω$X\,\Omega$, then the value of X is ________ Ω$\Omega$.
Bridge-like resistor circuit containing unknown resistance X and known resistances R and 2R.
A.(√(3)-1)R$(\sqrt{3}-1)R$
B.R$R$
C.2(√(3)-1)R$2(\sqrt{3}-1)R$
D.(√(3)+1)R$(\sqrt{3}+1)R$
Solution
Related Formula
For parallel combination: Req = (R₁ R₂)/(R₁ + R₂)$R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$
For series combination: Req = R₁ + R₂$R_{\text{eq}} = R_1 + R_2$
Core Logic
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The total equivalent resistance across A and B is given as X$X$.
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A meter bridge with two resistances R₁$R_{1}$ and R₂$R_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when 16~Ω$16~\Omega$ resistance is connected in parallel to R₂$R_{2}$. The values of resistances R₁$R_{1}$ and R₂$R_{2}$ are \_\_\_\_.
Meter bridge circuit with resistances R1 and R2 and balancing lengths.
Sees: Meter bridge balancing with parallel shunt resistance.
Shortcut: Set up length ratios for initial and modified states and solve linear simultaneous equations.
Check: Matches option (3). ✓
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More Current Electricity Questions — jee_main_2025_03_april_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
