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Question #43374

Difficulty: 8/10 | 🟢 Free

The Question

A metre bridge is set up to measure an unknown resistance. The left gap contains a standard resistor R=10 ΩR = 10\ \Omega and the right gap contains a resistor S=30 ΩS = 30\ \Omega. The uniform metre bridge wire has a length of 100 cm100\ \text{cm} and a total resistance Rw=10 ΩR_w = 10\ \Omega. The bridge is powered by a battery of electromotive force E=24 VE = 24\ \text{V} in series with an internal resistance r=4 Ωr = 4\ \Omega. The null detector consists of a galvanometer shunted by a dynamic non-linear resistor whose resistance varies with the galvanometer current IgI_g as Rsh=501+λ∣Ig∣ ΩR_{sh} = \frac{50}{1 + \lambda|I_g|}\ \Omega, where λ=100 A−1\lambda = 100\ \text{A}^{-1}. Due to local heating at the contacts, the end corrections of the metre bridge depend on the steady current IwI_w flowing through the bridge wire. The left end correction is given by α=(2.5 cm/A)Iw\alpha = (2.5\ \text{cm/A}) I_w and the right end correction is β=(1.25 cm/A)Iw\beta = (1.25\ \text{cm/A}) I_w. Determine the null point ll measured from the left end of the bridge wire.

Options

  • A.

    22.5 cm22.5\ \text{cm}

  • B.

    21.25 cm21.25\ \text{cm}

  • C.

    21.875 cm21.875\ \text{cm}

  • D.

    25.0 cm25.0\ \text{cm}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Metre Bridge End Correction|Shunt Resistance Conversion

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to decouple a non-linear component (dynamic shunt) from the fundamental balance condition of a Wheatstone bridge, while simultaneously coupling measurement uncertainty (end corrections) to the main circuit's state.

Students often incorrectly assume that a complex, current-dependent shunt in the galvanometer branch will shift the null point, failing to realize that at null, Ig=0I_g = 0 and the branch is effectively isolated.

Furthermore, students frequently miscalculate the current through the bridge wire by ignoring the battery's internal resistance or the parallel upper branch.

They must correctly model the entire circuit to find the true current IwI_w driving the dynamic end corrections, rather than blindly applying the standard static end correction formula or assuming Iw=E/RwI_w = E/R_w.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterCurrent Electricity - Class 12
ConceptMetre Bridge End Correction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree