Question
Download Solution PDFThe resistances of 125 Ω strain gauge changes by 1 Ω for 4000 micro-strain. The gauge factor for strain gauge is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The gauge factor is defined as the ratio of per unit change in resistance to per unit change in length.
Gauge factor \({G_f} = \frac{{{\rm{\Delta }}R/R}}{{{\rm{\Delta }}L/L}}\)
\(\frac{{{\rm{\Delta }}R}}{R} = {G_f}(\frac{{{\rm{\Delta }}L}}{L})= {G_f}~\varepsilon \)
Where ε = strain = ΔL/L
Calculation:
Given that, change in resistance (ΔR) = 1 Ω
Resistance (R) = 125 Ω
Micro strain (ε) = ΔL/L = 4000 μs
Gauge factor \( = \frac{{1/125}}{{4000 \times {{10}^{ - 6}}}} = 2\)
Last updated on Jun 23, 2025
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.