Question:

A metallic strain gauge with negligible piezoresistive effect is subjected to a strain of \(50\times10^{-6}\). For the metal, Young’s Modulus \(=80\) GPa and Poisson’s ratio \(=0.42\). What is the change in resistance (in m\(\Omega\)), if the unstrained resistance is \(200~\Omega\)? (Round off the answer to one decimal place.)

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- For metal strain gauges without piezoresistive effect, \(GF\approx 1+2\nu\).
- Convert microstrain \((\mu\varepsilon)\) to a pure number before multiplying.
Updated On: Aug 26, 2025
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Correct Answer: 18.4

Solution and Explanation

Step 1: With negligible piezoresistive effect, the gauge factor for a metallic gauge is \[ GF=1+2\nu=1+2(0.42)=1.84. \] Step 2: Relative resistance change: \[ \frac{\Delta R}{R}=GF\cdot\varepsilon=1.84\times 50\times10^{-6}=9.2\times10^{-5}. \] Step 3: Absolute change: \[ \Delta R=R\frac{\Delta R}{R}=200~\Omega \times 9.2\times10^{-5}=0.0184~\Omega=18.4~\text{m}\Omega. \]
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