GATE 2026 IN Q.25: solution and answer

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Half-bridge with two active gauges: the gauges occupy two adjacent arms of the Wheatstone bridge, and the other two arms are fixed resistors $R$. On the cantilever one gauge is in tension and the other in compression, so their resistances become $R+\Delta R$ and $R-\Delta R$.

2. Gauge factor: $G=\frac{\Delta R/R}{\epsilon }$, so $\frac{\Delta R}{R}=G\epsilon$.

3. The two active gauges form one voltage divider, whose output is $V_{ex}\frac{R+\Delta R}{(R+\Delta R)+(R-\Delta R)}=V_{ex}\frac{R+\Delta R}{2R}$. The two fixed resistors give a reference of $\frac{V_{ex}}{2}$. The bridge output is $$V_o=V_{ex}\left[\frac{R+\Delta R}{2R}-\frac{1}{2}\right]=\frac{V_{ex}}{2}\frac{\Delta R}{R}.$$

4. Substituting $\Delta R/R=G\epsilon$ gives $\frac{V_o}{V_{ex}}=\frac{G\epsilon }{2}$, so $$\epsilon =\frac{2V_o}{G{V}_{ex}}.$$

5. With $V_o=1\text{mV}=1\times 10^{-3}\text{V}$, $V_{ex}=10\text{V}$ and $G=2.5$: $$\epsilon =\frac{2\times 1\times 10^{-3}\text{V}}{2.5\times 10\text{V}}=8\times 10^{-5}.$$

6. $\epsilon =80\mu \text{strain}$, which is option (B).