GATE 2025 IN Q.15: solution and answer

GATE 2025 · INQ.15 · 1 mark · MCQ

In the force transducer shown in Figure (a), four identical strain gauges S1, S2, S3, and S4 are mounted on a cantilever at equal distance from its base. S1 and S2 are mounted on the top surface and S3 and S4 are mounted on the bottom surface, as shown in the Figure (a). These strain gauges are to be connected to form a Wheatstone bridge consisting of four arms A, B, C, and D, as shown in the Figure (b). From the following options, the correct order to maximize the measurement sensitivity is

GATE 2025 IN question 15: Figure (a): a cantilever beam fixed to a wall at the left, extending to the right with a downward Force arrow at the free… — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
A→S1, B→S2, C→S4, D→S3
(B)
A→S1, B→S4, C→S3, D→S2
(C)
A→S1, B→S2, C→S3, D→S4
(D)
A→S1, B→S4, C→S2, D→S3

Answer and solution

ANSWER

The correct order is A→S1, B→S4, C→S2, D→S3, which is option (D).

Official answer: DSource: Official answer key, GATE 2025 IN (IIT Roorkee)

A downward force on the free end bends the cantilever so the top surface is in tension and the bottom surface is in compression. The gauges are identical and at equal distance from the base, so the strain magnitude is the same: S1 and S2 have $Δ R / R = + Gε$ , and S3 and S4 have $Δ R / R = - Gε$ .
From Figure (b), $V_{s}$ is at the top node and ground at the bottom node, with A top-left, B top-right, C bottom-right and D bottom-left. The left output node is the A-D junction and the right output node is the B-C junction, so $V_{o} = V_{s} [\frac{D}{A + D} - \frac{C}{B + C}] .$
For small changes from a common nominal value $R_{0}$ in each arm: $Δ V_{o} \approx \frac{V _{s}}{4} [\frac{Δ D}{R _{0}} + \frac{Δ B}{R _{0}} - \frac{Δ A}{R _{0}} - \frac{Δ C}{R _{0}}] .$
The output is largest when $Δ B$ and $Δ D$ have one sign and $Δ A$ and $Δ C$ have the opposite sign. Then all four terms add and $Δ V_{o} = V_{s} Gε$ , which is 4 times the output of one active gauge.
So the arms A and C (opposite arms) must be given gauges of the same strain sign, B and D (the other opposite pair) must be given gauges of the other sign, and the two pairs must be opposite to each other.
(A) A on S1 ( $+$ ), C on S4 ( $-$ ): the A and C terms cancel. B on S2 ( $+$ ), D on S3 ( $-$ ): the B and D terms cancel. $Δ V_{o} = 0$ .
(B) A on S1 ( $+$ ), C on S3 ( $-$ ): cancel. B on S4 ( $-$ ), D on S2 ( $+$ ): cancel. $Δ V_{o} = 0$ .
(C) A on S1 ( $+$ ), C on S3 ( $-$ ): cancel. B on S2 ( $+$ ), D on S4 ( $-$ ): cancel. $Δ V_{o} = 0$ .
(D) A on S1 ( $+$ ) and C on S2 ( $+$ ) have the same sign. B on S4 ( $-$ ) and D on S3 ( $-$ ) have the opposite sign. Every term adds: $Δ V_{o} = V_{s} Gε$ , the maximum possible.
Hence the answer is (D).

KEY IDEA

In a Wheatstone bridge, $V_{o} \propto Δ D + Δ B - Δ A - Δ C$ . Put the tension gauges (S1, S2) in one pair of opposite arms (A, C) and the compression gauges (S3, S4) in the other pair (B, D). All four arms then add, which gives 4 times the single-gauge sensitivity.

WATCH OUT FOR

Mixing up which arms are adjacent and which are opposite. Putting tension gauges in adjacent arms makes their effects cancel. Arms A and C are opposite and arms B and D are opposite, so the signs of A and C must agree and the signs of B and D must agree.

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GATE 2025 IN Q.15: solution and answer

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