The chief value of a well-kept measurement record is that
Answer: B
When a component fails in service or a dispute arises, the record is what establishes whether the original acceptance was sound and what instrument and standard stood behind it. Nothing about the record changes the measurement, but a great deal depends on being able to reconstruct it.
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Which of the following most improves the trustworthiness of a measurement?
Answer: C
Traceability is what connects a reading to the definition of the unit, and without it no amount of resolution or repetition establishes correctness. Repetition reduces the random component only, and extra digits may show nothing but noise.
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The solution for the unknown resistance for a Wheatstone Bridge is \(R_{4} = \frac{R_{2}R_{3}}{R_{1}}\)
Where, R1 = 100 ± 0.5% Ω
R2 = 1000 ± 0.5% Ω
R3 = 842 ± 0.5% Ω
What will be the limiting error in percent for Resistance R4 ?
RSMSSB JE - 2025, RSMSSB - 2025,
Answer: D
The limiting error of \(R_{4}\) is 1.5%.
When a result is a product or quotient of quantities, the limiting (worst-case) relative errors add, whichever side of the fraction each quantity is on.
\[ \begin{aligned} \frac{\delta R_{4}}{R_{4}} &= \frac{\delta R_{1}}{R_{1}} + \frac{\delta R_{2}}{R_{2}} + \frac{\delta R_{3}}{R_{3}} \\ &= 0.5 + 0.5 + 0.5 = 1.5\% \end{aligned} \]
The nominal value is \(R_{4} = 1000 \times 842/100 = 8420\ \Omega\), so it lies within \(8420 \pm 126.3\ \Omega\). 0.50% counts only one resistor. 1.6% and 0.16% do not follow from adding three 0.5% errors.
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Match List I with List II.
| List I | List II |
|---|---|
| (a) Recording Instruments | (i) C.T. and P.T. |
| (b) Integrating Instruments | (ii) Measurement of current |
| (c) Galvanometer | (iii) Watt hour Meter |
| (d) Instrument Transformers | (iv) Magnetic Tape |
Choose the most appropriate answer from the options given below :
RSMSSB JE - 2025, RSMSSB - 2025,
Answer: B
The correct matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
The option ending (c)-(i), (d)-(ii) swaps the galvanometer and instrument transformers, and the options starting (a)-(iii) confuse recording with integrating instruments.
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In an experiment, the acceleration due to gravity is determined by measuring the time period of a simple pendulum. If there is an error of 1% in the measurement of time period, the error in the value of g is :
MH SET - 2019, SET Exam - 2019,
Answer: A
The error in g is 2%.
For a simple pendulum \(T = 2\pi\sqrt{L/g}\), so \(g = 4\pi^{2}L/T^{2}\). The relative error in a quantity raised to a power is that power times the relative error:
\[ \begin{aligned} \frac{\Delta g}{g} &= 2\frac{\Delta T}{T} \\ &= 2 \times 1\% = 2\% \end{aligned} \]
The option 1% ignores the square, and 0.5% treats g as depending on the square root of T.
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