101. Zero error in a vernier callipers is checked when the jaws are
ⓐ. used only for measuring mass
ⓑ. closed with no object between them
ⓒ. kept fully open and away from each other
ⓓ. replaced by a measuring tape
Correct Answer: closed with no object between them
Explanation: Zero error is checked by bringing the jaws of the vernier callipers into contact without any object between them. In this condition, the true reading should be zero. If the instrument shows a non-zero reading, the instrument has zero error. This error must be corrected before reporting the final measurement. The check is made at the zero position because it reveals whether the instrument starts from the proper reference point.
102. The relation between corrected reading, observed reading, and zero correction is
ⓐ. \(\text{corrected reading}=\text{observed reading}\times\text{zero correction}\)
ⓑ. \(\text{corrected reading}=\text{observed reading}+\text{zero correction}\)
ⓒ. \(\text{corrected reading}=\text{observed reading}-\text{zero correction}\)
ⓓ. \(\text{corrected reading}=\text{zero correction}-\text{observed reading}\)
Correct Answer: \(\text{corrected reading}=\text{observed reading}+\text{zero correction}\)
Explanation: A zero correction is the quantity added to the observed reading to remove zero error. The standard relation is \(\text{corrected reading}=\text{observed reading}+\text{zero correction}\). If the zero error is positive, the zero correction is negative. If the zero error is negative, the zero correction is positive. The word correction is important because it already carries the sign needed to repair the reading.
103. A vernier callipers has a positive zero error of \(+0.03\,\text{cm}\). Its zero correction is
ⓐ. \(+0.03\,\text{cm}\)
ⓑ. \(-0.03\,\text{cm}\)
ⓒ. \(-3.0\,\text{cm}\)
ⓓ. \(+0.30\,\text{cm}\)
Correct Answer: \(-0.03\,\text{cm}\)
Explanation: Positive zero error means the instrument shows a positive reading even when the true length is zero. To correct this, the extra positive reading must be subtracted from every observed reading. Therefore the zero correction has the opposite sign of the zero error. For a zero error of \(+0.03\,\text{cm}\), the zero correction is \(-0.03\,\text{cm}\). Applying the same sign again would increase the error instead of removing it.
104. A vernier callipers shows an observed length of \(3.46\,\text{cm}\). If its zero error is \(+0.02\,\text{cm}\), the corrected length is
ⓐ. \(3.46\,\text{cm}\)
ⓑ. \(3.50\,\text{cm}\)
ⓒ. \(3.48\,\text{cm}\)
ⓓ. \(3.44\,\text{cm}\)
Correct Answer: \(3.44\,\text{cm}\)
Explanation: \( \textbf{Observed reading:} \) \(3.46\,\text{cm}\).
\( \textbf{Zero error:} \) \(+0.02\,\text{cm}\).
A positive zero error means the instrument reads too much by \(0.02\,\text{cm}\).
So the zero correction is
\[
-0.02\,\text{cm}
\]
\( \textbf{Correction formula:} \)
\[
\text{corrected reading}=\text{observed reading}+\text{zero correction}
\]
\( \textbf{Substitution:} \)
\[
\text{corrected reading}=3.46\,\text{cm}+(-0.02\,\text{cm})
\]
\[
=3.44\,\text{cm}
\]
The correction is subtracted because the instrument was already reading high.
\( \textbf{Final answer:} \) The corrected length is \(3.44\,\text{cm}\).
105. A vernier callipers has a negative zero error of \(-0.04\,\text{cm}\). Its zero correction is
ⓐ. \(-0.40\,\text{cm}\)
ⓑ. \(+4.0\,\text{cm}\)
ⓒ. \(+0.04\,\text{cm}\)
ⓓ. \(-0.04\,\text{cm}\)
Correct Answer: \(+0.04\,\text{cm}\)
Explanation: A negative zero error means the instrument reads below zero when the true reading should be zero. To bring the reading back to the true value, a positive correction must be added. Hence the zero correction has the opposite sign of the zero error. For zero error \(-0.04\,\text{cm}\), the zero correction is \(+0.04\,\text{cm}\). This sign reversal is essential because correction is meant to remove the error, not repeat it.
106. The observed reading of a vernier callipers is \(2.58\,\text{cm}\). If the zero error is \(-0.03\,\text{cm}\), the corrected reading is
ⓐ. \(2.55\,\text{cm}\)
ⓑ. \(2.61\,\text{cm}\)
ⓒ. \(2.88\,\text{cm}\)
ⓓ. \(2.58\,\text{cm}\)
Correct Answer: \(2.61\,\text{cm}\)
Explanation: \( \textbf{Observed reading:} \) \(2.58\,\text{cm}\).
\( \textbf{Zero error:} \) \(-0.03\,\text{cm}\).
A negative zero error means the instrument reads less than it should.
So the zero correction is
\[
+0.03\,\text{cm}
\]
\( \textbf{Use the correction relation:} \)
\[
\text{corrected reading}=\text{observed reading}+\text{zero correction}
\]
\( \textbf{Substitute values:} \)
\[
\text{corrected reading}=2.58\,\text{cm}+0.03\,\text{cm}
\]
\[
=2.61\,\text{cm}
\]
Adding the correction is necessary because the instrument was under-reading.
\( \textbf{Final answer:} \) The corrected reading is \(2.61\,\text{cm}\).
107. A vernier callipers has least count \(0.01\,\text{cm}\). For an object, the main scale reading is \(4.20\,\text{cm}\), and the fifth vernier division coincides. The zero error is \(+0.02\,\text{cm}\). The corrected length is
ⓐ. \(4.23\,\text{cm}\)
ⓑ. \(4.45\,\text{cm}\)
ⓒ. \(4.27\,\text{cm}\)
ⓓ. \(4.25\,\text{cm}\)
Correct Answer: \(4.23\,\text{cm}\)
Explanation: \( \textbf{Given data:} \) \(\text{MSR}=4.20\,\text{cm}\), least count \(=0.01\,\text{cm}\), vernier coincidence \(=5\), zero error \(=+0.02\,\text{cm}\).
\( \textbf{Vernier contribution:} \)
\[
5\times0.01\,\text{cm}=0.05\,\text{cm}
\]
\( \textbf{Observed reading:} \)
\[
4.20\,\text{cm}+0.05\,\text{cm}=4.25\,\text{cm}
\]
A positive zero error means the zero correction is negative:
\[
\text{zero correction}=-0.02\,\text{cm}
\]
\( \textbf{Corrected reading:} \)
\[
4.25\,\text{cm}+(-0.02\,\text{cm})=4.23\,\text{cm}
\]
The observed reading is first built from scale parts, and only then is the zero correction applied.
\( \textbf{Final answer:} \) The corrected length is \(4.23\,\text{cm}\).
108. A vernier callipers has least count \(0.1\,\text{mm}\). The main scale reading is \(31\,\text{mm}\), the eighth vernier division coincides, and the zero error is \(-0.2\,\text{mm}\). The corrected reading is
ⓐ. \(32.8\,\text{mm}\)
ⓑ. \(31.8\,\text{mm}\)
ⓒ. \(31.6\,\text{mm}\)
ⓓ. \(32.0\,\text{mm}\)
Correct Answer: \(32.0\,\text{mm}\)
Explanation: \( \textbf{Given data:} \) \(\text{MSR}=31\,\text{mm}\), least count \(=0.1\,\text{mm}\), vernier coincidence \(=8\), zero error \(=-0.2\,\text{mm}\).
\( \textbf{Vernier reading:} \)
\[
8\times0.1\,\text{mm}=0.8\,\text{mm}
\]
\( \textbf{Observed reading:} \)
\[
31\,\text{mm}+0.8\,\text{mm}=31.8\,\text{mm}
\]
A negative zero error gives a positive zero correction:
\[
\text{zero correction}=+0.2\,\text{mm}
\]
\( \textbf{Corrected reading:} \)
\[
31.8\,\text{mm}+0.2\,\text{mm}=32.0\,\text{mm}
\]
The correction is added because the instrument was reading below zero in the zero-check position.
\( \textbf{Final answer:} \) The corrected reading is \(32.0\,\text{mm}\).
109. When the jaws of a vernier callipers are closed, the vernier zero lies to the right of the main scale zero. This situation indicates
ⓐ. zero correction equal to zero in all cases
ⓑ. negative zero error
ⓒ. no least count
ⓓ. positive zero error
Correct Answer: positive zero error
Explanation: If the vernier zero lies to the right of the main scale zero when the jaws are closed, the instrument shows a positive reading for zero length. That is called positive zero error. A positive zero error means the instrument is over-reading. The correction must therefore be negative when actual measurements are made. The sign is decided from the zero-check position, not from the object being measured.
110. When the jaws are closed, the vernier zero lies to the left of the main scale zero. The sign of zero correction is
ⓐ. positive
ⓑ. not related to zero error
ⓒ. negative
ⓓ. always zero
Correct Answer: positive
Explanation: When the vernier zero lies to the left of the main scale zero, the zero error is negative. A negative zero error means the instrument reads less than the true zero reading. To correct such a reading, a positive zero correction is added to the observed reading. Thus the sign of zero correction is opposite to the sign of zero error. The left-of-zero position is therefore linked with adding a correction, not subtracting one.
111. A measurement record says: observed reading \(=5.72\,\text{cm}\), zero correction \(=-0.04\,\text{cm}\). The corrected reading should be reported as
ⓐ. \(5.72\,\text{cm}\)
ⓑ. \(5.76\,\text{cm}\)
ⓒ. \(5.68\,\text{cm}\)
ⓓ. \(5.84\,\text{cm}\)
Correct Answer: \(5.68\,\text{cm}\)
Explanation: \( \textbf{Given data:} \) Observed reading \(=5.72\,\text{cm}\), zero correction \(=-0.04\,\text{cm}\).
\( \textbf{Correction relation:} \)
\[
\text{corrected reading}=\text{observed reading}+\text{zero correction}
\]
\( \textbf{Substitution:} \)
\[
\text{corrected reading}=5.72\,\text{cm}+(-0.04\,\text{cm})
\]
\( \textbf{Calculation:} \)
\[
5.72-0.04=5.68
\]
So,
\[
\text{corrected reading}=5.68\,\text{cm}
\]
The word correction already includes its sign, so it should be added algebraically rather than converted again into a zero error.
\( \textbf{Final answer:} \) The corrected reading is \(5.68\,\text{cm}\).
112. Two vernier records are shown for the same object.
| Case | Observed reading | Zero error |
|---|
| P | \(2.84\,\text{cm}\) | \(+0.01\,\text{cm}\) |
| Q | \(2.80\,\text{cm}\) | \(-0.03\,\text{cm}\) |
The corrected readings for cases P and Q are respectively
ⓐ. \(2.84\,\text{cm}\) and \(2.80\,\text{cm}\)
ⓑ. \(2.83\,\text{cm}\) and \(2.77\,\text{cm}\)
ⓒ. \(2.83\,\text{cm}\) and \(2.83\,\text{cm}\)
ⓓ. \(2.85\,\text{cm}\) and \(2.77\,\text{cm}\)
Correct Answer: \(2.83\,\text{cm}\) and \(2.83\,\text{cm}\)
Explanation: \( \textbf{Case P:} \) The zero error is \(+0.01\,\text{cm}\), so the zero correction is \(-0.01\,\text{cm}\).
\[
\text{corrected reading}_P=2.84\,\text{cm}-0.01\,\text{cm}=2.83\,\text{cm}
\]
\( \textbf{Case Q:} \) The zero error is \(-0.03\,\text{cm}\), so the zero correction is \(+0.03\,\text{cm}\).
\[
\text{corrected reading}_Q=2.80\,\text{cm}+0.03\,\text{cm}=2.83\,\text{cm}
\]
Both records can give the same corrected value even though their observed readings and zero errors are different.
The sign of the correction, not just the size of the zero error, decides the final value.
\( \textbf{Final answer:} \) The corrected readings are \(2.83\,\text{cm}\) and \(2.83\,\text{cm}\).
113. In a screw gauge, the pitch is the distance moved by the screw in
ⓐ. one smallest division of the main scale
ⓑ. one coinciding vernier division
ⓒ. one full rotation of the circular scale
ⓓ. one second of observation time
Correct Answer: one full rotation of the circular scale
Explanation: The pitch of a screw gauge tells how much the screw advances along its axis when the circular scale is turned through one complete rotation. It is a length, so it may be expressed in units such as \(\text{mm}\). Pitch is not the same as least count, because least count also depends on how many divisions are present on the circular scale. A screw gauge uses this controlled screw motion to measure small thicknesses and diameters. The idea is that rotation is converted into a small linear displacement.
114. A screw gauge has pitch \(0.5\,\text{mm}\) and \(50\) divisions on its circular scale. Its least count is
ⓐ. \(25\,\text{mm}\)
ⓑ. \(0.001\,\text{mm}\)
ⓒ. \(0.05\,\text{mm}\)
ⓓ. \(0.01\,\text{mm}\)
Correct Answer: \(0.01\,\text{mm}\)
Explanation: \( \textbf{Given data:} \) Pitch \(=0.5\,\text{mm}\), number of circular scale divisions \(=50\).
\( \textbf{Required quantity:} \) Least count of the screw gauge.
\( \textbf{Useful relation:} \)
\[
\text{least count}=\frac{\text{pitch}}{\text{number of circular scale divisions}}
\]
This applies because one complete rotation advances the screw by one pitch, and the circular scale divides that motion into equal smaller parts.
\( \textbf{Substitution:} \)
\[
\text{least count}=\frac{0.5\,\text{mm}}{50}
\]
\( \textbf{Calculation:} \)
\[
\text{least count}=0.01\,\text{mm}
\]
The least count is much smaller than the pitch because one rotation is subdivided into \(50\) equal circular-scale parts.
\( \textbf{Final answer:} \) The least count is \(0.01\,\text{mm}\).
115. The missing term in the screw-gauge relation \(\text{least count}=\frac{\text{pitch}}{\_\_\_\_}\) is
ⓐ. number of main scale divisions
ⓑ. main scale reading
ⓒ. zero correction
ⓓ. circular scale divisions
Correct Answer: circular scale divisions
Explanation: The least count of a screw gauge depends on how far the screw moves in one full rotation and how many equal divisions divide that rotation. The pitch gives the linear motion in one complete turn. The circular scale divisions divide that motion into smaller measurable parts. Therefore \(\text{least count}=\frac{\text{pitch}}{\text{number of circular scale divisions}}\). Zero correction may be needed later in the final reading, but it does not define the least count.
116. A screw gauge moves forward by \(1.0\,\text{mm}\) in \(2\) complete rotations. Its circular scale has \(100\) divisions. The least count is
ⓐ. \(0.020\,\text{mm}\)
ⓑ. \(0.005\,\text{mm}\)
ⓒ. \(0.500\,\text{mm}\)
ⓓ. \(0.010\,\text{mm}\)
Correct Answer: \(0.005\,\text{mm}\)
Explanation: \( \textbf{Given motion:} \) The screw advances \(1.0\,\text{mm}\) in \(2\) complete rotations.
\( \textbf{Find the pitch first:} \)
\[
\text{pitch}=\frac{1.0\,\text{mm}}{2}=0.5\,\text{mm}
\]
\( \textbf{Circular scale divisions:} \) \(100\).
\( \textbf{Least-count relation:} \)
\[
\text{least count}=\frac{\text{pitch}}{\text{number of circular scale divisions}}
\]
\( \textbf{Substitution:} \)
\[
\text{least count}=\frac{0.5\,\text{mm}}{100}
\]
\( \textbf{Calculation:} \)
\[
\text{least count}=0.005\,\text{mm}
\]
The first step is to find motion per one rotation; using \(1.0\,\text{mm}\) directly as the pitch would double the result.
\( \textbf{Final answer:} \) The least count is \(0.005\,\text{mm}\).
117. A screw gauge reading is obtained by adding the main scale reading to
ⓐ. circular scale reading multiplied by least count
ⓑ. circular scale reading divided by pitch
ⓒ. pitch multiplied by main scale reading
ⓓ. zero error multiplied by circular scale divisions
Correct Answer: circular scale reading multiplied by least count
Explanation: The main scale reading gives the larger directly visible part of the measurement. The circular scale reading gives the number of circular divisions that have passed the reference line. That number becomes a length only after multiplying by the least count. Thus the observed reading is \(\text{MSR}+(\text{CSR}\times\text{LC})\). Zero correction, if present, is applied after finding the observed reading.
118. A screw gauge has least count \(0.01\,\text{mm}\). The main scale reading is \(3.00\,\text{mm}\), and the circular scale reading is \(27\). The observed diameter is
ⓐ. \(3.54\,\text{mm}\)
ⓑ. \(3.27\,\text{mm}\)
ⓒ. \(5.70\,\text{mm}\)
ⓓ. \(3.17\,\text{mm}\)
Correct Answer: \(3.27\,\text{mm}\)
Explanation: \( \textbf{Given data:} \) Main scale reading \(=3.00\,\text{mm}\), circular scale reading \(=27\), least count \(=0.01\,\text{mm}\).
\( \textbf{Required quantity:} \) Observed screw-gauge reading.
\( \textbf{Circular scale contribution:} \)
\[
\text{CSR}\times\text{LC}=27\times0.01\,\text{mm}
\]
\[
=0.27\,\text{mm}
\]
\( \textbf{Observed reading:} \)
\[
\text{observed reading}=\text{MSR}+(\text{CSR}\times\text{LC})
\]
\[
=3.00\,\text{mm}+0.27\,\text{mm}
\]
\[
=3.27\,\text{mm}
\]
The circular scale number is not itself in \(\text{mm}\); it must be multiplied by the least count first.
\( \textbf{Final answer:} \) The observed diameter is \(3.27\,\text{mm}\).
119. A screw gauge record is shown below.
| Reading part | Value |
|---|
| Main scale reading | \(5.5\,\text{mm}\) |
| Circular scale reading | \(34\) |
| Least count | \(0.01\,\text{mm}\) |
The observed reading is
ⓐ. \(5.50\,\text{mm}\)
ⓑ. \(8.90\,\text{mm}\)
ⓒ. \(5.84\,\text{mm}\)
ⓓ. \(5.34\,\text{mm}\)
Correct Answer: \(5.84\,\text{mm}\)
Explanation: \( \textbf{Main scale part:} \) \(5.5\,\text{mm}\).
\( \textbf{Circular scale part:} \)
\[
34\times0.01\,\text{mm}=0.34\,\text{mm}
\]
\( \textbf{Observed reading:} \)
\[
5.5\,\text{mm}+0.34\,\text{mm}=5.84\,\text{mm}
\]
The main scale reading and circular scale contribution must be in the same unit before addition.
The option \(5.34\,\text{mm}\) ignores the given main scale value \(5.5\,\text{mm}\) and replaces it partly with the circular scale contribution.
\( \textbf{Final answer:} \) The observed reading is \(5.84\,\text{mm}\).
120. A screw gauge gives an observed wire diameter of \(0.48\,\text{mm}\). In metres, this is
ⓐ. \(4.8\times10^{-5}\,\text{m}\)
ⓑ. \(4.8\times10^{-4}\,\text{m}\)
ⓒ. \(4.8\times10^{4}\,\text{m}\)
ⓓ. \(4.8\times10^{-3}\,\text{m}\)
Correct Answer: \(4.8\times10^{-4}\,\text{m}\)
Explanation: \( \textbf{Given diameter:} \) \(0.48\,\text{mm}\).
\( \textbf{Unit relation:} \)
\[
1\,\text{mm}=10^{-3}\,\text{m}
\]
\( \textbf{Conversion setup:} \)
\[
0.48\,\text{mm}=0.48\times10^{-3}\,\text{m}
\]
Write \(0.48\) as \(4.8\times10^{-1}\):
\[
0.48\times10^{-3}=4.8\times10^{-1}\times10^{-3}
\]
\[
=4.8\times10^{-4}
\]
The value in metres is small because a millimetre is only \(10^{-3}\) of a metre.
\( \textbf{Final answer:} \) \(0.48\,\text{mm}=4.8\times10^{-4}\,\text{m}\).
