Vernier Calipers, Micrometer screw gauge Questions in English

(C) The least count of the screw gauge is $0.01\;mm$.
This means that any measurement taken with this instrument must be recorded to two decimal places in millimeters to reflect the precision of the device.
Among the given options,$2.00\;mm$ is the only value expressed with the correct precision corresponding to the least count of $0.01\;mm$.

(C) The most accurate length of the cylinder is the arithmetic mean of the given observations.
$\text{Mean length } (\overline{\ell}) = \frac{3.29 + 3.28 + 3.29 + 3.31 + 3.28 + 3.27 + 3.29 + 3.30}{8}$
$\overline{\ell} = \frac{26.31}{8} = 3.28875 \, cm$
Rounding off to two decimal places (consistent with the precision of the instrument),we get:
$\overline{\ell} \approx 3.29 \, cm$.

(B) Given that $50$ divisions of the vernier scale $(VSD)$ are equal to $49$ divisions of the main scale $(MSD)$.
First,we find the value of one main scale division $(MSD)$: $MSD = 7.05 \; cm - 7.00 \; cm = 0.05 \; cm$.
The least count $(LC)$ of the vernier calliper is defined as $LC = 1 \; MSD - 1 \; VSD$.
Since $50 \; VSD = 49 \; MSD$,we have $1 \; VSD = \frac{49}{50} \; MSD = 0.98 \; MSD$.
Therefore,$LC = 1 \; MSD - 0.98 \; MSD = 0.02 \; MSD$.
Substituting the value of $MSD$: $LC = 0.02 \times 0.05 \; cm = 0.001 \; cm$.
The main scale reading $(MSR)$ is the mark just before the vernier zero,which is $7.00 \; cm$.
The vernier scale reading $(VSR)$ is the coinciding division multiplied by the least count: $VSR = 23 \times 0.001 \; cm = 0.023 \; cm$.
The total measured thickness is $MSR + VSR = 7.00 \; cm + 0.023 \; cm = 7.023 \; cm$.

(C) The diameter of the ball is calculated using the formula: $\text{Diameter} = \text{MSR} + (\text{CSR} \times \text{LC}) - \text{Zero Error}$.
Given: $\text{MSR} = 5 \, mm = 0.5 \, cm$,$\text{CSR} = 25$,$\text{LC} = 0.001 \, cm$,and $\text{Zero Error} = -0.004 \, cm$.
Substituting the values:
$\text{Diameter} = 0.5 \, cm + (25 \times 0.001 \, cm) - (-0.004 \, cm)$.
$\text{Diameter} = 0.5 \, cm + 0.025 \, cm + 0.004 \, cm$.
$\text{Diameter} = 0.529 \, cm$.

(A) The pitch of the screw gauge is the distance moved in one full rotation. Since $2$ full turns cover $1 \ mm$,the pitch $= \frac{1 \ mm}{2} = 0.5 \ mm$.
The least count $(LC)$ is given by $\frac{\text{pitch}}{\text{total number of divisions}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$.
The observed reading is calculated as: $\text{Main Scale Reading} + (\text{Circular Scale Division} \times LC) = 3 \ mm + (35 \times 0.01 \ mm) = 3.35 \ mm$.
The actual diameter is calculated by subtracting the zero error from the observed reading: $\text{Diameter} = \text{Observed Reading} - (\text{Zero Error}) = 3.35 \ mm - (-0.03 \ mm) = 3.35 \ mm + 0.03 \ mm = 3.38 \ mm$.

(C) Given that $30$ divisions of the vernier scale coincide with $29$ divisions of the main scale.
Therefore,$1$ Vernier Scale Division $(VSD) = \frac{29}{30}$ Main Scale Division $(MSD)$.
The least count of the instrument is defined as the difference between one main scale division and one vernier scale division.
Least Count $= 1\,MSD - 1\,VSD$.
Least Count $= 1\,MSD - \frac{29}{30}\,MSD = \frac{1}{30}\,MSD$.
Given that $1\,MSD = 0.5^\circ$.
Least Count $= \frac{1}{30} \times 0.5^\circ = \frac{0.5}{30}^\circ = \frac{1}{60}^\circ$.
Since $1^\circ = 60$ minutes $(')$,
Least Count $= \frac{1}{60} \times 60' = 1'$.

(A) The Least Count $(LC)$ of the screw gauge is calculated as the value of one main scale division divided by the total number of circular scale divisions.
$LC = \frac{1 \ mm}{100} = 0.01 \ mm$.
The diameter of the wire is given by the formula: $\text{Diameter} = \text{Main Scale Reading} (MSR) + (\text{Circular Scale Reading} (CSR) \times LC)$.
Given $MSR = 0 \ mm$ and $CSR = 52 \ divisions$.
$\text{Diameter} = 0 \ mm + (52 \times 0.01 \ mm) = 0.52 \ mm$.
To convert the diameter from $mm$ to $cm$,we divide by $10$:
$\text{Diameter} = \frac{0.52}{10} \ cm = 0.052 \ cm$.

(D) The reading of a spectrometer is given by: $\text{Reading} = \text{Main scale reading} + (\text{Vernier scale reading} \times \text{Least count})$.
First,calculate the Least Count $(LC)$ of the Vernier scale:
$LC = \frac{\text{Value of 1 Main Scale Division}}{\text{Total number of Vernier divisions}} = \frac{0.5^{\circ}}{30}$.
Given:
Main scale reading $= 58.5^{\circ}$
Vernier scale reading $= 09$ divisions
Now,calculate the total reading $(R)$:
$R = 58.5^{\circ} + (9 \times \frac{0.5^{\circ}}{30})$
$R = 58.5^{\circ} + (9 \times 0.01667^{\circ})$
$R = 58.5^{\circ} + 0.15^{\circ}$
$R = 58.65^{\circ}$.

(A) The measurement $3.50\;cm$ indicates a precision up to two decimal places in centimeters,which corresponds to a least count of $0.01\;cm$.
For the vernier calliper described in option $A$:
Main Scale Division $(MSD)$ $= 1\;cm / 10 = 0.1\;cm$.
Given $10\;VSD = 9\;MSD$,so $1\;VSD = 0.9\;MSD = 0.9 \times 0.1\;cm = 0.09\;cm$.
Least Count $(LC)$ $= 1\;MSD - 1\;VSD = 0.1\;cm - 0.09\;cm = 0.01\;cm$.
For the screw gauge in option $B$:
$LC = \text{Pitch} / \text{Number of circular scale divisions} = 1\;mm / 100 = 0.01\;mm = 0.001\;cm$.
For the screw gauge in option $C$:
$LC = 1\;mm / 50 = 0.02\;mm = 0.002\;cm$.
Since the measurement $3.50\;cm$ has a precision of $0.01\;cm$,the vernier calliper described in option $A$ is the correct instrument.

(D) The Least Count $(LC)$ of the screw gauge is calculated as: $LC = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$.
Since the $45^{th}$ division coincides with the main scale line and the zero is barely visible,the zero error is negative. The number of divisions past the zero mark is $50 - 45 = 5$.
Zero Error $(ZE)$ = $-5 \times LC = -5 \times 0.01 \ mm = -0.05 \ mm$.
The observed reading is: $\text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC) = 0.5 \ mm + (25 \times 0.01 \ mm) = 0.5 \ mm + 0.25 \ mm = 0.75 \ mm$.
The corrected reading (thickness) is: $\text{Observed Reading} - ZE = 0.75 \ mm - (-0.05 \ mm) = 0.75 \ mm + 0.05 \ mm = 0.80 \ mm$.

(B) travelling microscope is a precision instrument used to measure small distances.
It consists of a main scale and a vernier scale attached to the microscope assembly.
When measuring the refractive index of a glass slab,the microscope is focused on a mark on the table,then on the mark through the glass slab,and finally on a dust particle on the top surface of the slab.
The vertical displacement is measured accurately using the vernier scale provided on the microscope.

(C) The least count $(L.C.)$ of the screw gauge is $L.C. = \frac{1 \, mm}{100} = 0.01 \, mm$.
Let the actual length of each rod be $\ell$ and the zero error be $x$.
From diagram $(I)$,the main scale reading is $2 \, mm$ and the circular scale reading is $12$. The observed reading is $2 + 12 \times 0.01 = 2.12 \, mm$. Thus,$\ell + x = 2.12 \, mm$.
From diagram $(II)$,the main scale reading is $4 \, mm$ and the circular scale reading is $10$. The observed reading is $4 + 10 \times 0.01 = 4.10 \, mm$. Thus,$2\ell + x = 4.10 \, mm$.
Subtracting the first equation from the second: $(2\ell + x) - (\ell + x) = 4.10 - 2.12$,which gives $\ell = 1.98 \, mm$.
Substituting $\ell$ into the first equation: $1.98 + x = 2.12$,so $x = 2.12 - 1.98 = 0.14 \, mm$.
Since the reading is positive,the zero error is $+0.14 \, mm$.

(A) The pitch of the screw gauge is defined as the distance moved by the spindle per complete rotation of the circular scale.
Given that $5$ complete rotations correspond to $1.5 \, mm$ on the linear scale,the pitch is calculated as:
$\text{Pitch} = \frac{1.5 \, mm}{5} = 0.3 \, mm$.
The least count of a screw gauge is defined as the ratio of the pitch to the total number of divisions on the circular scale.
Given that the circular scale has $50$ divisions,the least count is:
$\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.3 \, mm}{50} = 0.006 \, mm$.

(D) Given,least count $(LC)$ = $0.005 \ mm$ and total divisions on circular scale $(N)$ = $200$.
Pitch of the screw gauge = $LC \times N = 0.005 \times 200 = 1 \ mm$.
This means $1 \ MSD$ (Main Scale Division) = $1 \ mm$.
Zero error $(ZE)$ = $5 \times LC = 5 \times 0.005 = 0.025 \ mm$.
When the sphere is placed,the main scale reading $(MSR)$ = $4 \times 1 \ mm = 4 \ mm$.
The circular scale reading $(CSR)$ advances by $5$ times the initial reading,so final $CSR = 5 \times 5 = 25$.
Observed reading $(OR)$ = $MSR + (CSR \times LC) = 4 + (25 \times 0.005) = 4 + 0.125 = 4.125 \ mm$.
Corrected diameter $(D)$ = $OR - ZE = 4.125 - 0.025 = 4.10 \ mm$.
Radius $(r)$ = $D / 2 = 4.10 / 2 = 2.05 \ mm$.

(A) Main Scale Division $(MSD)$ = $1/20\;cm = 0.05\;cm$.
Vernier Scale Division $(VSD)$ = $(9/10) \times MSD = 0.9 \times 0.05\;cm = 0.045\;cm$.
Least Count $(LC)$ = $MSD - VSD = 0.05\;cm - 0.045\;cm = 0.005\;cm$.
Zero Error: Since the zero of the vernier scale is to the right of the main scale zero,it is a positive zero error. $Zero\;Error = + (6 \times LC) = + (6 \times 0.005\;cm) = +0.03\;cm$.
Measured Length = $Main\;Scale\;Reading (MSR) + (Vernier\;Coincidence \times LC) = 3.20\;cm + (8 \times 0.005\;cm) = 3.20 + 0.04 = 3.24\;cm$.
Corrected Length = $Measured\;Length - Zero\;Error = 3.24\;cm - 0.03\;cm = 3.21\;cm$.
Measured Diameter = $MSR + (Vernier\;Coincidence \times LC) = 1.50\;cm + (6 \times 0.005\;cm) = 1.50 + 0.03 = 1.53\;cm$.
Corrected Diameter = $Measured\;Diameter - Zero\;Error = 1.53\;cm - 0.03\;cm = 1.50\;cm$.

(B) The least count $(LC)$ of the screw gauge is calculated as: $LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$.
From figure $(i)$,the zero error is $+3$ divisions. Therefore,the zero error is $3 \times 0.01 \ mm = +0.03 \ mm$.
From figure $(ii)$,the main scale reading is $5.0 \ mm$ and the circular scale division coinciding with the reference line is $29$. The observed reading is $5.0 \ mm + (29 \times 0.01 \ mm) = 5.29 \ mm$.
The actual diameter is given by: $\text{Diameter} = \text{Observed reading} - \text{Zero error} = 5.29 \ mm - 0.03 \ mm = 5.26 \ mm$.

(C) The main scale division $(MSD)$ is $1 \text{ mm}$.
Given that $8$ vernier scale divisions $(VSD)$ coincide with $5$ main scale divisions $(MSD)$.
Therefore,$1 \text{ VSD} = \frac{5}{8} \text{ MSD} = \frac{5}{8} \text{ mm} = 0.625 \text{ mm}$.
The least count $(LC)$ of the vernier caliper is $1 \text{ MSD} - 1 \text{ VSD} = 1 \text{ mm} - 0.625 \text{ mm} = 0.375 \text{ mm}$.
The zero of the vernier scale is to the left of the $36^{th}$ division of the main scale,so the main scale reading $(MSR)$ is $35 \text{ mm}$.
The fourth division of the vernier scale coincides with the main scale,so the vernier scale reading $(VSR)$ is $4 \times \text{LC} = 4 \times 0.375 \text{ mm} = 1.5 \text{ mm}$.
The total reading is $\text{MSR} + \text{VSR} = 35 \text{ mm} + 1.5 \text{ mm} = 36.5 \text{ mm}$.
Converting to centimeters,$36.5 \text{ mm} = 3.65 \text{ cm}$.

(D) The main scale reading $(MSR)$ is $3.1 \ cm = 31 \ mm$.
The vernier scale divisions follow an $A.P.$ where the $n^{th}$ division $V_n$ is given by $V_n = a + (n-1)d$.
Here, $a = 0.95 \ mm$ and the common difference $d = 0.90 - 0.95 = -0.05 \ mm$.
The $4^{th}$ division $V_4 = 0.95 + (4-1)(-0.05) = 0.95 - 0.15 = 0.80 \ mm$.
The vernier constant $(VC)$ is the difference between the main scale division $(MSD)$ and the vernier scale division $(VSD)$.
For the $4^{th}$ division, the effective vernier reading is $n \times (MSD - V_n) = 4 \times (1.00 - 0.80) = 4 \times 0.20 = 0.80 \ mm = 0.08 \ cm$.
Total reading = $MSR + \text{Vernier Reading} = 3.1 \ cm + 0.08 \ cm = 3.18 \ cm$.

(B) The pitch of the screw gauge is $1.5\; mm$ and the number of circular scale divisions is $100$.
The Least Count $(L.C.)$ is calculated as: $L.C. = \frac{\text{Pitch}}{\text{Total circular scale divisions}} = \frac{1.5\; mm}{100} = 0.015\; mm$.
The main scale reading $(M.S.R.)$ is the last visible mark on the linear scale,which is $2\; mm$.
The circular scale reading $(C.S.R.)$ is the $76^{th}$ division.
The diameter is given by the formula: $Diameter = M.S.R. + (L.C. \times C.S.R.)$.
Substituting the values: $Diameter = 2\; mm + (0.015\; mm \times 76) = 2\; mm + 1.14\; mm = 3.14\; mm$.

(C) The Least Count $(LC)$ of a vernier callipers is given by: $LC = 1\,MSD - 1\,VSD$.
Given that $10\,VSD = 9\,MSD$,we have $1\,VSD = 0.9\,MSD$.
Since $1\,MSD = 0.1\,cm$,then $1\,VSD = 0.9 \times 0.1\,cm = 0.09\,cm$.
Therefore,$LC = 0.1\,cm - 0.09\,cm = 0.01\,cm$.
The diameter is calculated as: $Diameter = MSR + (LC \times VSR)$.
Here,the Main Scale Reading $(MSR)$ is $1.3\,cm$ and the Vernier Scale Reading $(VSR)$ is $2$.
$Diameter = 1.3\,cm + (0.01\,cm \times 2) = 1.3\,cm + 0.02\,cm = 1.32\,cm$.

(A) $1$. Calculate the Least Count $(LC)$ of the micrometer:
$LC = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{2 \, mm}{200} = 0.01 \, mm$.
$2$. Determine the zero error from the first figure:
The reference line of the main scale coincides with the $5^{th}$ division of the circular scale. Since the zero of the circular scale is above the reference line,the zero error is positive.
Zero Error $= +5 \times LC = +5 \times 0.01 \, mm = +0.05 \, mm$.
$3$. Determine the main scale and circular scale reading from the second figure:
Main scale reading $= 3 \, mm$.
Circular scale reading $= 46^{th}$ division coincides with the reference line.
Observed reading $= \text{Main scale reading} + (\text{Circular scale reading} \times LC) = 3 \, mm + (46 \times 0.01 \, mm) = 3.46 \, mm$.
$4$. Calculate the true thickness:
True thickness $= \text{Observed reading} - \text{Zero error} = 3.46 \, mm - 0.05 \, mm = 3.41 \, mm$.

(D) The pitch of the screw gauge is the distance moved in one complete rotation.
Pitch $= \frac{0.25\, cm}{5} = 0.05\, cm$.
The least count $(LC)$ is given by $\frac{\text{Pitch}}{\text{Number of circular scale divisions}}$.
$LC = \frac{0.05\, cm}{100} = 0.0005\, cm$.
The reading is calculated as: $\text{Main Scale Reading} + (\text{Circular Scale Reading} \times LC)$.
Main Scale Reading $= 4 \times 0.05\, cm = 0.20\, cm$.
Circular Scale Reading $= 30 \times 0.0005\, cm = 0.0150\, cm$.
Total thickness $= 0.20\, cm + 0.0150\, cm = 0.2150\, cm$.

(B) The least count $(LC)$ of the Vernier callipers is given by:
$LC = \frac{\text{Value of 1 MS division}}{\text{Total number of VS divisions}} = \frac{0.1\,cm}{10} = 0.01\,cm$.
The observed diameter $d$ is calculated as: $d = \text{Main Scale Reading} + (VS \text{ division} \times LC)$.
Given zero error $= -0.03\,cm$,the correction is $-(\text{zero error}) = +0.03\,cm$.
Corrected diameter $d' = d + 0.03\,cm$.
For measurement $(1)$: $d_1 = 0.5 + (8 \times 0.01) + 0.03 = 0.5 + 0.08 + 0.03 = 0.61\,cm$.
For measurement $(2)$: $d_2 = 0.5 + (4 \times 0.01) + 0.03 = 0.5 + 0.04 + 0.03 = 0.57\,cm$.
For measurement $(3)$: $d_3 = 0.5 + (6 \times 0.01) + 0.03 = 0.5 + 0.06 + 0.03 = 0.59\,cm$.
Mean corrected diameter $= \frac{0.61 + 0.57 + 0.59}{3} = \frac{1.77}{3} = 0.59\,cm$.

(D) The least count $(L.C.)$ of a screw gauge represents the smallest length that can be measured accurately with the instrument.
Given that the least count is $0.001\, cm$,which is equivalent to $10^{-3}\, cm$.
This implies that the instrument is precise up to the third decimal place.
Therefore,any measurement taken with this screw gauge must be recorded up to $3$ decimal places to maintain the correct precision.
Among the given options,$5.320\, cm$ is the only value recorded to $3$ decimal places.

(D) Given that $N$ divisions of the main scale coincide with $(N + 1)$ divisions of the vernier scale.
Let the value of one main scale division be $a$.
Let the value of one vernier scale division be $v$.
According to the problem,$(N + 1)v = Na$.
Therefore,the value of one vernier scale division is $v = \frac{Na}{N + 1}$.
The least count of a vernier calliper is defined as the difference between one main scale division and one vernier scale division.
Least Count = $a - v$.
Substituting the value of $v$:
Least Count = $a - \frac{Na}{N + 1} = a \left( 1 - \frac{N}{N + 1} \right)$.
Least Count = $a \left( \frac{N + 1 - N}{N + 1} \right) = \frac{a}{N + 1}$.

(C) The Least Count $(LC)$ of the screw gauge is given by:
$LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5\,mm}{100} = 0.005\,mm$.
Since the zero of the circular scale lies $3$ divisions below the mean line,there is a positive zero error:
$\text{Zero Error} = +3 \times LC = 3 \times 0.005\,mm = 0.015\,mm$.
The observed reading is given by:
$\text{Observed Reading} = \text{Main Scale Reading} (MSR) + (\text{Circular Scale Reading} (CSR) \times LC)$.
$\text{Observed Reading} = 5.5\,mm + (48 \times 0.005\,mm) = 5.5\,mm + 0.240\,mm = 5.740\,mm$.
The actual thickness is calculated by subtracting the zero error from the observed reading:
$\text{Thickness} = \text{Observed Reading} - \text{Zero Error}$.
$\text{Thickness} = 5.740\,mm - 0.015\,mm = 5.725\,mm$.

(B) The least count $(LC)$ of a screw gauge is defined as the ratio of the pitch of the screw to the total number of divisions on the circular scale $(N)$.
Given:
Pitch = $1\, mm = 10^{-3}\, m$
$LC = 5\,\mu m = 5 \times 10^{-6}\, m$
Using the formula:
$LC = \frac{\text{Pitch}}{N}$
$5 \times 10^{-6} = \frac{10^{-3}}{N}$
$N = \frac{10^{-3}}{5 \times 10^{-6}}$
$N = \frac{1000}{5} = 200$
Therefore,the minimum number of divisions required is $200$.

(A) The thickness of the wall of a hollow cylinder is given by the difference between the external radius $(R)$ and the internal radius $(r)$.
$t = R - r$
Given: $R = (4.23 \pm 0.01) \, cm$ and $r = (3.87 \pm 0.01) \, cm$.
The mean value of the thickness is $t = 4.23 - 3.87 = 0.36 \, cm$.
When quantities are subtracted,the absolute errors are added.
Therefore,the absolute error in thickness is $\Delta t = \Delta R + \Delta r = 0.01 + 0.01 = 0.02 \, cm$.
Thus,the thickness of the wall is $(0.36 \pm 0.02) \, cm$.

(C) Given that the main scale has $n$ divisions per $cm$,the value of $1 \text{ MSD}$ (Main Scale Division) is $\frac{1}{n} \text{ cm}$.
According to the problem,$n \text{ VSD}$ (Vernier Scale Divisions) coincide with $(n-1) \text{ MSD}$.
Therefore,$1 \text{ VSD} = \frac{n-1}{n} \text{ MSD}$.
The least count of a vernier calliper is defined as $1 \text{ MSD} - 1 \text{ VSD}$.
Least Count $= 1 \text{ MSD} - \left( \frac{n-1}{n} \right) \text{ MSD} = \left( 1 - \frac{n-1}{n} \right) \text{ MSD} = \frac{1}{n} \text{ MSD}$.
Substituting the value of $1 \text{ MSD} = \frac{1}{n} \text{ cm}$,we get:
Least Count $= \frac{1}{n} \times \frac{1}{n} \text{ cm} = \frac{1}{n^2} \text{ cm}$.

(B) The pitch of the screw gauge is the distance moved by the screw in one complete rotation.
Given that $6$ rotations correspond to a movement of $3\; mm$ on the main scale.
Therefore,the pitch $= \frac{3\; mm}{6} = 0.5\; mm$.
The least count $(LC)$ of a screw gauge is defined as the ratio of the pitch to the total number of divisions on the circular scale.
$LC = \frac{\text{Pitch}}{\text{Number of circular scale divisions}}$.
Given the number of divisions $= 50$.
$LC = \frac{0.5\; mm}{50} = 0.01\; mm$.
Converting this to centimeters,$0.01\; mm = 0.001\; cm$.

(C) device with the smallest least count is the most precise for measuring length.
$1.$ Least count of vernier callipers: Assuming $1$ main scale division $(MSD) = 1 \; mm$,and $20$ vernier divisions $(VD)$ coincide with $19$ $MSD$,the least count $= 1 \; MSD - 1 \; VD = 1 \; mm - \frac{19}{20} \; mm = 0.05 \; mm = 0.005 \; cm$.
$2.$ Least count of screw gauge $= \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{1 \; mm}{100} = 0.01 \; mm = 0.001 \; cm$.
$3.$ Least count of an optical device $\approx$ Wavelength of light $\approx 10^{-5} \; cm = 0.00001 \; cm$.
Comparing the values: $0.00001 \; cm < 0.001 \; cm < 0.005 \; cm$.
Since the optical instrument has the smallest least count,it is the most precise device.

(N/A) Part $(a)$: Wrap the thread closely around a uniform smooth rod such that the coils are touching each other without overlapping. Measure the total length $(L)$ of the coiled portion using a metre scale. If $(n)$ is the number of turns,the diameter $(d)$ of the thread is given by $d = \frac{L}{n}$.
Part $(b)$: No,it is not possible to increase the accuracy arbitrarily. While increasing the number of divisions decreases the least count,the accuracy is limited by other factors such as the mechanical errors of the instrument,the flexibility of the screw,and the precision of the human observer.
Part $(c)$: $A$ set of $100$ measurements is more reliable because the random errors in measurement follow a statistical distribution. As the number of observations $(N)$ increases,the random error in the mean value decreases by a factor of $\frac{1}{\sqrt{N}}$. Thus,$100$ measurements provide a much smaller uncertainty compared to $5$ measurements.

(N/A) The measurement of length varies significantly depending on the scale. Below is a table summarizing the devices and methods used for different orders of magnitude:

Order of Length $(m)$	Device (Instrument)
$10^{-10}$ to $10^{-8}$	Optical/Electron Microscope
$10^{-5}$ to $10^{-4}$	Screw gauge,Spherometer,Vernier Calipers
$10^{-3}$ to $10^{2}$	Meter scale
$> 10^{2}$	Telescope,Radar,Laser,Sonar

(A) The minimum inaccuracy in a vernier instrument is equal to its least count.
Given that $50$ vernier scale divisions $(VSD)$ coincide with $49$ main scale divisions $(MSD)$.
$1 \ VSD = \frac{49}{50} \ MSD$
Least Count $(LC)$ = $1 \ MSD - 1 \ VSD$
$LC = 1 \ MSD - \frac{49}{50} \ MSD = \frac{1}{50} \ MSD$
Given that $1 \ MSD = 0.5 \ mm$.
$LC = \frac{1}{50} \times 0.5 \ mm = \frac{0.5}{50} \ mm = 0.01 \ mm$.
Therefore,the minimum inaccuracy is $0.01 \ mm$.

(B) The least count of a measuring instrument is the smallest value that can be measured by it.
$(1)$ A Traveling Microscope typically has a least count of $0.001\,cm$ (or $10\,\mu m$).
$(2)$ A standard Screw Gauge has a least count of $0.001\,cm$ (or $0.01\,mm$).
However, based on the provided options and standard laboratory equipment specifications:
- Traveling Microscope: $0.001\,cm$ (Option $b$)
- Screw Gauge: $0.001\,cm$ (Option $b$)
Given the structure of the question, the most accurate match is $(1-b)$ and $(2-b)$ if they share the same precision, or $(1-c)$ if the microscope is high-precision. Re-evaluating standard textbook values: A standard Screw Gauge is $0.001\,cm$. A Traveling Microscope is often $0.001\,cm$. If we must choose unique mappings from the provided list: $(1-b)$ and $(2-b)$ are both $0.001\,cm$. Given the options, the correct mapping is $(1-b, 2-b)$.

(D) $1$. Calculate the Least Count $(LC)$: $LC = 1\, MS D - 1\, VSD$. Given $10\, VSD = 9\, MSD$,so $1\, VSD = 0.9\, MSD = 0.9\, mm$. Thus,$LC = 1\, mm - 0.9\, mm = 0.1\, mm = 0.01\, cm$.
$2$. Calculate Zero Error: Since the zero of the vernier scale is to the right of the main scale zero,the error is positive. The $7^{th}$ division coincides,so $Zero\, Error = + (7 \times LC) = + (7 \times 0.01\, cm) = +0.07\, cm$.
$3$. Calculate Observed Reading: $Main\, Scale\, Reading (MSR) = 3.1\, cm$. $Vernier\, Scale\, Reading (VSR) = 4 \times LC = 4 \times 0.01\, cm = 0.04\, cm$. $Observed\, Reading = MSR + VSR = 3.1\, cm + 0.04\, cm = 3.14\, cm$.
$4$. Calculate Corrected Length: $Corrected\, Length = Observed\, Reading - Zero\, Error = 3.14\, cm - 0.07\, cm = 3.07\, cm$.

(D) The Least Count $(LC)$ of the screw gauge is calculated as:
$LC = \frac{\text{pitch}}{\text{number of circular scale divisions}} = \frac{0.1 \ cm}{50} = 0.002 \ cm$.
Any measurement taken with this instrument must be an integer multiple of the least count $(0.002 \ cm)$.
Checking the options:
$A) \ 2.123 / 0.002 = 1061.5$ (Not an integer)
$B) \ 2.125 / 0.002 = 1062.5$ (Not an integer)
$C) \ 2.121 / 0.002 = 1060.5$ (Not an integer)
$D) \ 2.124 / 0.002 = 1062$ (This is an integer).
Therefore,the correct measurement is $2.124 \ cm$.

(B) The least count $(LC)$ of a screw gauge is calculated as:
$LC = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
Given,$\text{Pitch} = 0.5\, mm$ and $\text{Number of divisions} = 50$.
$LC = \frac{0.5\, mm}{50} = 0.01\, mm = 10\, \mu m$.
Since the circular scale is $4$ units ahead of the pitch scale marking (the zero of the circular scale is above the reference line),the zero error is positive.
Therefore,the nature of zero error is positive and the least count is $10\, \mu m$.

(D) The least count $(L.C.)$ of a screw gauge is defined by the formula:
$L.C. = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
Given:
$L.C. = 0.01\, mm$
Number of divisions = $50$
Substituting the values into the formula:
$0.01\, mm = \frac{\text{Pitch}}{50}$
Therefore,the pitch is:
$\text{Pitch} = 0.01\, mm \times 50 = 0.5\, mm$.

(C) Spherometer is a precision instrument designed to measure the sagitta (height) of a spherical surface. By knowing the distance between the legs of the Spherometer and measuring the sagitta,we can calculate the radius of curvature $(R)$ of the curved surface using the formula $R = \frac{a^2}{6h} + \frac{h}{2}$,where $a$ is the distance between the legs and $h$ is the sagitta. Therefore,a Spherometer is used to measure the radius of curvature of the curved surface.

(D) Given that $1 \text{ Main Scale Division (MSD)} = a \ cm$.
Let $1 \text{ Vernier Scale Division (VSD)} = a' \ cm$.
According to the problem,the $n^{\text{th}}$ division of the vernier scale coincides with the $(n-1)^{\text{th}}$ division of the main scale:
$n \times a' = (n-1) \times a$
$a' = \frac{(n-1)a}{n} \ cm$.
The least count ($L$.$C$.) is defined as the difference between one main scale division and one vernier scale division:
$L.C. = 1 \text{ MSD} - 1 \text{ VSD} = (a - a') \ cm$.
Substituting the value of $a'$:
$L.C. = a - \frac{(n-1)a}{n} = \frac{na - na + a}{n} = \frac{a}{n} \ cm$.
To convert the least count into $mm$,we multiply by $10$ (since $1 \ cm = 10 \ mm$):
$L.C. = \left(\frac{a}{n}\right) \times 10 \ mm = \frac{10a}{n} \ mm$.

(B) Given,positive zero error $= 0.2\, mm = 0.02\, cm$.
Main scale reading $(MSR) = 8.5\, cm$.
Vernier scale reading $(VSR) = \text{Vernier coincidence} \times \text{Least count} = 6 \times 0.01\, cm = 0.06\, cm$.
Observed reading $= MSR + VSR = 8.5\, cm + 0.06\, cm = 8.56\, cm$.
Correct reading $= \text{Observed reading} - \text{Zero error} = 8.56\, cm - 0.02\, cm = 8.54\, cm$.

(B) Least count $(LC)$ $= \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{1\, mm}{100} = 0.01\, mm$.
Zero error $= +8 \times LC = +8 \times 0.01\, mm = +0.08\, mm$.
Observed reading $= \text{Main scale reading} + (\text{Circular scale reading} \times LC) = 1\, mm + (72 \times 0.01\, mm) = 1.72\, mm$.
True diameter $= \text{Observed reading} - \text{Zero error} = 1.72\, mm - 0.08\, mm = 1.64\, mm$.
Radius of the wire $= \frac{\text{Diameter}}{2} = \frac{1.64\, mm}{2} = 0.82\, mm$.

(C) $1$. Calculate the Least Count $(LC)$: The pitch is $0.5 \, mm$ and the number of circular scale divisions is $50$. Thus,$LC = \frac{0.5 \, mm}{50} = 0.01 \, mm$.
$2$. Determine the Zero Error: The $5^{th}$ division coincides with the reference line when the ratchet is closed. Since it is positive (implied by standard convention for this type of problem),Zero Error $= +5 \times LC = 5 \times 0.01 = 0.05 \, mm$.
$3$. Calculate the Observed Reading: Observed Reading = Main Scale Reading + (Circular Scale Division $\times LC$) = $5 \, mm + (20 \times 0.01 \, mm) = 5.20 \, mm$.
$4$. Calculate the True Reading: True Reading = Observed Reading - Zero Error = $5.20 \, mm - 0.05 \, mm = 5.15 \, mm$.

(D) Given: $9 \, {MSD} = 10 \, {VSD}$.
Since $1 \, {MSD} = 1 \, {mm}$,then $10 \, {VSD} = 9 \, {mm}$,so $1 \, {VSD} = 0.9 \, {mm}$.
Least Count $({LC})$ = $1 \, {MSD} - 1 \, {VSD} = 1 \, {mm} - 0.9 \, {mm} = 0.1 \, {mm} = 0.01 \, {cm}$.
Observed Diameter = ${MSR} + ({VSR} \times {LC}) = 10 \, {mm} + (8 \times 0.1 \, {mm}) = 10.8 \, {mm} = 1.08 \, {cm}$.
Positive zero error = $0.04 \, {cm}$.
Corrected Diameter = $\text{Observed Reading} - \text{Zero Error} = 1.08 \, {cm} - 0.04 \, {cm} = 1.04 \, {cm}$.
Radius = $\frac{\text{Diameter}}{2} = \frac{1.04 \, {cm}}{2} = 0.52 \, {cm}$.
Expressing in $10^{-2} \, {cm}$: $0.52 \, {cm} = 52 \times 10^{-2} \, {cm}$.
Thus,the value is $52$.

(D) The Least Count $(L.C.)$ of the screw gauge is calculated as:
$L.C. = \frac{\text{Pitch}}{\text{Total circular scale divisions}} = \frac{1 \, mm}{100} = 0.01 \, mm$.
Since $1 \, mm = 0.1 \, cm$,the $L.C.$ in $cm$ is:
$L.C. = 0.01 \, mm = 0.001 \, cm$.
The diameter $(D)$ is given by the formula:
$D = \text{Main scale reading} + (\text{Circular scale reading} \times L.C.)$
Substituting the given values:
$D = 0 \, mm + (52 \times 0.01 \, mm) = 0.52 \, mm$.
Converting the diameter into $cm$:
$D = \frac{0.52}{10} \, cm = 0.052 \, cm$.

(A) The least count $(LC)$ of the screw gauge is given by:
$LC = \frac{\text{Pitch}}{\text{Total circular divisions}} = \frac{0.1 \, \text{cm}}{100} = 0.001 \, \text{cm}$.
For Student $A$:
Zero error $= +5 \times LC = +0.005 \, \text{cm}$.
Observed reading $= \text{MSR} + (\text{CSR} \times LC) = 0.322 \, \text{cm}$.
Assuming the main scale reading $(MSR)$ is $0.300 \, \text{cm}$,then:
$0.300 + (\text{CSR}_A \times 0.001) - 0.005 = 0.322 \implies \text{CSR}_A \times 0.001 = 0.027 \implies \text{CSR}_A = 27$.
For Student $B$:
Zero error $= -8 \times LC = -0.008 \, \text{cm}$ (since $92$ is aligned,error is $92-100 = -8$).
Assuming the main scale reading $(MSR)$ is $0.300 \, \text{cm}$,then:
$0.300 + (\text{CSR}_B \times 0.001) - (-0.008) = 0.322 \implies \text{CSR}_B \times 0.001 = 0.014 \implies \text{CSR}_B = 14$.
Difference in circular scale readings $= |27 - 14| = 13$.

(B) The pitch of a screw gauge is defined as the distance moved by the spindle per complete rotation of the circular scale.
Given that in $5$ complete rotations,the distance moved on the main scale is $5 \, mm$.
Therefore,the pitch $= \frac{5 \, mm}{5} = 1 \, mm$.
The least count is defined as: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$.
Given total divisions $= 50$.
So,$\text{Least Count} = \frac{1 \, mm}{50} = 0.02 \, mm$.
Converting to centimeters: $0.02 \, mm = 0.002 \, cm$.
Since the calculated least count is $0.002 \, cm$ and the assertion states $0.001 \, cm$,assertion $A$ is incorrect.
Reason $R$ is the standard definition of the least count of a screw gauge,which is correct.

(D) Given that $50 \; VSD = 49 \; MSD$.
Therefore,$1 \; VSD = \frac{49}{50} \; MSD$.
The least count of the travelling microscope is defined as $LC = 1 \; MSD - 1 \; VSD$.
$LC = (1 - \frac{49}{50}) \; MSD = \frac{1}{50} \; MSD$.
Since $40$ divisions are present in $1 \; cm$ on the main scale,$1 \; MSD = \frac{1}{40} \; cm$.
Substituting this value,$LC = \frac{1}{50} \times \frac{1}{40} \; cm = \frac{1}{2000} \; cm$.
Converting to meters: $LC = \frac{1}{2000} \times 10^{-2} \; m = 0.5 \times 10^{-5} \; m$.
Expressing in the required format: $LC = 5 \times 10^{-6} \; m$.
Thus,the value is $5$.

(B) Given that $20 \; MSD = 1 \; cm$.
Therefore,$1 \; MSD = \frac{1}{20} \; cm = 0.05 \; cm = 0.5 \; mm$.
We are given that $10 \; VSD = 9 \; MSD$.
Thus,$1 \; VSD = \frac{9}{10} \; MSD = 0.9 \times 0.5 \; mm = 0.45 \; mm$.
The Vernier Constant $(VC)$ is defined as $VC = 1 \; MSD - 1 \; VSD$.
$VC = 0.5 \; mm - 0.45 \; mm = 0.05 \; mm$.
Expressing this in the form $\dots \times 10^{-2} \; mm$,we get $VC = 5 \times 10^{-2} \; mm$.
Thus,the value is $5$.