# Screw Gauge

The screw gauge is used to measure diameter of a wire or thickness of a sheet. It consists of a screw, a stud, and two scales (1) main scale and (2) circular scale, as shown in figure. The linear distance moved by the screw in one complete rotation is called its pitch ($p$). It is equal to the distance between two consecutive divisions on the main scale. The circular scale is divided into $N$ equal divisions. The linear distance moved by the screw when circular scale is rotated by one division is called least count (LC). It is given by $\mathrm{LC}=p/N$.

In the absence of zero error, both main scale reading ($\mathrm{MSR_0}$) and circular scale reading ($\mathrm{CSR_0}$) are zeros i.e., 0th mark on main scale coincides with 0th mark on the circular scale when the screw touches the stud. If any or both of $\mathrm{MSR_0}$ and $\mathrm{CSR_0}$ are non-zero then the screw gauge has zero error which is given by \begin{align} \mathrm{Zero\ Error}=\mathrm{MSR_0}+\mathrm{CSR_0}\times \mathrm{LC}.\nonumber \end{align} The zero error can be positive or negative (see figures). In case of negative zero error, the 0th mark on the main scale is barely visible.

If MSR and CSR are readings when the object is placed between the screw and the stud, then the size (diameter or thickness) of the object is given by \begin{align} D=\mathrm{MSR}+\mathrm{CSR}\times \mathrm{LC}-\mathrm{Zero\ Error}.\nonumber \end{align}

## Solved Problems on Screw Gauge

### Problem from IIT JEE 2011

The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has relative error of 2%, the percentage error in density is

1. 0.9%
2. 2.4%
3. 3.1%
4. 4.2%

Solution: The density of a ball of mass $m$ and diameter $D$ is given by \begin{align} \rho={6m}/\left(\pi D^3\right).\nonumber \end{align} Differentiate to get \begin{align} \mathrm{d}\rho=\frac{6}{\pi}\left(\frac{D^3 \mathrm{d} m -3 m D^2 \mathrm{d} D}{D^6}\right). \nonumber \end{align} Divide the second equation by the first to get \begin{align} \frac{\mathrm{d} \rho}{\rho}=\frac{\mathrm{d} m}{m}-3\frac{\mathrm{d} D}{D}. \nonumber \end{align} In error analysis, measured mass $m_\text{measured}$ and actual mass $m_\text{actual}$ are related by \begin{align} m_\text{actual}=m_\text{measured}\pm \Delta m,\nonumber \end{align} where $\Delta m$ is a small positive number representing measurement error. Let $\Delta m$ and $\Delta D$ be the measurement errors (both positive) in $m$ and $D$. From above expression, $\mathrm{d}\rho$ is maximum when $\mathrm{d}m=\Delta m$ and $\mathrm{d}D=-\Delta D$. Thus, the error in $\rho$ is \begin{align} \frac{\Delta \rho}{\rho}=\frac{\Delta m}{m}+3\frac{\Delta D}{D}.\nonumber \end{align} The least count (LC) of a screw gauge, with pitch $p$ and total number of divisions on the circular scale $N$, is given by \begin{align} &\text{LC}={p}/{N}={0.5}/{50}={0.01}\;\mathrm{mm}.\nonumber \end{align} LC is error in measurement of the ball diameter $D$ i.e., $\Delta D={0.01}\;\mathrm{mm}$. The measured diameter for the given main scale reading (MSR) and circular scale reading $(n)$ is \begin{align} &D=\text{MSR}+\left(n\right)\text{LC}=2.5+(20)0.01={2.70}\;\mathrm{mm}.\nonumber \end{align} Given $\frac{\Delta m}{m}=\frac{2}{100}=0.02$. Substitute the values to get \begin{align} &\frac{\Delta \rho}{\rho}=0.02+3\times\frac{0.01}{2.70}=0.031=3.1\%.\nonumber \end{align}

### Problem from IIT JEE 2004

The pitch of a screw gauge is 1 mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm) of the wire in appropriate number of significant figures.

Solution: The distance moved on the linear scale when circular scale makes one complete rotation is $p=1\;\mathrm{mm}$ (pitch). The number of divisions on the circular scale is $N=100$. Thus, one division on the circular scale is $\text{LC}=p/N={1}/{100}=0.01 \;\mathrm{mm}$. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is \begin{align} d&=\text{LSR}+\text{CSR}\times\text{LC}\nonumber\\ &=1+47\times 0.01=1.47 \;\mathrm{mm}=0.147\;\mathrm{cm}. \nonumber \end{align} The curved surface area of the cylinder is \begin{align} A=\pi d l=3.14(0.147)(5.6)=2.5848 \;\mathrm{cm^2}. \nonumber \end{align} Rounding off to two significant digits gives $A=2.6\;\mathrm{cm^2}$.

## Questions on Screw Gauge

Question 1: Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then,

A. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
B. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
C. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 cm.
D. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.