Question

Student $A$ and Student $B$ used two screw gauges of equal pitch and $100$ circular divisions each, to measure the radius of a given wire. The actual value of the radius of the wire is $0.322\text{cm}.$ The absolute value of the difference between the final circular scale readings observed by the students $A$ and $B$ is
[Figure shows position of reference ${}^{\prime }{O}^{\prime }$ when jaws of screw gauge are closed]
Given pitch = $0.1\text{cm}.$

• A. 13.0
• B. 13.00
• C. 13

Answer 1

Answer: (A), (B), (C)

The least count of each screw gauge is,

$LC=\frac{\text{pitch}}{\text{total divisions on circular scale}}$

$\therefore LC=\frac{0.1}{100}=0.001\text{cm}$

For $\text{A}$, the error is on the positive side, so,

$\text{Actual value}=\text{Reading}-\text{Error}$

$\Rightarrow \text{Reading}=\text{Actual value}+\text{Error}$

$MSR+CSR=0.322+5\times 0.001$
$0.300+CSR=0.327$
$CSR=0.027\text{cm}$

For $\text{B}$ the error is on the negative side, so,

$\text{Actual value}=\text{Reading}+\text{Error}$

$\Rightarrow \text{Reading}=\text{Actual value}-\text{Error}$

$0.300+CSR=0.322-0.008$
$CSR=0.014\text{cm}$

Difference in $CSR=0.027-0.014=0.013\text{cm}$

Division on circular scale $=\frac{0.013}{0.001}=13$