Q46 / 54 · JEE Advanced 2026 (Paper 2) · Physics
Two thin wires, Wire-1 of diameter $0.650$ mm and Wire-2 of unknown diameter $d$ are given. To obtain the value of $d$, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of $0.5$ mm and there are $100$ divisions on the circular scale (CS). The smallest division on the linear scale (LS) is $0.5$ mm. The table shows the readings of LS and CS for the measurements. The value of $d$ (in $\mu$m) is:
Readings LS (mm) CS Wire-1 $0.5$ $42$ Wire-2 $1.5$ $95$
| Readings | ||
|---|---|---|
| LS (mm) | CS | |
| Wire-1 | $0.5$ | $42$ |
| Wire-2 | $1.5$ | $95$ |
Correct answer: 1915
Solution
The least count (LC) of the screw gauge is given by:
$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on CS}} = \dfrac{0.5}{100} = 0.005 \text{ mm}$
For Wire-1, the measured reading is:
$\text{Reading}_1 = \text{LS} + \text{CS} \times \text{LC}$
$\text{Reading}_1 = 0.5 + 42 \times 0.005 = 0.5 + 0.210 = 0.710 \text{ mm}$
The actual diameter of Wire-1 is $0.650 \text{ mm}$. The zero error of the screw gauge is:
$\text{Zero Error} = \text{Measured Reading} - \text{Actual Value} = 0.710 - 0.650 = +0.060 \text{ mm}$
For Wire-2, the measured reading is:
$\text{Reading}_2 = \text{LS} + \text{CS} \times \text{LC}$
$\text{Reading}_2 = 1.5 + 95 \times 0.005 = 1.5 + 0.475 = 1.975 \text{ mm}$
The actual diameter $d$ of Wire-2 is obtained by subtracting the zero error from its measured reading:
$d = \text{Reading}_2 - \text{Zero Error} = 1.975 - 0.060 = 1.915 \text{ mm}$
Converting the diameter into micrometers ($\mu$m):
$d = 1.915 \times 1000 \text{ } \mu\text{m} = 1915 \text{ } \mu\text{m}$
Answer: $1915$
$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on CS}} = \dfrac{0.5}{100} = 0.005 \text{ mm}$
For Wire-1, the measured reading is:
$\text{Reading}_1 = \text{LS} + \text{CS} \times \text{LC}$
$\text{Reading}_1 = 0.5 + 42 \times 0.005 = 0.5 + 0.210 = 0.710 \text{ mm}$
The actual diameter of Wire-1 is $0.650 \text{ mm}$. The zero error of the screw gauge is:
$\text{Zero Error} = \text{Measured Reading} - \text{Actual Value} = 0.710 - 0.650 = +0.060 \text{ mm}$
For Wire-2, the measured reading is:
$\text{Reading}_2 = \text{LS} + \text{CS} \times \text{LC}$
$\text{Reading}_2 = 1.5 + 95 \times 0.005 = 1.5 + 0.475 = 1.975 \text{ mm}$
The actual diameter $d$ of Wire-2 is obtained by subtracting the zero error from its measured reading:
$d = \text{Reading}_2 - \text{Zero Error} = 1.975 - 0.060 = 1.915 \text{ mm}$
Converting the diameter into micrometers ($\mu$m):
$d = 1.915 \times 1000 \text{ } \mu\text{m} = 1915 \text{ } \mu\text{m}$
Answer: $1915$
Hence, the correct answer is 1915.