Q51 / 75 · JEE Main 2026 (04 April Shift 1) · Physics
In a screw gauge when the circular scale is given five complete rotations it moves linearly by $2.5$ mm. If the circular scale has $100$ divisions, the least count of screw gauge is _____ mm.
- A$1 \times 10^{-2}$
- B$1 \times 10^{-3}$
- C$5 \times 10^{-2}$
- Correct answerD$5 \times 10^{-3}$
Solution
Pitch of the screw gauge is the linear distance moved in one complete rotation.
$\text{Pitch} = \dfrac{2.5}{5} = 0.5 \text{ mm}$
Least count of the screw gauge is given by the ratio of the pitch to the total number of divisions on the circular scale.
$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
$\text{Least count} = \dfrac{0.5}{100} = 0.005 \text{ mm}$
$\text{Least count} = 5 \times 10^{-3} \text{ mm}$
Answer: $5 \times 10^{-3}$
$\text{Pitch} = \dfrac{2.5}{5} = 0.5 \text{ mm}$
Least count of the screw gauge is given by the ratio of the pitch to the total number of divisions on the circular scale.
$\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$
$\text{Least count} = \dfrac{0.5}{100} = 0.005 \text{ mm}$
$\text{Least count} = 5 \times 10^{-3} \text{ mm}$
Answer: $5 \times 10^{-3}$
Hence, the correct answer is option: 'D'.