A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively :

Question

A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5 mm is noticed on the pitch scale. The nature of zero error involved, and the least count of the screw gauge, are respectively : (A) Negative, 2 μ m (B) Positive, 10 μ m (C) Positive, 0.1 μ m (D) Positive, 0.1 mm

Tardigrade Admin · Accepted Answer

Least count of screw gauge =( text Pitch / text no. of division on circular scale ) =(0.5/50) mm =1 × 10-5 m =10 μ m Zero error in positive

Negative, $2 μ m$

Positive, $10 μ m$

Positive, $0.1 μ m$

Positive, $0.1 mm$

Least count of screw gauge
$= \frac{Pitch}{no. of division on circular scale}$
$= \frac{0.5}{50} mm = 1 \times 1 0^{- 5} m$
$= 10 μ m$
Zero error in positive

Negative, 2μm

Positive, 10μm

Positive, 0.1μm

Positive, 0.1mm

Least count of screw gauge = no. of division on circular scale Pitch ​ =500.5​mm=1×10−5m =10μm Zero error in positive

Negative, $2 μ m$

Positive, $10 μ m$

Positive, $0.1 μ m$

Positive, $0.1 mm$

Least count of screw gauge
$= \frac{Pitch}{no. of division on circular scale}$
$= \frac{0.5}{50} mm = 1 \times 1 0^{- 5} m$
$= 10 μ m$
Zero error in positive