The picture shown below a picture of vernier screw gauge. Main scale or pitch is 0.5 mm . There are 50 divisions on circular scale. 9 circular scale divisions correspond to 10 vernier scale divisions. What is the reading R (in μm ) ?

Question

The picture shown below a picture of vernier screw gauge. Main scale or pitch is 0.5 mm . There are 50 divisions on circular scale. 9 circular scale divisions correspond to 10 vernier scale divisions. What is the reading R (in μm ) ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jf0sl6ykuwi65y2u.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

10VSd=9CSd 2VSd=0.9CSd× 2 Reading =5.5 mm+((. 30 - 2 × 0 . 9 .) CSd/50 CSd)× 0.5 mm =5.5+0.282 mm =5.782 mm

Q. The picture shown below a picture of vernier screw gauge. Main scale or pitch is $0.5 mm$ . There are $50$ divisions on circular scale. $9$ circular scale divisions correspond to $10$ vernier scale divisions. What is the reading $R$ (in $μ m$ ) ?

$10 V S d = 9 CS d$
$2 V S d = 0.9 CS d \times 2$
Reading
$= 5.5 mm + \frac{( 30 - 2 \times 0.9 ) CS d}{50 CS d} \times 0.5 mm$
$= 5.5 + 0.282 mm$
$= 5.782 mm$

Q. The picture shown below a picture of vernier screw gauge. Main scale or pitch is 0.5mm . There are 50 divisions on circular scale. 9 circular scale divisions correspond to 10 vernier scale divisions. What is the reading R (in μm ) ?

10VSd=9CSd 2VSd=0.9CSd×2 Reading =5.5mm+50CSd(30−2×0.9)CSd​×0.5mm =5.5+0.282mm =5.782mm

Q. The picture shown below a picture of vernier screw gauge. Main scale or pitch is $0.5 mm$ . There are $50$ divisions on circular scale. $9$ circular scale divisions correspond to $10$ vernier scale divisions. What is the reading $R$ (in $μ m$ ) ?

$10 V S d = 9 CS d$
$2 V S d = 0.9 CS d \times 2$
Reading
$= 5.5 mm + \frac{( 30 - 2 \times 0.9 ) CS d}{50 CS d} \times 0.5 mm$
$= 5.5 + 0.282 mm$
$= 5.782 mm$