Vernier scales
Vernier Scale:
These scales, like diagonal scales, are used to read to a very small unit with great accuracy. It consists of two parts - a primary scale and a vernier. The primary scale is a plain scale fully divided into minor divisions.
As it would be difficult to sub-divide the minor division in an ordinary way, it is done with the help of the vernier. The graduations on vernier are derived from those on the primary scale.
The figure shows a part of a plain scale in which length A-O represents 10 cm. if we divide A-O into ten equal parts, each will be of 1 cm. now it would not be easy to divide each of these parts into ten equal divisions to get measurements in millimeter.
Now if we teak a length BO equal to 10 + 1 = 11 such equal parts, thus representing 11 cm, and divide it into ten equal divisions, each of these divisions will represent 11 / 10 -1.1 cm.
The difference between one part of AO and One division of BO will be equal 1.1-1.0=0.1 cm or mm.
This difference is called the least count of the scale. Minimum this distance can be measured by this scale. The upper-scale BO is the vernier. The combination of plain scale and the vernier is vernier scale.
1. Drew a vernier scale of RF=1/25 to read centimeter up to 4 meters and on it, show length 2.39 m and 0.91 m
SOLUTION:
Length of scale=RF x max. distance
=1/25 x 4 x 100
=16 cm
CONSTRUCTION:(Main scale)
Drew a line 16 cm long. Divide it into 4 equal parts. (each will represent meter) Sub-divide each part in 10 equal parts. (each will represent decimetre) Name those properly.
CONSTRUCTION:(Vernier)
Take 11 parts of Dm length and divide it into 10 equal parts Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle covering these of vernier.
TO MEASURE GIVEN LENGTHS:
- For 2.39 m : subtract 0.99 from 2.39 i.e. 2.39 - 0.99 = 1.4 m The distance between 0.99 (left of zero) and 1.4 (right of zero) is 2.39 m
- For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91-0.11 =0.80 m The distance between 0.11 and 0.80 (both left side of zero) is 0.91 m
2. A map of size 500 cm x 50 cm wide represents an area of 6250 sq. km. Constructs a vernier scale to measure kilometer, hectometre, and decametre and long enough to measure up to 7 km. Indicate on it (a) 5.33 km (b) 59 decametres.
SOLUTION:
RF=(Area of Drawing/actual area)1/2
=(500 x 50 cm sq. / 6250 km sq.)1/2
=2/105
Length of scale=RF x max. distance
=2/ 105 x 7 km
=14 cm
CONSTRUCTION:(Main scale)
Drew a line 14 cm long. Divide it into 7 equal parts. (each will represent km) Sub-divide each part in 10 equal parts. (each will represent hectometre) Name those properly.
CONSTRUCTION:(Vernier )
Take 11 parts of the hectometre part length and divide it into 10 equal parts. Each will show 1.1 hm or 11 dm and Covering in a rectangle complete scale.
To Measure Given Length:
- For 5.33 km : subtract 0.33 from 5.33 i.e. 5.33 - 0.33 = 5.00, The distance between 33 dm (left of zero)and 5.00(right of zero) is 5.33 km
- For 59 dm : Subtract 0.99 from 0.59 i.e. 0.59 - 0.99 = -0.4 km (-ve sign means left of zero) The distance between 99 dm and -0.4 km is 59 dm(both left of zero)
