Question

Which of the following has the largest least count?
A. Spherometer
B. Vernier callipers
C. Screw gauge
D. Metre scale

Answer 1

Hint: Generally for normal measurements daily we will be using metre scale. If we want to measure things which are very minute we use vernier calipers and for things smaller than that we will be using screw gauge. Here we calculate what is 1 main scale division(1MSD) and what is one vernier scale division and we subtract 1VSD from 1MSD to get the least count.
Formula used:
Least count(LC) = 1MSD – 1VSD
1 cm = 10 mm

Complete answer:
Least count is the least value one can measure using an instrument. For vernier callipers we have
If It is given that N divisions of vernier scale coincide with N – 1 divisions of main scale and initially when both jaws are meeting
From this information we can understand that
$\begin{array}{rl}& N(VSD)=N-1(MSD)\\ & \Rightarrow 1VSD={\displaystyle \frac{N-1}{N}}(MSD)\\ & \therefore 1VSD=1-{\displaystyle \frac{1}{N}}(MSD)\end{array}$
We got the value of 1 VSD
Now Least count(LC) = 1MSD – 1VSD
Least count will be
$\begin{array}{rl}& 1(MSD)-[1-{\displaystyle \frac{1}{N}}](MSD)\\ & \therefore {\displaystyle \frac{1}{N}}(MSD)\end{array}$
Usually the least count of vernier calipers will be 0.1mm. since 1 cm = 10 mm the value is 0.01cm
Least count of meter scale is known to all which will be 1mm
Least count of screw gauge and spherometer will be 0.01mm
Hence the measuring tool which has the highest value of least count is meter scale.

Hence option D will be the answer.

Note:
For screw gauge the least count will be measured by using different formulas. It is the ratio of pitch of the screw gauge and the number of divisions on the circular scale. For the meter scale it is clearly visible on the scale itself that the least value one can measure is 1 millimeter. We had given the clear procedure of how to determine the least count of vernier calipers.