Question

15. In a vernier callipers each main scale division is of
one mm. Now if 10 vernier scale divisions coincide
with 9 main scale divisions then the least count of
the vernier callipers is​

Answer 1

Given :

In a vernier callipers each main scale division is of 1mm.

10 vernier scale divisions coincide with 9 main scale divisions$.$

To Find :

Least count of the vernier callipers.

Solution :

As given that : 10 VSD = 9 MSD

• VSD = Vernier scale division
• MSD = Main scale division

∴ 1 VSD = (9/10) MSD .......... (I)

Least count of vernier callipers is given by

➙ LC = least count = 1 MSD - 1 VSD

From the first equation;

➙ LC = 1 MSD - (9/10) MSD

➙ LC = [(10 - 9) / 10] MSD

➙ LC = 1/10 MSD

We are given that, 1 MSD = 1mm

➙ LC = 1/10 × 1mm

➙ LC = 0.1 mm

Knowledge BoosteR :

• Least count errors are random errors but within a limited size; they occur with both random and systematic errors.
• The accuracy of measurement is related to the systematic errors but its precision is related to the random errors, which include least count error also.

Answer 2

$\huge{\underline{\underline {\mathtt{\purple{Hlw}\pink{Mate}}}}}$

$\Large\bold\green{Given:}$

Let,

VSD = Vernier scale division

MSD = Main scale division

• 10 vernier scale divisions coincide with 9 main scale divisions. [10 VSD = 9 MSD]

• Each main scale division is of 1mm.

$\Large\bold\green{To\:Find:}$

Least count of the vernier callipers.

$\Large\bold\green{Solution :}$

Given value: 10 VSD = 9 MSD

∴ 1 VSD = (9/10) MSD be 1st equation

Least count of vernier callipers is given by

➙ LC = least count = 1 MSD - 1 VSD

From the first equation;

➙ LC = 1 MSD - (9/10) MSD

➙ LC = [(10 - 9) / 10] MSD

➙ LC = 1/10 MSD

From given value that, 1 MSD = 1mm

➙ LC = 1/10 × 1mm

➙ LC = 0.1 mm

$\Large\fbox\red{Mark\:As\:Brainliest}$