Least count of a vernier calliper is cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Least count of a vernier calliper is $\frac{1}{20 N}$ cm. The value of one division on the main scale is 1 mm. Then the number of divisions of main scale that coincide with N divisions of vernier scale is [2024]

1 $(\frac{2 N - 1}{2 N})$

2 $(2 N - 1)$

3 $(\frac{2 N - 1}{2})$

4 $(\frac{2 N - 1}{20 N})$

Ans.
(3)

$Least count of vernier callipers = \frac{1}{20 N} cm$

   $∵$ $Least count = 1$ $M S D - 1$ $V S D$

   $Let x number of divisions of the main scale coincides with N divisions of the vernier scale, then$

   $1$ $V S D = \frac{x \times 1 mm}{N}$

$∴ \frac{1}{20 N} c m = 1 m m - \frac{x \times 1 m m}{N}$

$\frac{1}{2 N}$ $m m = 1$ $m m - \frac{x}{N} m m$

   $x = (1 - \frac{1}{2 N}) N = \frac{2 N - 1}{2}$

Want Us to Email you About Special Offers & Updates?