JEE Vernier callipers

Classical Mechanics Level 2

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

5.136 5.112 5.148 5.124

Anand Raj by Anand Raj , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Anand Raj
Apr 5, 2014

Main scale division (s) = .05 cm

Vernier scale division (v) = .049

Least count = .05 – .049 = .001 cm

Diameter: 5.10 + 24 × .001

= 5.124 cm

34 Helpful 1 Interesting 0 Brilliant 0 Confused

Why main scale division become .050cm

Deepa Bisht - 5 years, 2 months ago

Log in to reply

here, we consider the given msr to be adjacent divisions ie. the 5.10 and 5.15 , so msd=5.15-5.10=0.05 cm

Anne Bolt - 4 years, 11 months ago

Why vsd become .49

Vivek Teja - 4 years, 11 months ago

Log in to reply

Here 1 VSD =2.45/50 since total length is 2.45 and there are 50 divisions .

Manish Kumar - 4 years, 2 months ago

hey!Can you just elaborate your solution?It's not easy to understand.How you got vernier scale division as 0.049??

Manasi Singh - 2 years, 9 months ago

Log in to reply

Since 50 VSD = 2.45cm, Thus 1VSD = 0.049cm

Anand Raj - 2 years, 6 months ago

Log in to reply

Oh,ok!Thanks.

Manasi Singh - 2 years, 4 months ago

BRO CAN YOU GIVE MORE JEE QUESTIONS U SOLVED AND WITH THEIR SOLUTIONS OF YOURS PLZ. I CAN UNDERSTAND UR SOLUTIONS SO CLEARY THX BRO .

dev father - 9 months, 4 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...