Question

vernier calliper has 20 divisions or vernier scale and its M.S.D is 0.5 mm. when a hollow cylinder is held by internal jaws the M.S.R. and V.C.D. of the caliper are 1.2 cm and 10, respectively find the radius of the cross section of the cylinder.
If the caliper used in the above problem is faulty and the positive zero error coinciding division is 2, then find out the actual radius of the cylinder.




please somebody give the answer.​

Answer 1

Given,

• Main Scale Reading (MSR) = 1.2 cm

• VCD = 10

• Main Scale Reading (MSD) = 0.5 mm = 0.05 cm

• Vernier Caliper Divisions = 20

$\boxed{\bold{Radius \:of \:Cylinder = \frac{(MSR\times VCD)+MSD}{Number \:of \:Divisions}}}$

$\bold{Radius = \frac{(1.2\times 10)+0.05}{20}}$

$\boxed{\bold{\therefore{Radius=0.6025\:cm}}}$

Answer 2

Answer:

Given:

1 M.S.D =. 0.5 mm=0.05 cm

V.C.D = 10

M.S.R= 1.2 cm

Solution:

L.C= 1 M.S.D/ No.Of V.C.D

= 0.05 cm/ 20

= 0.0025 cm

Diameter= M.S.R + (V.S.R*L.C)

= 1.2 cm + 10* 0.0025 cm

= 1.2 cm+ 0.025 cm

= 1.225 cm

Radius= Diameter/2

= 1.225 cm/2

= 0.6125 cm