Consider a vernier calliper in which each 1cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the vernier calliper, 5 divisions of the vernier scale coincides with 4 divisions of main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

If the least count of screw gauge is 0.005 mm, then the ratio of the pitch of screw gauge to least count of the vernier calliper is 2

If the least count of screw gauge is 0.01 mm, then the ratio of pitch of the screw gauge and least count of vernier calliper is 2

If the least count of screw gauge is 0.01 mm then the ratio of least count of the linear scale of the screw gauge to least count of vernier calliper is 2

If the least count of screw gauge is 0.005 mm, then the ratio of the least count of the linear scale of the screw gauge to least count of the vernier calliper is 2

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$1 MSD = \frac{1}{8} cm; 5 VSD = 4 MSD$

$1 VSD = \frac{4}{5} MSD = \frac{4}{5} \times \frac{1}{8} = \frac{1}{10} cm$

Least count of vernier caliper $= 1 MSD - 1 VSD = \frac{1}{8} cm - \frac{1}{10} cm = 0 .025 cm$
Pitch of scres gauge $P = (L \cdot C) 100$
If $L \cdot C = 0 .005 mm$ , then $P = 0 .5 mm$
Now $\frac{P}{L \cdot C of vernier callipers} = \frac{0 .5 mm}{0 .025 cm} = 2$
So option A is correct
If L.C of screw gauge = 0.01 mm, then P = 1 mm
Now L.C of linear scale of screw gauge $= \frac{P}{2} = 0 .5 mm$
$\frac{P / 2}{L \cdot C of V \cdot C} = \frac{0 .5 mm}{0 .025 cm} = 2$
So option C is correct