Question

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

• A. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm
• B. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.01$ mm
• C. If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm
• D. If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm

Answer 1

Answer: (A), (D)

If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is $0.005$ mm

For vernier,

$5VSD=4MSD4\times \frac{1}{8}\text{cm}$

$\Rightarrow 5VSD=5mm$

$\Rightarrow 1VSD=1mm$

Least count= $1MSD–1VSD$

= $\frac{10}{8}\text{mm}-1\text{mm}$

$\Rightarrow$ Least count $=\frac{1}{4}mm$

(B) Now, Pitch = $2\times \frac{1}{4}\text{mm}$

Least count = $\frac{\text{Pitch}}{\text{CSD}}=\frac{1}{2\times 100}=0.005\text{mm}$

(C) When least count of linear scale = $2\times \frac{1}{4}\text{mm}$

$\Rightarrow \text{Pitch}=2\times \frac{1}{2}=1\text{mm}$

Now, least count = $\frac{1}{100}=0.01\text{mm}$