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. [2015]

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: [2015]

Smart Achievers · Accepted Answer

(2, 3)

Vernier callipers

$1 MSD = \frac{1 cm}{8} = 0.125 cm$

$5 VSD = 4 MSD$ $∴ 1 VSD = \frac{4}{5} \times 0.125 = 0.1 cm$

$L.C. = 1 MSD - 1 VSD = 0.125 cm - 0.1 cm = 0.025 cm$

Screw gauge

If the pitch of screw gauge is twice the L.C. of vernier callipers, then pitch = $2 \times$ L.C. of vernier calliper $= 2 \times 0.025 = 0.05 cm$

$L.C. of screw gauge = \frac{Pitch}{Total no. of divisions of circular scale}$

$= \frac{0.05}{100} cm = 0.0005 cm = 0.005 mm$

Now if the least count of the linear scale of the screw gauge is twice the least count of vernier callipers,

then L.C. of linear scale of screw gauge $= 2 \times 0.025 = 0.05 cm$

$Then pitch = 2 \times 0.05 = 0.1 cm$

$∴$ $L.C. of screw gauge = \frac{0.1}{100} cm = 0.001 cm = 0.01 mm$

Solution

1 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

2 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

3 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

4 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

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