A student measured the length of a rod and wrote it as 3.50 cm . Which instrument did he used to measure it? (1) a meter scale with least count 0.1 cm (2) a vernier calliper where the 10 division in vernier scale matches with 9 division in main scale and main scale has 10 division in 1 cm (3) a screw gauge having 100 division in the circular scale and pitch as 1 mm (4) a screw gauge having 50 divisions in the circular scale and pitch as 1 mm Explain.

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Q. Let

A

represents the least count of a vernier caliper whose 10 main scale divisions are equal in length to 12 vernier scale divisions, and one main scale division is equal to 1 mm. Say

B

represents the least count of a screw gauge with 25 circular scale divisions and a pitch of 1 mm,

A - B

in mm is

Q. In a screw gauge, the main scale division is

1 mm

and there are

100

divisions on circular scale. The least count of screw gauge is (Pitch

= 1 mm

)

Q. On the basis of details given about two measuring instruments, select the correct statement .

(i) Vernier callipers having main scale division=0.05 cm and Vernier scale division

= \frac{2.45}{50}

cm

(ii) Screw gauge having pitch 0.5 mm and its circular scale has 50 divisions

Q. Consider a Vernier calipers 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:

Consider a Vernier calipers in which each $1 c m$ 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

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