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 -2-45-50 cmii Screw gauge having pitch 0-5 mm and its circular scale has 50 divisionsA- Least count of screw gauge is lesser than that of Vernier callipers-B- Both the instrument have same least countC- Both the instrument have same least count but screw gauge can be more precise-D- Both the instrument have different least count and screw gauge is more precise-

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Question

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

Both the instrument have same least count

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Least count of screw gauge is lesser than that of Vernier callipers.

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Both the instrument have same least count but screw gauge can be more precise.

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Both the instrument have different least count and screw gauge is more precise.

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Solution

The correct option is C Both the instrument have same least count but screw gauge can be more precise.
L.C of vernier =M.S.D – V.S.D $= [5 - 4.9] 10^{- 4} m = 0.01 m m$

L.C. of screw gauge =0.01mm

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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 =2.4550 cm (ii) Screw gauge having pitch 0.5 mm and its circular scale has 50 divisions

The correct option is C Both the instrument have same least count but screw gauge can be more precise.L.C of vernier =M.S.D – V.S.D =[5−4.9]10−4 m=0.01 mm L.C. of screw gauge =0.01mm

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

The correct option is C Both the instrument have same least count but screw gauge can be more precise.
L.C of vernier =M.S.D – V.S.D $= [5 - 4.9] 10^{- 4} m = 0.01 m m$

L.C. of screw gauge =0.01mm