Question 265: Units , Dimensions and Error Analysis

Question

The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

Accepted Answer

Least count of Vernier calipers (L.C) = $$\left( {1 - {9 \over {10}}} ight)0.1$$ = 0.01 cm
We know that main scale reading (MSR) is the first reading on the main scale immediately to
the left of the zero of the Vernier scale. But there are
no marks on the main scale before zero of the Vernier
scale. We claim that MSR = Ã¢ÂÂ0.1 cm. The
Vernier scale reading is VSR = 6 and the least count
is LC = 0.01 cm.
Substitute these values to get,
Zero Error = MSR + VSR $$ 	imes $$ LC
= -0.1 + 6 $$ 	imes $$ 0.01 = -0.04 cm
Now in the second figure, the reading from main scale is 3.1 cm will be added to 1
st
matching
division of vernier so,
Reading = 3.1 cm + 1 $$ 	imes $$ L.C
= 3.1 cm + 0.01
= 3.11 cm
So correct diameter of the sphere
= 3.11 - (Zero Error)
= (3.11 + 0.04) cm
= 3.15 cm

Answer

3.07 cm

Answer

3.11 cm

Answer

3.15 cm

Answer

3.17 cm

Units , Dimensions and Error Analysis

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