A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the $45^{\text{th}}$ division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the $25^{\text{th}}$ division coincides with the main scale line?

JEE Main · 2016 · 03 Apr 2016 · Offline · Units & Measurement

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the

4 5^{th}

division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the

2 5^{th}

division coincides with the main scale line?

(A)0.80 mm✓ Correct
(B)0.70 mm
(C)0.50 mm
(D)0.75 mm

Answer: (A)

Solution

The least count is

\frac{pitch}{divisions} = \frac{0.5}{50} = 0.01

mm. With the jaws closed the 45th division lines up instead of the zero, and as the main-scale zero is only just visible this is a negative zero error of

- (50 - 45) = - 5

divisions, i.e.

- 0.05

mm. The raw reading is

0.5 + 25 \times 0.01 = 0.75

mm. Correcting, the true thickness is the observed reading minus the zero error,

0.75 - (- 0.05) = 0.80

mm.

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option

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JEE Main · 2016 · 03 Apr 2016 · Offline · Units & Measurement

Solution

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