The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

Question

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is : (A) 5.755 m (B) 5.725 mm (C) 5.740 m (D) 5.950 mm

Tardigrade Admin · Accepted Answer

LC = ( textPitch/ textNO textof textdivision) LC = 0.5 ×10-2 mm +ve error = 3 × 0.5 × 10-2 mm = 1.5 × 10-2 mm = 0.015 mm Reading = MSR + CSR - (+ve error) = 5.5 mm + (48 × 0.5 × 10-2) - 0.015 = 5.5 + 0.24 - 0.015 = 5.725 mm

$5.755 m$

$5.725 mm$

$5.740 m$

$5.950 mm$

$L C = \frac{Pitch}{NO of division}$
$L C = 0.5 \times 1 0^{- 2} mm$
+ve error = $3 \times 0.5 \times 1 0^{- 2} mm$
$= 1.5 \times 1 0^{- 2} mm = 0.015 mm$
Reading = MSR + CSR - (+ve error)
$= 5.5 mm + (48 \times 0.5 \times 1 0^{- 2}) - 0.015$
= $5.5 + 0.24 - 0.015 = 5.725 mm$

5.755m

5.725mm

5.740m

5.950mm

LC=NOofdivisionPitch​ LC=0.5×10−2mm +ve error = 3×0.5×10−2mm =1.5×10−2mm=0.015mm Reading = MSR + CSR - (+ve error) =5.5mm+(48×0.5×10−2)−0.015 = 5.5+0.24−0.015=5.725mm

$5.755 m$

$5.725 mm$

$5.740 m$

$5.950 mm$

$L C = \frac{Pitch}{NO of division}$
$L C = 0.5 \times 1 0^{- 2} mm$
+ve error = $3 \times 0.5 \times 1 0^{- 2} mm$
$= 1.5 \times 1 0^{- 2} mm = 0.015 mm$
Reading = MSR + CSR - (+ve error)
$= 5.5 mm + (48 \times 0.5 \times 1 0^{- 2}) - 0.015$
= $5.5 + 0.24 - 0.015 = 5.725 mm$