A screw gauge has the pitch of 1 mm and 50 divisions on its circular scale. When an object is fitted between its studs, it reads 2 main scale divisions and 32 divisions on a circular scale. There is no zero error. The reading of screw gauge is:

2.32 mm

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2.64 mm

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1.32 mm

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1.64 mm

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Solution

The correct option is B 2.64 mm
$l e a s t c o u n t = \frac{m i n i m u m r e a d i n g o n m a i n s c a l e}{t o t a l d i v i s i o n o n v e r n i e r s c a l e}$

$2 \times 1 m m + 32 \times \frac{1 m m}{50} = 2.64 m m$

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A screw gauge has the pitch of 1 mm and 50 divisions on its circular scale. When an object is fitted between its studs, it reads 2 main scale divisions and 32 divisions on a circular scale. There is no zero error. The reading of screw gauge is:

The correct option is B 2.64 mmleast count=minimum reading on main scaletotal division on vernier scale2×1mm+32×1mm50=2.64mm

The correct option is B 2.64 mm
$l e a s t c o u n t = \frac{m i n i m u m r e a d i n g o n m a i n s c a l e}{t o t a l d i v i s i o n o n v e r n i e r s c a l e}$

$2 \times 1 m m + 32 \times \frac{1 m m}{50} = 2.64 m m$