The main scale of a vernier caliper reads 10mm in 10 divisions. 10 divisions of vernier scale coincide with 9 divisions of the main scale. When the two jaws of the calipers touch each other, the fifth division of the vernier coincides with a main scale division and the zero of the vernier is to the right of zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly to the left of 3.2cm and the fourth vernier division coincides with the main scale division. The diameter of the cylinder is,

Question

The main scale of a vernier caliper reads 10mm in 10 divisions. 10 divisions of vernier scale coincide with 9 divisions of the main scale. When the two jaws of the calipers touch each other, the fifth division of the vernier coincides with a main scale division and the zero of the vernier is to the right of zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly to the left of 3.2cm and the fourth vernier division coincides with the main scale division. The diameter of the cylinder is, (A) 3.09cm (B) 3.14cm (C) 3.19cm (D) none of these

Tardigrade Admin · Accepted Answer

LC=(1/10)mm=0.01cm and e=VSD× LC=0.05cm Therefore, TSR=MSR+(.VSD× LC.)-e ⇒ TSR=3.1+(.4× 0.01.)-0.05 ⇒ TSR=3.14-0.05=3.09cm

$3.09 c m$

$3.14 c m$

$3.19 c m$

none of these

$L C = \frac{1}{10} mm = 0.01 c m$
and $e = V S D \times L C = 0.05 c m$
Therefore,
$TSR = MSR + (V S D \times L C) - e$
$\Rightarrow TSR = 3.1 + (4 \times 0.01) - 0.05$
$\Rightarrow TSR = 3.14 - 0.05 = 3.09 c m$

3.09cm

3.14cm

3.19cm

none of these

LC=101​mm=0.01cm and e=VSD×LC=0.05cm Therefore, TSR=MSR+(VSD×LC)−e ⇒TSR=3.1+(4×0.01)−0.05 ⇒TSR=3.14−0.05=3.09cm

$3.09 c m$

$3.14 c m$

$3.19 c m$

$L C = \frac{1}{10} mm = 0.01 c m$
and $e = V S D \times L C = 0.05 c m$
Therefore,
$TSR = MSR + (V S D \times L C) - e$
$\Rightarrow TSR = 3.1 + (4 \times 0.01) - 0.05$
$\Rightarrow TSR = 3.14 - 0.05 = 3.09 c m$