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JEE Main
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Physics
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Experimental Physics

JEE Main · Physics · Experimental Physics

The figure shows two diagrams of the same screw gauge. In the first...

The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second case a small ball is kept between the jaws for which the diameter is required to be measured. The number of circular divisions on the shown screw gauge is 5050. It moves 0.5mm0 . 5mm on the main scale for one complete rotation and the main scale has 12mm\dfrac{1}{2}mm marks. The diameter of the ball is

The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second

The figure shows two diagrams of the same screw gauge. In the first case, nothing is kept in its jaw and in the second

Options

  1. A2.252.25 mm
  2. B2.202.20 mm
  3. C1.201.20 mm
  4. D1.251.25 mm

Correct answer

C. 1.201.20 mm

Step-by-step solution

Pitch=0.5 mm\text{Pitch} = \text{0.5 mm}

L.C=0.550\text{L.C} = \dfrac{\text{0.5}}{\text{50}}

=0.01 mm= \text{0.01 mm}

M.S.R = 2\text{M.S.R = 2}

Neck reading = 25

d=2×(0.5)+25×0.01\text{d} = 2 \times \left(\text{0.5}\right) + \text{25} \times \text{0.01}

=1.25 mm= \text{1.25 mm}

Considering zero error

Here zero error is

=+5×0.01= + 5 \times \text{0.01}

=+0.05 mm= + \text{0.05 mm}

so actual diameter

=d−0.05= \text{d} - \text{0.05}

=1.20 mm= \text{1.20 mm}
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