To determine the diameter of the wire, we need to understand the readings given by the screw gauge:
The screw gauge uses two scales: the main scale and the circular scale.
The given data is:
According to the question, \(1\, \text{mm}\) on the main scale corresponds to \(100\) divisions on the circular scale. Therefore, each division on the circular scale corresponds to:
\(\frac{1}{100}\, \text{mm} = 0.01\, \text{mm}\)
Given the circular scale reading is \(52\) divisions, the value it represents in millimeters is:
\(52 \times 0.01\, \text{mm} = 0.52\, \text{mm}\)
Since the main scale reading is \(0\, \text{mm}\), the total reading (which is the diameter of the wire) is solely due to the circular scale:
\(0 + 0.52\, \text{mm} = 0.52\, \text{mm}\)
Converting \(0.52\, \text{mm}\) to centimeters (since the options are in cm):
\(0.52\, \text{mm} = 0.052\, \text{cm}\)
Therefore, the diameter of the wire is \(0.052\, \text{cm}\).
Thus, the correct answer is 0.052 cm.
The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10–4 is ____ kJ/m3. Assume that material is elastic up to the linear strain of 5×10–4
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The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.