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
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is : 
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
The current passing through the battery in the given circuit, is: 