A screw gauge with a pitch of 0.5mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement it is found that when the two jaws of the screw gauge are brought in contact the 45th division concides with the main scale line and that the zero of the main scale is barely visible. what is the thickness of the sheet if the main scale reading is 0.5mm and the 25th division coincides with the main scale line?
$$LC=\dfrac { \text{pitch} }{ \text{circular scale divisions} } =\dfrac { 0.5 }{ 50 } mm=0.01mm$$
Negative zero error = $$-5\times (0.01)mm=-0.05mm$$
Measured value = $$0.5mm+{ 25×0.01–(–0.05) }mm=0.8mm$$
A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is :
A student measured the diameter of a small steel ball using a screw gauge of least count $$0.001$$ cm. The main scale reading is $$5$$mm and zero of circular scale division coincides with $$25$$ divisions above the reference level. If screw gauge has a zero error of $$-0.004$$cm, the correct diameter of the ball is?
Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
The thimble of screw gauge has 50 divisions. The spindle advances 1mm when the screw is turned through two revolutions. (i) What is the pitch of screw gauge? (ii) What is the least count of the screw gauge?
The pitch of a screw gauge is 1mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2mm and $$45^{th}$$ mark on circular scale coincides with the base line, Find: (i) The lease count, (ii) The diameter of the wire
A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two rotations. When the flat end of the screw is in contact with the stud, the zero of circular scale lies below the base line and $$4^{th}$$ division of circular scale is in line with the base lime. Find (i) the pitch, (ii) the least count and (iii) the zero error, of the screw gauge.
Figure below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on main scale when circular head is rotated once. Find: (i) pitch of the screw gauge, (ii) Least count of the screw gauge, and (iii) The diameter of the wire.