Question

Question: The pit and the number of divisions on the circular scale, for a given Screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean lines. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 M and 48 respectively, the thickness of its sheet is:
a) 5.755 mm b) 5.725 mm c) 5.740 mm d) 5.950 mm

Answer 1

Hint In this type of problem related to screw gauge, the least count plays the most important role in the solution . At first, we find the least count for the given question. Mainly, the least count of the screw gauge is 0.01 mm but here it will be calculated according to the value of pitch given in the question.
Least count: $=\dfrac{\left( \text{pitch} \right)}{\left( \text{total number of divisions on circular scale} \right)}$
Now we would find the error present in the question, and hence find the thickness using the formula:-
 Reading MSRCSR$-\left( +\text{ve}\ \text{error} \right)$

Complete step-by-step solution
The screw gauge is an instrument used for measuring exactly the diameter of a thin wire or the width of a sheet of metal. It comprises a U-shaped mount which is fixed with a screwed pin which is fixed to a thimble.
Working principle micrometer screw gauge works on the simple principle of converting small distances into larger ones by measuring the rotation of the screw. This “screw”
Principal facilitates reading of smaller distances on a scale after amplifying them. To simplify it further, we can take in a normal screw with threads.
There are two parameters used in every screw gauge. They are pitch and least count of screw gauge.
*Pitch: Pitch of the screw gauge is defined as the distance moved by the spindle per revolution which is measured by moving the head scale over the pitch scale in order to complete one rotation.
Pitch of the screw gauge \[\text{=}\dfrac{\left( \text{distance moved by the screw} \right)}{\left( \text{number of rotations given} \right)}\]
*Least count: The Least count of screw is defined as the distance moved by the tip of the screw when turned through one division of the head scale.
 Least count(LC) of screw gauge $=\dfrac{\left( \text{pitch} \right)}{\left( \text{total number of divisions on circular scale} \right)}$
     $=\dfrac{\left( 1\ \text{mm} \right)}{\left( 100 \right)}$
     $=0.01\ \text{mm}$
Given in equation:- Pitch $=0.5\ \text{mm}$;and number of divisions 100.
So, Least count(LC) $=\dfrac{\text{Pitch}}{\text{No of division}}$
       $=\dfrac{0.5\ \text{mm}}{100}$
                               $=0.5\times {{10}^{-2}}\ \text{mm}$
It's given that the zero of its circular scale is ‘3’ divisions below its mean line.
So positive error\[\text{=}\ \text{3}\times \text{0}\text{.5}\times \text{1}{{\text{0}}^{-\text{2}}}\ \text{mm}\]
  \[\text{=}\ \text{1}\text{.5}\times \text{1}{{\text{0}}^{-\text{2}}}\ \text{mm}\]
  $=\ 0.015\ \text{mm}$
Now, Main scale reading (M.S.R) $=\ 5.5\ \text{mm}$ and, circular scale reading (C.S.R)\[=\ \left( \text{48}\times \text{0}\text{.5}\times \text{1}{{\text{0}}^{-\text{2}}} \right)\text{ mm}\]
[C.S.R) given reading L.C ]
$\therefore $ Reading $=$MSR$+$CSR$-\left( +\text{ve}\ \text{error} \right)$
      \[\text{=}\ \text{5}\text{.5 mm + }\left( \text{48}\times \text{0}\text{.5}\times \text{1}{{\text{0}}^{-\text{2}}}\text{mm} \right)\ -\text{0}\text{.015}\ \text{mm }\]
           \[\text{=}\ \text{5}\text{.5 mm +}\ \text{0}\text{.24}\ \text{mm}\ -\text{0}\text{.015}\ \text{mm }\]
     \[\text{=}\ \text{5}\text{.725}\ \text{mm}\]

Note In this type of question the least count is very important to solve the question further. Least count of the screw gauge\[\text{=}\ \text{0}\text{.01}\ \text{mm}\]
Thickness of the sheet can be calculated using the formula:-
Reading (Main Scale Reading)(Circular Scale Reading) $-\left( +\text{ve}\ \text{error} \right)$