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  • Can someone help me with this, A screw gauge with a pitch of 0.5 mm 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

A screw gauge with a pitch of 0.5 mm 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 coincides 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 0.5 mm and the 25th division coincides with  the main scale line

  • Option 1)

    0.75 mm

  • Option 2)

     0.80 mm

  • Option 3)

     0.70 mm

  • Option 4)

     0.50 mm

 

Answers (1)

best_answer

As we learnt in 

To measure the thickness of the given sheet using screw gauge -

 

1.    Note the number of divisions on the circular scale.

2.    Give five complete rotations to the screw.

3.    Note the linear distance moved by the screw.

4.    Find the pitch and L.C. of screw gauge.

5.    Find the zero error and zero correction by moving the screw only in one direction in such a way that studs A and B just touch each other.

6.    Now grip the given sheet in the gap A and B of the screw gauge.

7.    Turn the screw head till the ratchet arrangement gives a click.

8.    Note the readings of linear scale and circular scale and find the observed thickness using the relation, observation thickness = L.S.R. + C.S.R.

9.    Add the zero correction to the observed thickness to find the corrected diameter.

10.    Repeat steps 6 to 9 to find the thickness from four more different places.

 

- wherein

C.S.R= n \times L.C

Total reading 

= L.S.R + C.S.R

= N\times n +N.C.

L.C = least count

C.S.R = Circular scale reading

N= Nth division of linear scale 

 

 

 Least \: count=\frac{pitch}{no. \: of \: division \: on \: circular \: scale}

                     =\frac{0.5mm}{50}

L.C.=0.001mm

-ve \ zero \: error=-5\times LC= -0.005mm

Measured value=main scale reading + screw gauge reading - zero error

=0.5mm+(25\times 0.001-(-0.05))mm=0.80mm

 


Option 1)

0.75 mm

This option is incorrect

Option 2)

 0.80 mm

This option is correct

Option 3)

 0.70 mm

This option is incorrect

Option 4)

 0.50 mm

This option is incorrect

Posted by

divya.saini

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