JEE MAIN - Physics (2011 - No. 27)
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 date is:
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 date is:
0.052 cm
0.026 cm
0.005 cm
0.52 cm
Explanation
Least count of screw gauge
= $${{Pitch} \over {Number\,of\,division\,on\,circular\,scale}}$$
= $${1 \over {100}}mm$$
= 0.01 mm
Diameter of the wire = M.S.R + C.S.R $$ \times $$ L.C
= 0 + 52 $$ \times $$ 0.01
= 0.52 mm
= 0.052 cm
= $${{Pitch} \over {Number\,of\,division\,on\,circular\,scale}}$$
= $${1 \over {100}}mm$$
= 0.01 mm
Diameter of the wire = M.S.R + C.S.R $$ \times $$ L.C
= 0 + 52 $$ \times $$ 0.01
= 0.52 mm
= 0.052 cm
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