JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 9)
The pitch 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 line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
5.755 mm
5.950 mm
5.725 mm
5.740 mm
Explanation
We know,
Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$
$$ \therefore $$ LC = $${{0.5} \over {100}}$$
= 0.5 $$ \times $$ 10$$-$$2 mm
Reading = MSR + CSR $$-$$ positive error
Given, Main scale reading (MSR) = 5.5 mm
Circular scale reading (CSR)
= 48 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm
= 0.24
As zero of its circular scale lines 3 division below the mean line, it means error is position error.
$$ \therefore $$ positive error
= 3 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm
= 0.015 mm
$$ \therefore $$ Reading = 5.5 + 0.24 $$-$$ 0.015
= 5.725 mm
Least count (LC) = $${{Pitch} \over {no.\,of\,divisions}}$$
$$ \therefore $$ LC = $${{0.5} \over {100}}$$
= 0.5 $$ \times $$ 10$$-$$2 mm
Reading = MSR + CSR $$-$$ positive error
Given, Main scale reading (MSR) = 5.5 mm
Circular scale reading (CSR)
= 48 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm
= 0.24
As zero of its circular scale lines 3 division below the mean line, it means error is position error.
$$ \therefore $$ positive error
= 3 $$ \times $$ 0.5 $$ \times $$ 10$$-$$2 mm
= 0.015 mm
$$ \therefore $$ Reading = 5.5 + 0.24 $$-$$ 0.015
= 5.725 mm
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