JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 11)
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :
1.80 mm
0.90 mm
0.82 mm
1.64 mm
Explicació
Least count = $${{1mm} \over {100}} = 0.01mm$$
zero error = + 8 $$\times$$ LC = + 0.08 mm
True reading (Diameter)
= (1 mm + 72 $$\times$$ LC) $$-$$ (Zero error)
= (1 mm + 72 $$\times$$ 0.01 mm) $$-$$ 0.08 mm
= 1.72 mm $$-$$ 0.08 mm
= 1.64 mm
Therefore, radius = $${{1.64} \over 2}$$ = 0.82 mm.
zero error = + 8 $$\times$$ LC = + 0.08 mm
True reading (Diameter)
= (1 mm + 72 $$\times$$ LC) $$-$$ (Zero error)
= (1 mm + 72 $$\times$$ 0.01 mm) $$-$$ 0.08 mm
= 1.72 mm $$-$$ 0.08 mm
= 1.64 mm
Therefore, radius = $${{1.64} \over 2}$$ = 0.82 mm.