JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Units & Measurement

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In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and

5 0^{th}

division of circular scale coincides with the reference line of main scale. The diameter of sphere is ____ mm.

(A)5.045✓ Correct
(B)5.055
(C)5.450
(D)5.550

Answer: (A)

Solution

The least count of the screw gauge is

\frac{pitch}{circular divisions} = \frac{0.1}{100} = 0.001 mm

. With the studs in contact the circular scale reads

5

divisions instead of zero, a positive zero error of

+ 5 \times 0.001 = + 0.005 mm

. While measuring, the uncorrected reading is

MSR + (CSD \times LC) = 5 + 50 \times 0.001 = 5.050 mm

. Subtracting the zero error, the diameter of the sphere is

5.050 - 0.005 = 5.045 mm

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Units & Measurement

Solution

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