Experimental Physics

JEE Main level, single correct, medium

4 of 70

Question 4 of 70

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, 4th 4^{\text {th }} mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and 5th 5^{\text {th }} division of vernier scale coincides with a main scale division. Measured length of cylinder is ____\_\_\_\_ mm.
(Least count of Vernier calliper
=0.1 mm=0.1 \mathrm{~mm})

Show answer and solution

Answer: (C) 15.1

Given Least Count (LCLC) = 0.10.1 mm.
When jaws touch, the zero of the vernier scale is to the right of the main scale zero, indicating a positive zero error.

Zero error (
ee) = +(n×LC)+ (n \times LC), where nn is the coinciding vernier division.
e=+(4×0.1)=+0.4e = +(4 \times 0.1) = +0.4 mm.
While measuring the cylinder:

Main Scale Reading (
MSRMSR) = 1515 divisions.
Assuming
11 MSD = 11 mm (standard for LC=0.1LC = 0.1 mm), MSR=15MSR = 15 mm.
Vernier Scale Reading (
VSRVSR) = 5×LC=5×0.1=0.55 \times LC = 5 \times 0.1 = 0.5 mm.
Observed Reading =
MSR+VSR=15+0.5=15.5MSR + VSR = 15 + 0.5 = 15.5 mm.
True Measured Length = Observed Reading
−- Zero Error
True Length =
15.5−0.4=15.115.5 - 0.4 = 15.1 mm.

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