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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Experimental Physics

JEE Advanced · Physics · Experimental Physics

In ordinary Vernier calipers, 10^ th division of the Vernier scale...

In ordinary Vernier calipers, 10th 10^{\text {th }} division of the Vernier scale coincides with 9mit 9^{\text {mit }} division of the main scale. In a specially designed Vernier calipers the Vernier scale is so constructed that 10lh 10^{\text {lh }} division on it coincides with 11th 11^{\text {th }} division on the main scale. Each division on the main scale equals to 1 mm . The calipers have a zero error as shown in the figure-I. When the Vernier caliper is used to measure a length, the concerned portion of its scale is shown in figure-II. In ordinary Vernier calipers, 10^ th division of the Vernier scale coincides with 9^ mit division of the main scale. In
In ordinary Vernier calipers, 10^ th division of the Vernier scale coincides with 9^ mit division of the main scale. In

Options

  1. AZero error in the calipers has magnitude 0.7 mm .
  2. BThe length being measured is 1.08 cm .
  3. CThe length being measured is 1.22 cm .
  4. DThough the given Vernier scale does not follow the principle of Vernier scale, yet can be used satisfactorily.

Correct answer

C. The length being measured is 1.22 cm .

Step-by-step solution

VSD=1.1 mm\mathrm{VSD}=1.1 \mathrm{~mm} and 1MSD=1.0 mm1 \mathrm{MSD}=1.0 \mathrm{~mm} Least count ==\mathrm{LC}=\mathrm{MSD}∼VSD=0.1 mm\sim \mathrm{VSD}=0.1 \mathrm{~mm} For (A):(\mathbf{A}): Zero error =−(=-( No. of divisions of vernier scale coincident )×LC=−7×0.1=−0.7 mm) \times \mathrm{LC}=-7 \times 0.1=-0.7 \mathrm{~mm} For (C)(\mathbf{C}) : Reading = (No. of divisions coincident on main scale) ×\times MSD - (No. of divisions coincident on vernier scale ) LC- zero error =1.22 cm=1.22 \mathrm{~cm}
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