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JEE Advanced 2016 · Paper 2, Physics question 6

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1)(C_1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2)(C_2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1C_1 and C2C_2, respectively, are

Two Vernier caliper readings. C₁: main scale marked 2, 3, 4 cm with mm divisions; the Vernier zero lies just past 2.8 cm and its 7th division lines up with a main-scale mark. C₂: same main scale; the Vernier zero lies just past 2.8 cm (Vernier divisions longer than main divisions, numbered 0, 5, 10), with its 7th division lining up with a main-scale mark.
Choose one option
Answer and worked solution

Official answer: B

Least counts. 1 MSD = 0.1 cm.

  • C₁: 10 VSD = 9 MSD, so 1 VSD = 0.09 cm and the least count is +0.01 cm.
  • C₂: 10 VSD = 11 MSD, so 1 VSD = 0.11 cm and the least count is −0.01 cm.

C₁ reading. The zero is just past 2.8 cm and the 7th division coincides: 2.8+7×0.01=2.872.8+7\times0.01=2.87 cm.

C₂ reading. Let the zero be at x0x_0 (between 2.8 and 2.9). The 7th mark, at x0+0.77x_0+0.77, coincides with the 3.6 cm main mark, so x0=3.60−0.77=2.83x_0=3.60-0.77=2.83 cm.

(B)

Topics:Vernier callipers and screw gaugeLeast count
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