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NEET
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Physics
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Experimental Physics

NEET · Physics · Experimental Physics

Match List-I with List-II regarding the specifications of different...

Match List-I with List-II regarding the specifications of different vernier callipers and their least counts.

List-I (Specifications)List-II (Least Count)
A. 10 main scale divisions per cm, 10 VSD = 9 MSDI. 0.05 mm
B. 20 main scale divisions per cm, 20 VSD = 19 MSDII. 0.02 mm
C. 20 main scale divisions per cm, 25 VSD = 24 MSDIII. 0.025 mm
D. 10 main scale divisions per cm, 20 VSD = 19 MSDIV. 0.1 mm

Choose the correct answer from the options given below:

Options

  1. AA-IV, B-II, C-III, D-I
  2. BA-I, B-III, C-II, D-IV
  3. CA-IV, B-III, C-II, D-I
  4. DA-I, B-II, C-III, D-IV

Correct answer

C. A-IV, B-III, C-II, D-I

Step-by-step solution

For a vernier calliper where nn VSD = (n−1)(n-1) MSD, the Least Count (LC) is given by 1n MSD\dfrac{1}{n} \text{ MSD}.

A. 10 divisions/cm⇒1 MSD=110 cm=1 mm10 \text{ divisions/cm} \Rightarrow 1 \text{ MSD} = \dfrac{1}{10} \text{ cm} = 1 \text{ mm}.
LC=1 mm10=0.1 mm\text{LC} = \dfrac{1 \text{ mm}}{10} = 0.1 \text{ mm}. (Matches IV)

B. 20 divisions/cm⇒1 MSD=120 cm=0.5 mm20 \text{ divisions/cm} \Rightarrow 1 \text{ MSD} = \dfrac{1}{20} \text{ cm} = 0.5 \text{ mm}.
LC=0.5 mm20=0.025 mm\text{LC} = \dfrac{0.5 \text{ mm}}{20} = 0.025 \text{ mm}. (Matches III)

C. 20 divisions/cm⇒1 MSD=120 cm=0.5 mm20 \text{ divisions/cm} \Rightarrow 1 \text{ MSD} = \dfrac{1}{20} \text{ cm} = 0.5 \text{ mm}.
LC=0.5 mm25=0.02 mm\text{LC} = \dfrac{0.5 \text{ mm}}{25} = 0.02 \text{ mm}. (Matches II)

D. 10 divisions/cm⇒1 MSD=110 cm=1 mm10 \text{ divisions/cm} \Rightarrow 1 \text{ MSD} = \dfrac{1}{10} \text{ cm} = 1 \text{ mm}.
LC=1 mm20=0.05 mm\text{LC} = \dfrac{1 \text{ mm}}{20} = 0.05 \text{ mm}. (Matches I)

Therefore, the correct matching is A-IV, B-III, C-II, D-I.

Answer: A-IV, B-III, C-II, D-I
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