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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A student uses a Vernier callipers to measure the length of a metal...

A student uses a Vernier callipers to measure the length of a metal cylinder. The main scale is graduated in millimeters (1 M.S.D.=1 mm1 \text{ M.S.D.} = 1 \text{ mm}), and 2020 Vernier scale divisions (V.S.D.) coincide with 1919 main scale divisions.

When the jaws of the callipers are brought into contact with each other, the zero of the Vernier scale lies to the left of the main scale zero, and the 8th8^{\text{th}} Vernier division coincides with a main scale mark.

While measuring the cylinder, the main scale reading is 3.4 cm3.4 \text{ cm} and the 14th14^{\text{th}} Vernier division coincides with a main scale mark. The true length of the cylinder is:

Options

  1. A3.43 cm3.43 \text{ cm}
  2. B3.51 cm3.51 \text{ cm}
  3. C3.47 cm3.47 \text{ cm}
  4. D3.53 cm3.53 \text{ cm}

Correct answer

D. 3.53 cm3.53 \text{ cm}

Step-by-step solution

11 M.S.D. = 11 mm = 0.10.1 cm

2020 V.S.D. = 1919 M.S.D.

11 V.S.D. = 1920\dfrac{19}{20} M.S.D. = 0.950.95 mm

Least count (L.C.) = 11 M.S.D. - 11 V.S.D. = 11 mm - 0.950.95 mm = 0.050.05 mm = 0.0050.005 cm

Since the zero of the Vernier scale lies to the left of the main scale zero, the zero error is negative.

Zero error = −(N−n)×L.C.- (N - n) \times \text{L.C.}

Zero error = −(20−8)×0.005- (20 - 8) \times 0.005 cm = −12×0.005-12 \times 0.005 cm = −0.06-0.06 cm

Measured value = M.S.R. + n′×L.C.n' \times \text{L.C.}

Measured value = 3.43.4 cm + 14×0.00514 \times 0.005 cm = 3.43.4 cm + 0.070.07 cm = 3.473.47 cm

True length = Measured value - Zero error

True length = 3.473.47 cm - (−0.06-0.06 cm) = 3.533.53 cm

Answer: 3.53 cm3.53 \text{ cm}
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