Q. No.Consider the diameter of a spherical object being measured with the help of a Vernier callipers. Suppose its \(10\) Vernier Scale Divisions (V.S.D.) are equal to its \(9\) Main Scale Divisions (M.S.D.). The least division in the M.S. is \(0.1\text{ cm}\) and the zero of V.S. is at \(x=0.1~ \text{cm}\) when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is \({M}=5~\text{cm}\) and the number of coinciding vernier division is \(8 ,\) the measured diameter after zero error correction, is:
1. \(4.98 ~\text{cm}\)
2. \(5.00 ~\text{cm}\)
3. \(5.18 ~\text{cm}\)
4. \(5.08 ~\text{cm}\)
1. \(4.98 ~\text{cm}\)
2. \(5.00 ~\text{cm}\)
3. \(5.18 ~\text{cm}\)
4. \(5.08 ~\text{cm}\)
Subtopic: Â Measurement & Measuring Devices |
From NCERT
NEET - 2025
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In a vernier callipers, \((N+1)\) divisions of the vernier scale coincide with \(N\) divisions of the main scale. If \(1\) \(\text{MSD}\) represents \(0.1~\text{mm}\), the vernier constant (in \(\text{cm}\)) is:
| 1. | \(\dfrac{1}{100(N+1)} \) | 2. | \(100N\) |
| 3. | \(10(N+1) \) | 4. | \(\dfrac{1}{10 N}\) |
Subtopic: Â Measurement & Measuring Devices |
 71%
From NCERT
NEET - 2024
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A vernier calipers has \(1~\text{mm}\) marks on the main scale. It has \(20\) equal divisions on the vernier scale which match with \(16\) main scale divisions. The least count of this vernier calipers is:
| 1. | \(0.02~\text{mm}\) | 2. | \(0.05~\text{mm}\) |
| 3. | \(0.10~\text{mm}\) | 4. | \(0.20~\text{mm}\) |
Subtopic: Â Measurement & Measuring Devices |
 70%
From NCERT
NEET - 2024
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The pitch of an error-free screw gauge is \(1~\text{mm},\) and there are \(100\) divisions on the circular scale. While measuring the diameter of a thick wire, the pitch scale reads \(1~\text{mm},\) and \(63^\text{rd}\) division on the circular scale coincides with the reference line. The diameter of the wire is:
| 1. | \(1.63 ~\text{cm}\) | 2. | \(0.163 ~\text{cm}\) |
| 3. | \(0.163~\text m\) | 4. | \(1.63 ~\text m\) |
Subtopic: Â Measurement & Measuring Devices |
 72%
From NCERT
NEET - 2024
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The diameter of a spherical bob, when measured with vernier callipers yielded the values: \(3.33\) cm, \(3.32\) cm, \(3.34\) cm, \(3.33\) cm and \(3.32\) cm. The mean diameter to appropriate significant figures is:
1. \(3.328\) cm
2. \(3.3\) cm
3. \(3.33\) cm
4. \(3.32\) cm
1. \(3.328\) cm
2. \(3.3\) cm
3. \(3.33\) cm
4. \(3.32\) cm
Subtopic: Â Measurement & Measuring Devices |
 59%
From NCERT
NEET - 2023
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When the circular scale of a screw gauge completes \(2\) rotations, it covers \(1\) mm over the pitch scale. The total number of circular scale divisions is \(50.\) The least count of the screw gauge in metres is:
1. \(10^{-4}\)
2. \(10^{-5}\)
3. \(10^{-2}\)
4. \(10^{-3}\)
Subtopic: Â Measurement & Measuring Devices |
 57%
From NCERT
NEET - 2022
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A screw gauge gives the following readings when used to measure the diameter of a wire:
Main scale reading: \(0\) mm
Circular scale reading: \(52\) divisions
Given that \(1\) mm on the main scale corresponds to \(100\) divisions on the circular scale, the diameter of the wire that can be inferred from the given data is:
| 1. | \(0.26\) cm | 2. | \(0.052\) cm |
| 3. | \(0.52\) cm | 4. | \(0.026\) cm |
Subtopic: Â Measurement & Measuring Devices |
 69%
From NCERT
NEET - 2021
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A screw gauge has the least count of \(0.01~\text{mm}\) and there are \(50\) divisions in its circular scale. The pitch of the screw gauge is:
| 1. | \(0.25~\text{mm}\) | 2. | \(0.5~\text{mm}\) |
| 3. | \(1.0~\text{mm}\) | 4. | \(0.01~\text{mm}\) |
Subtopic: Â Measurement & Measuring Devices |
 85%
From NCERT
NEET - 2020
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The angle of \(1'\) (minute of an arc) in radian is nearly equal to:
1. \(2.91 \times 10^{-4}~\text{rad} \)
2. \(4.85 \times 10^{-4}~\text{rad} \)
3. \(4.80 \times 10^{-6} ~\text{rad} \)
4. \(1.75 \times 10^{-2}~\text{rad} \)
1. \(2.91 \times 10^{-4}~\text{rad} \)
2. \(4.85 \times 10^{-4}~\text{rad} \)
3. \(4.80 \times 10^{-6} ~\text{rad} \)
4. \(1.75 \times 10^{-2}~\text{rad} \)
Subtopic: Â Measurement & Measuring Devices |
 53%
From NCERT
NEET - 2020
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The main scale of a vernier calliper has \(n\) divisions/cm. \(n\) divisions of the vernier scale coincide with \((n-1)\) divisions of the main scale. The least count of the vernier calliper is:
1. \(\dfrac{1}{(n+1)(n-1)}\) cm
2. \(\dfrac{1}{n}\) cm
3. \(\dfrac{1}{n^{2}}\) cm
4. \(\dfrac{1}{(n)(n+1)}\) cm
Subtopic: Â Measurement & Measuring Devices |
 53%
From NCERT
NEET - 2019
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