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Chapter 29: Problem 2

On measuring the diameter of a spherical body using vernier callipers, main scale reading \(=1.3 \mathrm{~cm}, 5\) th vernier scale division is coinciding with any main scale division and zero error is \(-0.03 \mathrm{~cm}\), what will be corrected reading? (a) \(1.38 \mathrm{~cm}\) (b) \(1.32 \mathrm{~cm}\) (c) \(1.35 \mathrm{~cm}\) (d) \(-1.38 \mathrm{~cm}\)

Short Answer

Expert verified
The corrected reading is 1.38 cm.

Step by step solution

01

Identify the Main Scale Reading

The main scale reading given is 1.3 cm. This is the linear measurement on the main scale of the vernier calipers.
02

Calculate Vernier Scale Reading

Since the 5th vernier scale division coincides with a main scale division, the vernier scale reading is 0.05 cm (each scale division is assumed 0.01 cm).
03

Calculate Actual Measurement

The actual measurement is found by adding the main scale reading and the vernier scale reading: \(1.3 + 0.05 = 1.35\) cm.
04

Account for Zero Error

Since the zero error is \(-0.03\) cm, add the absolute value of the zero error to the actual measurement to correct for it: \(1.35 + 0.03 = 1.38\) cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Vernier Scale
Vernier calipers are an essential tool in precision measurement, which use two scales known as the main scale and the vernier scale. The vernier scale is a smaller, auxiliary scale that slides along the main scale. It's crucial because it allows us to measure lengths with much finer precision, beyond the limitations of the main scale alone.

Here's how it works:
  • Each division on the vernier scale is a fraction smaller than a division on the main scale.
  • The idea is to find a point where a vernier division aligns perfectly with a division on the main scale.
  • In this problem, the 5th division of the vernier scale aligns with a division on the main scale, which translates to 0.05 cm when you multiply 5 by 0.01 cm (assuming each vernier division is 0.01 cm).
This method enhances measurement precision because it reduces the chance of human error in estimating small divisions on the main scale.
The Importance of Zero Error Correction
When using vernier calipers, zero error is a common occurrence. Zero error happens when the zero marks on the vernier scale and main scale do not coincide when the jaws are fully closed. If unnoticed, it could lead to inaccurate readings.

Here's what you should know:
  • Zero error can be positive or negative. If the zero of the vernier scale is to the right of the main scale zero, the error is positive. If it鈥檚 to the left, it鈥檚 negative.
  • In this exercise, the given zero error is \(-0.03\) cm, meaning you need to add this value to the measurement as a mathematical correction.
  • To correct for this in our measurement, we take the actual measurement found and add this zero error correction:\((1.35 + 0.03 = 1.38\) cm).
Correcting for zero error ensures that the measurement starts from the true zero point of the measuring device.
Main Scale Reading Explained
The main scale of the vernier calipers provides the primary measurement in centimeters or millimeters, as applicable. It is the linear, clearly marked scale on the body of the calipers and typically determines tenths or whole units of a measurement.

While taking a measurement:
  • Read the main scale measurement before noting any vernier scale alignment.
  • In this exercise, the main scale reading is clearly marked as 1.3 cm.
  • This basic reading serves as the foundation to which any additional precision from the vernier scale is added.
Understanding the main scale reading is the first crucial step in achieving a precise measurement with vernier calipers, as it provides the significant figures before factoring in the vernier's enhanced precision.

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Most popular questions from this chapter

When a glass capillary tube of radius \(0.015 \mathrm{~cm}\) is dipped in water, the water rises to a height of \(15 \mathrm{~cm}\) within it. Assuming contact angle between water and glass to be \(0^{\circ}\), the surface tension of water is \(\left[\rho_{\text {water }}=1000 \mathrm{~kg} \mathrm{~m}^{-3,} \mathrm{~g}=9.81 \mathrm{~ms}^{-2}\right]\) (a) \(0.11 \mathrm{Nm}^{-1}\) (b) \(0.7 \mathrm{Nm}^{-1}\) (c) \(0.072 \mathrm{Nm}^{-1}\) (d) None of the above

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In an experiment the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a-degree (i.e., \(0.5^{\circ}\) ), then the least count of the instrument is (a) one minute (b) half minute (c) one degree (d) half degree

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In determination of refractive index of glass slab using travelling microscope, first of all we take a reading when the microscope is focused on a mark. This reading comes out to be \(s_{1}\), then we place a glass slab on the surface covering the mark. Now, the microscope is re-adjusted to focus the mark through the slab and this time reading comes out to be \(s_{2}\). Then, we place an opaque object on the glass slab and adjust the microscope to focus on opaque object, this time the reading of microscope is \(s_{3} .\) The refractive index of the glass slab is (a) \(\frac{s_{3}-s_{1}}{s_{2}-s_{1}}\) (b) \(\frac{s_{3}-s_{2}}{s_{2}-s_{1}}\) (c) \(\frac{s_{3}-s_{1}}{s_{3}-s_{2}}\) (d) \(\frac{s_{3}}{s_{3}-s_{2}}\)
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