Numerical/problem-solving

1. TITLE

How to solve Screw Gauge with Zero Error numericals in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This category of numerical tests your understanding of measurement instruments and error correction. We will need the formulas for calculating Least Count (LC) and True Reading. The most typical mistake students make is applying the incorrect sign for the zero error correction, leading to an inaccurate final measurement.

3. MCQ 

Question:
A screw gauge has a pitch of $1.0\text{ mm}$ and $200$ divisions on the circular scale. When there is nothing between the jaws, the zero of the circular scale lies $4$ divisions above the reference line. When measuring the diameter of a wire, the linear scale reads $3\text{ mm}$ and the $60^{\text{th}}$ division of the circular scale coincides with the reference line. What is the true diameter of the wire?

(a) $3.320\text{ mm}$
(b) $3.280\text{ mm}$
(c) $3.300\text{ mm}$
(d) $3.600\text{ mm}$

4. STEP-BY-STEP SOLUTION

SOLUTION
Given:
* $\text{Pitch} = 1.0\text{ mm}$
* $\text{Number of circular divisions} = 200$
* $\text{Initial zero alignment} = 4\text{ divisions above reference line}$ (Negative Zero Error)
* $\text{Linear Scale Reading (LSR)} = 3\text{ mm}$
* $\text{Circular Scale Reading (CSR)} = 60$

To Find:
* The true diameter of the wire in $\text{mm}$.

Formula:
* $\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of circular divisions}}$
(Use this to find the precision of the instrument.)
* $\text{True Reading} = \text{LSR} + (\text{CSR} \times \text{LC}) - \text{Zero Error}$
(Use this to calculate the final dimension while compensating for the initial instrument error.)

Step 1: Calculate the Least Count (LC)

$$\text{LC} = \frac{1.0\text{ mm}}{200} = 0.005\text{ mm}$$

 

Step 2: Calculate the Zero Error
Since the zero is above the reference line, the error is negative.

$$\text{Zero Error} = - (4 \times \text{LC}) = - (4 \times 0.005\text{ mm}) = -0.020\text{ mm}$$

 

Step 3: Calculate the True Reading

$$\text{Measured Reading} = \text{LSR} + (\text{CSR} \times \text{LC})$$

 

$$\text{Measured Reading} = 3\text{ mm} + (60 \times 0.005\text{ mm}) = 3\text{ mm} + 0.300\text{ mm} = 3.300\text{ mm}$$

 

$$\text{True Reading} = \text{Measured Reading} - \text{Zero Error}$$

 

$$\text{True Reading} = 3.300\text{ mm} - (-0.020\text{ mm})$$

 

$$\text{True Reading} = 3.300\text{ mm} + 0.020\text{ mm} = 3.320\text{ mm}$$

Answer: $3.320\text{ mm}$ (Option a)

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Using the wrong sign for zero error (Resulting in Option b: $3.280\text{ mm}$). Remember that negative zero errors must be subtracted, which mathematically means adding the value to your measured reading!
  • ❌ Mistake 2: Forgetting the zero error correction entirely (Resulting in Option c: $3.300\text{ mm}$).
  • ❌ Mistake 3: Miscalculating the least count by assuming a standard $100$ divisions instead of the given $200$ divisions (Resulting in Option d: $3.600\text{ mm}$). Always read the pitch and division count carefully.

6. QUICK FORMULA REFERENCE

Formulas Used:

  • Formula 1 — $\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Number of circular divisions}}$ (Use when determining the precision of the specific screw gauge).
  • Formula 2 — $\text{True Reading} = \text{LSR} + (\text{CSR} \times \text{LC}) - \text{Zero Error}$ (Use when calculating the final dimension of an object compensating for instrument flaws).

7. SIMILAR PROBLEM HINT

If NEET changes the values, the method stays the same. Practice with: A scenario with a positive zero error where the circular scale zero is below the reference line, meaning you will need to subtract a positive value at the end.

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 1, Exercise 1.8

9. CTA

More solved numericals → Physics and Measurement Complete Guide
Practice problems → Mock exams