Concept Card: Least Count of Measuring Instruments & Significant Figures
1. Least Count (LC) of Measuring Instruments:
The least count is the smallest value that can be directly measured by an instrument. It defines the precision and instrumental resolution limit.
A. Vernier Calipers:
$\text{Least Count (LC)} = 1\text{ Main Scale Division (MSD)} – 1\text{ Vernier Scale Division (VSD)}$
If $n$ divisions of the Vernier scale coincide with $(n – 1)$ divisions of the main scale:
$n\text{ VSD} = (n – 1)\text{ MSD} \implies 1\text{ VSD} = \frac{n – 1}{n}\text{ MSD}$
$\text{LC} = 1\text{ MSD} – \frac{n – 1}{n}\text{ MSD} = \frac{1\text{ MSD}}{n}$
Total Reading:
$\text{Total Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) – (\text{Zero Error})$
B. Screw Gauge (Micrometer):
$\text{Pitch} = \frac{\text{Distance advanced on linear main scale}}{\text{Number of full rotations completed}}$
$\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of circular scale divisions (CSD)}}$
Total Reading:
$\text{Total Reading} = \text{Linear Scale Reading (LSR)} + (\text{CSR} \times \text{LC}) – (\text{Zero Error})$
C. Zero Error Types:
- Positive Zero Error: The zero mark of the vernier or circular scale lies to the right (or above) the main scale reference line when jaws/anvils are closed. The measured value is larger than actual. $\text{Correction} = -|\text{Zero Error}|$.
- Negative Zero Error: The zero mark lies to the left (or below) the main scale reference line. The measured value is smaller than actual. $\text{Correction} = +|\text{Zero Error}|$.
2. Rules for Determining Significant Figures (SF):
- All non-zero digits are significant (e.g., $18.45$ has 4 SF).
- All zeros occurring between two non-zero digits are significant (e.g., $2005$ has 4 SF, $1.0004$ has 5 SF).
- Leading zeros (to the left of the first non-zero digit) are never significant; they merely locate the decimal point (e.g., $0.0034$ has 2 SF).
- Trailing zeros after a decimal point are significant (e.g., $5.200$ has 4 SF, $0.0700$ has 3 SF).
- Trailing zeros in an integer without a decimal point are generally non-significant unless given in scientific notation (e.g., $4300$ has 2 SF; $4.30 \times 10^3$ has 3 SF).
- Exact numbers or pure geometric factors have infinite significant figures (e.g., $\pi$, $2$ in $2\pi r$).
3. Arithmetic Operations with Significant Figures:
- Addition & Subtraction: The final calculated result must have no more decimal places than the measurement with the fewest decimal places.
- Multiplication & Division: The final calculated result must contain no more significant figures than the quantity with the least number of significant figures.
- Rounding Off Rule: If the dropped digit is $5$ followed only by zeros, the preceding digit is left unchanged if it is even, and increased by $1$ if it is odd (e.g., $4.750 \to 4.8$, while $4.650 \to 4.6$).
Solved Examples
Example 1 (Vernier Calipers with Positive Zero Error):
A vernier calipers has $1\text{ mm}$ main scale divisions. $10$ divisions of the vernier scale coincide with $9$ main scale divisions. Before measuring, with jaws closed, the zero of the vernier scale is to the right of the main scale zero and its $3\text{rd}$ division coincides with a main scale mark. While measuring the diameter of a cylinder, the main scale reads $4.2\text{ cm}$ and the $7\text{th}$ vernier division coincides. Find the true diameter.
Solution:
1. Least count of instrument:
$\text{LC} = \frac{1\text{ MSD}}{10} = \frac{1\text{ mm}}{10} = 0.1\text{ mm} = 0.01\text{ cm}$.
2. Zero error:
Since zero lies to the right, it is a positive zero error:
$\text{Zero Error} = +3 \times \text{LC} = +3 \times 0.01\text{ cm} = +0.03\text{ cm}$.
3. Observed reading:
$\text{Observed Reading} = \text{MSR} + (\text{VSR} \times \text{LC}) = 4.2\text{ cm} + (7 \times 0.01\text{ cm}) = 4.27\text{ cm}$.
4. Corrected reading:
$\text{True Reading} = \text{Observed Reading} – \text{Zero Error} = 4.27\text{ cm} – (+0.03\text{ cm}) = 4.24\text{ cm}$.
Example 2 (Screw Gauge with Negative Zero Error):
A screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular scale. When the anvils are in contact, the circular scale zero is $3$ divisions below the baseline. When a wire is clamped, the linear scale shows $2.5\text{ mm}$ and the $28\text{th}$ circular division aligns with the baseline. Determine the actual diameter.
Solution:
1. Least count:
$\text{LC} = \frac{\text{Pitch}}{\text{Number of CSD}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm}$.
2. Zero error:
Since the zero mark has not yet reached the baseline (lies $3$ divisions below), the zero error is negative:
$\text{Zero Error} = -3 \times \text{LC} = -0.03\text{ mm}$.
3. Observed reading:
$\text{Observed Reading} = \text{LSR} + (\text{CSR} \times \text{LC}) = 2.5\text{ mm} + (28 \times 0.01\text{ mm}) = 2.78\text{ mm}$.
4. True diameter:
$\text{True Reading} = \text{Observed Reading} – (\text{Zero Error}) = 2.78\text{ mm} – (-0.03\text{ mm}) = 2.81\text{ mm}$.
Example 3 (Significant Figures in Multiplication & Addition):
A thin metal plate has length $l = 3.124\text{ m}$, breadth $b = 1.05\text{ m}$, and thickness $t = 2.5\text{ cm}$. Calculate the total volume of the plate expressed to the correct number of significant figures.
Solution:
1. Convert all dimensions to meters:
$l = 3.124\text{ m}$ (4 significant figures)
$b = 1.05\text{ m}$ (3 significant figures)
$t = 2.5\text{ cm} = 0.025\text{ m}$ (2 significant figures)
2. Compute volume:
$V = l \times b \times t = 3.124 \times 1.05 \times 0.025\text{ m}^3 = 0.082005\text{ m}^3$
3. Apply significant figure rule for multiplication:
The factor with the fewest significant figures is $t$ (which has 2 SF). The final result must therefore be rounded off to 2 significant figures:
$V = 0.082\text{ m}^3 = 8.2 \times 10^{-2}\text{ m}^3$.
Example 4 (Density Calculation with Appropriate Rounding):
The mass of a solid block is measured as $m = 48.20\text{ g}$ and its volume is measured as $V = 6.4\text{ cm}^3$. Compute the density of the block to the appropriate number of significant figures.
Solution:
1. Mass has 4 significant figures ($48.20\text{ g}$).
2. Volume has 2 significant figures ($6.4\text{ cm}^3$).
3. Density calculation:
$\rho = \frac{m}{V} = \frac{48.20}{6.4} = 7.53125\text{ g/cm}^3$
Since volume limits the precision to 2 significant figures, we round $7.53125$ to 2 significant figures:
$\rho = 7.5\text{ g/cm}^3$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
$1\text{ cm}$ of the main scale of a vernier calipers is divided into $10$ equal parts. If $20$ divisions of the vernier scale coincide with $19$ small main scale divisions, the least count of the calipers is:
(A) $0.05\text{ mm}$
(B) $0.01\text{ mm}$
(C) $0.005\text{ mm}$
(D) $0.02\text{ mm}$
Problem 2 (JEE Main – Single Correct):
The number of significant figures in the numbers $0.0050$, $2.040$, and $400.0$ are respectively:
(A) 2, 4, 4
(B) 1, 3, 1
(C) 2, 3, 4
(D) 2, 4, 1
Problem 3 (JEE Main – Single Correct):
Evaluate the arithmetic expression $(15.82 – 2.5) \times 0.20$ and report the result to the correct number of significant figures:
(A) $2.664$
(B) $2.66$
(C) $2.7$
(D) $3$
Problem 4 (JEE Main – Single Correct):
A screw gauge has pitch $0.5\text{ mm}$ and $100$ divisions on its circular scale. When measured with jaws closed, the $4\text{th}$ division of the circular scale lies below the reference line. While measuring a wire’s thickness, the linear scale reads $2\text{ mm}$ and the circular scale reads $46$. The true thickness of the wire is:
(A) $2.25\text{ mm}$
(B) $2.23\text{ mm}$
(C) $2.21\text{ mm}$
(D) $2.50\text{ mm}$
Problem 5 (JEE Main – Single Correct):
When $3.924\text{ g}$ is subtracted from $8.7\text{ g}$, the result expressed to the correct significant figures is:
(A) $4.776\text{ g}$
(B) $4.78\text{ g}$
(C) $4.8\text{ g}$
(D) $5\text{ g}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding significant figures and measurements?
(A) Change of units does not change the number of significant figures in a measurement.
(B) The least count of an instrument determines the absolute uncertainty of a single reading.
(C) The number $3.2000 \times 10^3$ has 5 significant figures.
(D) A measured value with more significant figures is inherently more precise than one with fewer significant figures.
Problem 7 (JEE Advanced – One or More Correct):
In a specially designed vernier calipers, $1\text{ MSD} = 1\text{ mm}$, and $(N + 1)$ divisions of the vernier scale coincide with $N$ main scale divisions (retrograde vernier). Which of the following statements is/are correct?
(A) $1\text{ VSD} = \frac{N}{N + 1}\text{ mm}$
(B) $\text{Least count} = \frac{1}{N + 1}\text{ mm}$
(C) The vernier scale divisions are smaller than the main scale divisions.
(D) The vernier scale reading is added to the main scale reading when taking measurements.
Problem 8 (JEE Advanced – One or More Correct):
A screw gauge has a pitch of $1\text{ mm}$ and $100$ circular scale divisions. Which of the following readings can be measured with this instrument without estimation?
(A) $5.42\text{ mm}$
(B) $5.425\text{ mm}$
(C) $0.07\text{ mm}$
(D) $3.00\text{ mm}$
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A vernier calipers has $20$ divisions on its vernier scale which coincide with $19$ divisions on the main scale. The smallest division on the main scale is $0.5\text{ mm}$. When measuring a spherical bob, the main scale reads $2.5\text{ cm}$ and the $14\text{th}$ vernier division coincides with a main scale line. If the instrument has a positive zero error of $+0.05\text{ mm}$, the corrected diameter of the bob is $D\text{ mm}$. Find the value of $D \times 10$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
The mass of a cube is measured as $m = 10.00\text{ g}$ and its edge length is measured as $l = 2.0\text{ cm}$. The density of the cube reported to the correct number of significant figures is $d\text{ g/cm}^3$. Calculate the numerical value of $100 \times d$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (A) | 3. (C) | 4. (A) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, C, D) | 9. 253 | 10. 130
Solution 1:
$1\text{ MSD} = \frac{1\text{ cm}}{10} = 1\text{ mm} = 0.1\text{ cm}$.
$20\text{ VSD} = 19\text{ MSD} \implies 1\text{ VSD} = \frac{19}{20}\text{ MSD}$.
$\text{LC} = 1\text{ MSD} – 1\text{ VSD} = 1\text{ MSD} – \frac{19}{20}\text{ MSD} = \frac{1}{20}\text{ MSD} = \frac{1\text{ mm}}{20} = 0.05\text{ mm}$.
Correct Answer: (A)
Solution 2:
– $0.0050$: Leading zeros are not significant; trailing zero after decimal is significant $\implies 2$ SF.
– $2.040$: Embedded zero and trailing zero after decimal are significant $\implies 4$ SF.
– $400.0$: Decimal is present, so trailing zeros are significant $\implies 4$ SF.
Correct Answer: (A)
Solution 3:
First step (subtraction): $15.82 – 2.5 = 13.32$.
Since $2.5$ has $1$ decimal place, the result of subtraction is limited to $1$ decimal place ($13.3$).
Second step (multiplication): $13.3 \times 0.20$.
$13.3$ has $3$ SF and $0.20$ has $2$ SF. The final product must have $2$ significant figures:
$13.3 \times 0.20 = 2.66 \implies \text{rounded to 2 SF} = 2.7$.
Correct Answer: (C)
Solution 4:
$\text{LC} = \frac{\text{Pitch}}{\text{CSD}} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm}$.
Zero error: The $4\text{th}$ division lies below the reference line, so the error is negative:
$\text{Zero Error} = -4 \times \text{LC} = -4 \times 0.005\text{ mm} = -0.020\text{ mm}$.
Observed reading:
$\text{Observed} = \text{LSR} + (\text{CSR} \times \text{LC}) = 2\text{ mm} + (46 \times 0.005\text{ mm}) = 2\text{ mm} + 0.230\text{ mm} = 2.230\text{ mm}$.
True thickness:
$\text{True Reading} = 2.230\text{ mm} – (-0.020\text{ mm}) = 2.250\text{ mm} = 2.25\text{ mm}$.
Correct Answer: (A)
Solution 5:
$8.7\text{ g} – 3.924\text{ g} = 4.776\text{ g}$.
In subtraction, the result can only have as many decimal places as the number with the least decimal places ($8.7$ has 1 decimal place).
Rounding $4.776$ to 1 decimal place gives $4.8\text{ g}$.
Correct Answer: (C)
Solution 6:
(A) True: Converting $2.4\text{ m}$ to $240\text{ cm}$ retains 2 SF ($2.4 \times 10^2\text{ cm}$).
(B) True: Absolute resolution limit is bounded by instrument least count.
(C) True: In $3.2000 \times 10^3$, all digits before the exponent are significant (5 SF).
(D) True: More significant figures reflect smaller relative uncertainty and higher precision.
Correct Answer: (A, B, C, D)
Solution 7:
$(N + 1)\text{ VSD} = N\text{ MSD} \implies 1\text{ VSD} = \frac{N}{N + 1}\text{ MSD} = \frac{N}{N + 1}\text{ mm}$.
$\text{LC} = 1\text{ MSD} – 1\text{ VSD} = 1 – \frac{N}{N + 1} = \frac{1}{N + 1}\text{ mm}$.
Since $\frac{N}{N + 1} < 1$, $1\text{ VSD} < 1\text{ MSD}$.
Total reading is still formed by $\text{MSR} + (\text{VSR} \times \text{LC})$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
$\text{LC} = \frac{1\text{ mm}}{100} = 0.01\text{ mm}$.
An instrument with $\text{LC} = 0.01\text{ mm}$ can directly measure any value that is an integer multiple of $0.01\text{ mm}$ up to two decimal places in mm.
– $5.42\text{ mm}$ is $542 \times 0.01\text{ mm}$ → Valid.
– $0.07\text{ mm}$ is $7 \times 0.01\text{ mm}$ → Valid.
– $3.00\text{ mm}$ is $300 \times 0.01\text{ mm}$ → Valid.
– $5.425\text{ mm}$ requires estimation of half a division (third decimal place) → Not directly measurable.
Correct Answer: (A, C, D)
Solution 9:
$\text{LC} = \frac{1\text{ MSD}}{20} = \frac{0.5\text{ mm}}{20} = 0.025\text{ mm}$.
Main scale reading: $\text{MSR} = 2.5\text{ cm} = 25.0\text{ mm}$.
Vernier scale reading: $\text{VSR} = 14 \times 0.025\text{ mm} = 0.35\text{ mm}$.
Observed reading: $25.0 + 0.35 = 25.35\text{ mm}$.
Zero error is $+0.05\text{ mm}$.
Corrected diameter: $D = 25.35 – 0.05 = 25.30\text{ mm} = 25.3\text{ mm}$.
Therefore, $D \times 10 = 25.3 \times 10 = 253$.
Correct Answer: 253
Solution 10:
Volume of cube $V = l^3 = (2.0\text{ cm})^3 = 8.0\text{ cm}^3$ (2 significant figures).
Mass $m = 10.00\text{ g}$ (4 significant figures).
Density $d = \frac{m}{V} = \frac{10.00}{8.0} = 1.25\text{ g/cm}^3$.
Rounding to 2 significant figures (since $V$ has 2 SF):
Following rounding off rules for $5$ preceded by an even number ($2$ is even), $1.25 \to 1.2$ or rounding normally to nearest even gives $1.2\text{ g/cm}^3$ (or standard rounding $1.3\text{ g/cm}^3$). With standard nearest rounding: $d = 1.3\text{ g/cm}^3$.
Thus, $100 \times d = 100 \times 1.3 = 130$.
Correct Answer: 130