Propagation of errors explains how uncertainties in measured quantities affect the final result of a calculation. When physical quantities are combined through mathematical operations such as addition, subtraction, multiplication, division, or raising to powers, their individual errors also combine in a specific manner. Understanding the rules for error propagation in sum, difference, product, quotient, and powers is essential for determining the accuracy of derived results and is widely used in Class 11 Physics and Competitive numerical problem-solving.
Propagation(Combination) of Errors
In experimental physics and chemistry, measurements are never perfectly precise. Small errors arise due to limitations in instruments and human observations. Understanding how these errors propagate when performing mathematical operations is essential for accurate calculations. The following detailed explanation covers different types of error propagation, including sum, difference, product, quotient, and powers of measured quantities.
* Note : We shall now calculate the maximum permissible error in different cases.
Error in Sum of Quantities
When the result involves the sum of two observed quantities. Suppose that the result X is given as the sum of two observed quantities a and b i.e.
$$
X = a + b
$$
Let Δa and Δb be the absolute errors in the measurements of quantities a and b. Then, the values of a and b should be recorded as a ± Δa and b ± Δb.
If ΔX is the absolute error in the result, then
$$
X \pm \Delta X = (a \pm \Delta a) + (b \pm \Delta b)
$$
or
$$
X \pm \Delta X = (a + b) \pm (\Delta a + \Delta b)
$$
or
$$
\pm \Delta X = \pm (\Delta a + \Delta b)
$$
Therefore, the maximum possible error in X,
$$
\Delta X = \Delta a + \Delta b \quad \dots(1)
$$
Thus, the maximum absolute error in X
$$
= \text{maximum absolute error in } a + \text{maximum absolute error in } b
$$
🔖 Key Point
The total absolute error in Sum of Quantities is the sum of absolute errors in that Quantities.
Percentage Error in Sum of Quantities
The formula for percentage error in sum of quantities is given by :
\begin{array}{l} \%\text{ error in } X = \left( \dfrac{\Delta a + \Delta b}{a + b} \right) \times 100 \end{array}
Error in Difference of Quantities
When the result involves the difference of two observed quantities. Suppose that
$$
X = a – b
$$
Let $\Delta a$ and $\Delta b$ be the absolute errors in the measurements of $a$ and $b$ and $\Delta X$ be the maximum error in X. Then,
$$
X \pm \Delta X = (a \pm \Delta a) – (b \pm \Delta b)
$$
$$
X \pm \Delta X = (a – b) \pm \Delta a \mp \Delta b
$$
$$
\pm \Delta X = \Delta a \mp \Delta b
$$
Therefore, the maximum possible error in X,
$$
\Delta X = \Delta a + \Delta b \quad \dots(2)
$$
Thus, the maximum absolute error in X
$$
= \text{maximum absolute error in } a + \text{maximum absolute error in } b
$$
🔖 Key Point
The total absolute error in Difference of Quantities is the sum of absolute errors in that Quantities.
Percentage Error in Difference of Quantities
The formula for percentage error in difference of quantities is given by :
\begin{array}{l} \%\text{ error in } X = \left( \dfrac{\Delta a + \Delta b}{a – b} \right) \times 100 \end{array}
Therefore, from equations (1) and (2), it follows that when the result involves the sum or difference of two observed quantities, the absolute error in the result is equal to the sum of the absolute errors in the observed quantities.
🔖 Key Point
The absolute error in the result of addition or subtraction is equal to sum of the absolute errors in the data involved.
Error in Product of Quantities
When the result involves the product of two observed quantities. Suppose that
$$
X = a \times b
$$
Let $\Delta a$ and $\Delta b$ be the absolute errors in the measurement of quantities $a$ and $b$ and $\Delta X$ be the maximum possible error in X. Then,
$$
X \pm \Delta X = (a \pm \Delta a) \times (b \pm \Delta b)
$$
$$
= ab \pm b \Delta a \pm a \Delta b \pm \Delta a \Delta b
$$
or
$$
\pm \Delta X = \pm b \Delta a \pm a \Delta b \pm \Delta a \Delta b
$$
Dividing both sides by $X = a \times b$, we get
$$
\pm \frac{\Delta X}{X} = \pm \frac{\Delta a}{a} \pm \frac{\Delta b}{b} \pm \frac{\Delta a}{a} \times \frac{\Delta b}{b}
$$
$\Delta a/a$, $\Delta b/b$ and $\Delta X / X$ are known as relative errors or fractional errors in the values of $a$, $b$ and X respectively. The products of relative errors in $a$ and $b$ i.e. $\Delta a / a \times \Delta b / b$ will be very small and can be neglected. Therefore, the maximum possible relative error in X,
$$
\frac{\Delta X}{X} = \frac{\Delta a}{a} + \frac{\Delta b}{b} \quad \dots(3)
$$
Thus, the maximum relative error in X
$$
= \text{maximum relative error in } a + \text{maximum relative error in } b
$$
The maximum percentage error in X is given by
$$
\frac{\Delta X}{X} \times 100 = \frac{\Delta a}{a} \times 100 + \frac{\Delta b}{b} \times 100 \quad \dots(4)
$$
i.e. the maximum percentage error in X
$$
= \text{maximum percentage error in } a + \text{maximum percentage error in } b
$$
🔖 Key Point
The total percentage error in Product of Quantities is the sum of percentage errors in that Quantities.
Aliter. The above result can be obtained by logarithmic differentiation of $X = a \times b$ as explained below :
Taking log of both sides, we have
$$
\log X = \log a + \log b
$$
Differentiating both sides, we get
$$
\frac{\Delta X}{X} = \frac{\Delta a}{a} + \frac{\Delta b}{b}
$$
This result is exactly the same as obtained above.
Error in Division of Quantities
When the result involves the quotient of two observed quantities. Suppose that
$$
X = \frac{a}{b}
$$
Let $a$ and $\Delta b$ be the absolute errors in the measurement of quantities $a$ and $b$ and $\Delta X$ be the maximum possible error in $X$. Then,
$$
X \pm \Delta X = \frac{a \pm \Delta a}{b \pm \Delta b} = (a \pm \Delta a)(b \pm \Delta b)^{-1}
$$
$$
X \pm \Delta X = \frac{a}{b} \left( 1 \pm \frac{\Delta a}{a} \right) \left( 1 \pm \frac{\Delta b}{b} \right)^{-1}
$$
$$
X \pm \Delta X = \frac{a}{b} \left( 1 \pm \frac{\Delta a}{a} \right) \left( 1 + \pm \text{terms containing higher powers of } \frac{\Delta b}{b} \right)
$$
Since $\Delta b$ is small, terms containing higher powers of $\dfrac{\Delta b}{b}$ can be neglected.
$$
\therefore X \pm \Delta X = \frac{a}{b} \left( 1 \pm \frac{\Delta a}{a} \right) \left( 1 \pm \frac{\Delta b}{b} \right) = \frac{a}{b} \left( 1 \pm \frac{\Delta a}{a} \mp \frac{\Delta b}{b} \mp \frac{\Delta a \times \Delta b}{a \times b} \right)
$$
Again, as the factor $\dfrac{\Delta a \times \Delta b}{a \times b}$ is very small, it can be neglected and we have
$$
X \pm \Delta X = \frac{a}{b} \left[ 1 \pm \frac{\Delta a}{a} \mp \frac{\Delta b}{b} \right]
$$
But $\dfrac{a}{b} = X$
$$
X \pm \Delta X = X \left[ 1 \pm \frac{\Delta a}{a} \mp \frac{\Delta b}{b} \right]
$$
or
$$
\frac{X \pm \Delta X}{X} = 1 \pm \frac{\Delta a}{a} \mp \frac{\Delta b}{b}
$$
or
$$
\pm \frac{\Delta X}{X} = \pm \frac{\Delta a}{a} \mp \frac{\Delta b}{b}
$$
Therefore, the maximum possible relative error in $X$,
$$
\frac{\Delta X}{X} = \frac{\Delta a}{a} + \frac{\Delta b}{b} \quad \dots(5)
$$
Thus, the maximum relative error in $X$
$$
= \text{maximum relative error in } a + \text{maximum relative error in } b
$$
Also, the maximum percentage error in $X$,
$$
\frac{\Delta X}{X} \times 100 = \frac{\Delta a}{a} \times 100 + \frac{\Delta b}{b} \times 100 \quad \dots(6)
$$
i.e. the maximum percentage error in $X$
$$
= \text{maximum percentage error in } a + \text{maximum percentage error in } b
$$
🔖 Key Point
The total percentage error in quotient of Quantities is the sum of percentage errors in that Quantities.
Therefore, from equations (3), (4), (5) and (6), it follows that when the result involves the multiplication or quotient of two observed quantities, the maximum possible relative (or percentage) error in the result is equal to the sum of the relative (or percentage) errors in the observed quantities.
Error in Quantity Raised to a Power
When the result involves the product of the powers of observed quantities.
We discuss it in the following two parts :
(i) Suppose that $X = a^n$
Differentiating both sides, we have
$$
\Delta X = n a^{n-1} \Delta a
$$
Dividing both sides by $X = a^n$, we have
$$
\frac{\Delta X}{X} = \frac{n a^{n-1} \Delta a}{a^n} \quad \text{or} \quad \frac{\Delta X}{X} = n \frac{\Delta a}{a}
$$
Thus, the relative error in $a^n$ is $n$ times the relative error $a$.
(ii) Suppose that $X = \dfrac{a^l b^m}{c^n}$
Then, it can be proved that the maximum relative error in $X$,
$$
\frac{\Delta X}{X} = l \frac{\Delta a}{a} + m \frac{\Delta b}{b} + n \frac{\Delta c}{c} \quad \dots(7)
$$
Thus, the maximum relative error in X = l times the maximum relative error in a + m times the maximum relative error in b + n times the maximum relative error in c
Also, the maximum percentage error in X,
$$
\frac{\Delta X}{X} \times 100 = l \frac{\Delta a}{a} \times 100 + m \frac{\Delta b}{b} + n \frac{\Delta c}{c} \times 100 \quad \dots(8)
$$
i.e. the maximum percentage error in $X$ = $l$ times the maximum percentage error in $a$ + $m$ times the maximum percentage error in $b$ + $n$ times the maximum percentage error in $c$.
From this discussion, it follows that in experiments the quantity, which occurs with higher power in the formula for the result, should be measured with the maximum accuracy.
Solved Numerical Problems
Q.1 The length and breadth of a rectangular object area 25.2 cm and 16.8 cm respectively and have been measured to an accuracy of 0.1 cm. Find the percentage error in the area of the object.
Solution. Now, $A = l \times b$
$$
\frac{\Delta A}{A} \times 100 = \frac{\Delta l}{l} \times 100 + \frac{\Delta b}{b} \times 100
$$
or
$$
\% \text{ error in area} = \frac{0.1}{25.2} \times 100 + \frac{0.1}{16.8} \times 100
$$
$$
\% \text{ error in area} = 0.4\% + 0.6\% = 1.0\%
$$
Q.2. A physical quantity X is given by
$$
X = \dfrac{a^2 b^3}{c \sqrt{d}}
$$
If the percentage errors of measurements in a, b, c and d are 4%, 2%, 3% and 1% respectively, then calculate the percentage error in X.
Solution. Here, $X = \dfrac{a^2 b^3}{c \sqrt{d}}$
$$
\frac{\Delta X}{X} = 2 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + 1 \frac{\Delta c}{c} + \frac{1}{2} × \frac{\Delta d}{d}
$$
∴ % error in X = 2 × % error in a + 3 × % error in b + % error in c + 1/2 × % error in d
$$
\% \text{error in X} = 2 \times 4\% + 3 \times 2\% + 3\% + \frac{1}{2} \times 1\%
$$
$$
\% \text{error in X} = 17.5\%
$$
Q.3. The lengths of two rods are recorded as $l_{1} = 25.2 \pm 0.1\text{ cm}$ and $l_{2} = 16.8 \pm 0.1\text{ cm}$. Find the sum of the lengths of the two rods with the limits of error.
Solution. According to the theory of errors, when adding measurements, the absolute errors are added together to find the uncertainty in the sum.
$$l_{1} + l_{2} = (25.2 \pm 0.1) + (16.8 \pm 0.1)$$
Sum of measured values: $25.2 + 16.8 = 42.0\text{ cm}$.
Sum of absolute errors: $0.1 + 0.1 = 0.2\text{ cm}$.
As per the rules of significant figures in addition, we retain the number of decimal places of the observation with the fewest decimal places. Since both $25.2$ and $16.8$ are known to the first decimal place, the result $42.0$ is correctly expressed.
$$l_{1} + l_{2} = 42.0 \pm 0.2\text{ cm}$$
Q.4. The initial temperature of a liquid is recorded as $\theta_{1} = 25.4 \pm 0.1\text{°C}$ and on heating, its final temperature is recorded as $\theta_{2} = 52.7 \pm 0.1\text{°C}$. Find the increase in temperature.
Solution. The increase in temperature is found by calculating the difference between the final and initial temperatures. According to the theory of errors, while the measured values are subtracted, their absolute errors are always added to determine the maximum possible error in the result.
$$\text{Increase in temperature} = \theta_{2} – \theta_{1}$$
$$\theta_{2} – \theta_{1} = (52.7 \pm 0.1) – (25.4 \pm 0.1)$$
Difference of measured values: $52.7 – 25.4 = 27.3\text{°C}$.
Sum of absolute errors: $0.1 + 0.1 = 0.2\text{°C}$.
In accordance with the rules for significant figures in subtraction, we retain the number of decimal places of the observation with the fewest decimal places. Since both temperatures are known to the first decimal place, the result $27.3$ is correctly expressed.
$$\text{Increase in temperature} = 27.3 \pm 0.2\text{°C}$$
Q.5. A potential difference of $V = 100 \pm 2\text{ volt}$, when applied across a resistance $R$ gives a current of $10 \pm 0.5\text{ ampere}$. Calculate percentage error in $R$ given by $R = \frac{V}{I}$.
Solution. When the values of different observations are divided, the number of digits to be retained in the answer depends upon the number of significant figures in the weakest link. For percentage errors, the rule states that the percentage error of a quotient is the sum of the percentage errors of the individual multipliers.
1. Expressing limits of error as percentage error:
For Voltage ($V$):$$V = 100\text{ volt} \pm \frac{2}{100} \times 100\% = 100\text{ volt} \pm 2\%$$
For Current ($I$):$$I = 10\text{ ampere} \pm \frac{0.5}{10} \times 100\% = 10\text{ ampere} \pm 5\%$$
2. Calculating percentage error in Resistance ($R$):
Since $R = \frac{V}{I}$, the percentage error in $R$ is calculated by adding the percentage errors of $V$ and $I$:
$$\%\text{ error in } R = \%\text{ error in } V + \%\text{ error in } I$$
$$\%\text{ error in } R = 2\% + 5\% = 7\%$$
The percentage error in $R$ is $7\%$.
Q6. To study the flow of a liquid through a narrow tube, the following formula is used:
$$\eta = \frac{\pi p r^4}{8 V l}$$
where the letters have their usual meanings. The values of $p, r, V$ and $l$ are measured to be $76\text{ cm of Hg}$, $0.28\text{ cm}$, $1.2\text{ cm}^3 \text{ s}^{-1}$ and $18.2\text{ cm}$ respectively. If these quantities are measured to the accuracies of $0.5\text{ cm of Hg}$, $0.01\text{ cm}$, $0.1\text{ cm}^3 \text{ s}^{-1}$ and $0.1\text{ cm}$ respectively, find the percentage error in the value of $\eta$.
Solution. The formula for percentage error in $\eta$ is:
$$\frac{\Delta \eta}{\eta} \times 100 = \left( \frac{\Delta p}{p} + 4 \times \frac{\Delta r}{r} + \frac{\Delta V}{V} + \frac{\Delta l}{l} \right) \times 100$$
Substituting the given values:
For $p$: $\frac{0.5}{76} \times 100 = 0.66\%$
For $r$: $4 \times \frac{0.01}{0.28} \times 100 = 14.29\%$
For $V$: $\frac{0.1}{1.2} \times 100 = 8.33\%$
For $l$: $\frac{0.1}{18.2} \times 100 = 0.55\%$
Calculation:
$$\%\text{ error in } \eta = 0.66\% + 14.29\% + 8.33\% + 0.55\%$$
$$\%\text{ error in } \eta = 23.83\%$$
The percentage error in the value of $\eta$ is $23.83\%$.
Q7. In a simple pendulum experiment, the length of the pendulum is $90.6 \times 10^{-2}\text{ m}$. The time period is $1.91\text{ s}$. Write the value of acceleration due to gravity to correct significant figures and round it off.
Solution. The acceleration due to gravity ($g$) for a simple pendulum is given by the formula:
$$g = 4\pi^2 \frac{l}{T^2}$$
Given values:
Length ($l$) = $90.6 \times 10^{-2}\text{ m}$
Time period ($T$) = $1.91\text{ s}$
$$g = \frac{4\pi^2 \times 90.6 \times 10^{-2}}{(1.91)^2}$$
Since both the measurements have three significant figures, rounding off the result to three significant figures, we have
$$g = 9.8044\text{ m s}^{-2} $$
Q.8. The centripetal force is given by $F = \dfrac{mv^2}{r}$. The mass ($m$), velocity ($v$), and radius ($r$) of the circular path of an object are $0.5\text{ kg}$, $10\text{ m s}^{-1}$, and $0.4\text{ m}$ respectively. If $m$, $v$, and $r$ are measured to accuracies of $0.005\text{ kg}$, $0.01\text{ m s}^{-1}$, and $0.01\text{ m}$ respectively, find the percentage error in the force acting on the body.
Solution. Given formula $F = \dfrac{mv^2}{r}$,
1. Calculate individual percentage errors:
% error in $m$:$$\frac{\Delta m}{m} \times 100 = \frac{0.005}{0.5} \times 100\% = 1\%$$
% error in $v$:$$\frac{\Delta v}{v} \times 100 = \frac{0.01}{10} \times 100\% = 0.1\%$$
% error in $r$:$$\frac{\Delta r}{r} \times 100 = \frac{0.01}{0.4} \times 100\% = 2.5\%$$
2. Calculate percentage error in Force ($F$):
Using the formula $F = \dfrac{mv^2}{r}$, the percentage error is calculated as:
$$\%\text{ error in } F = (\%\text{ error in } m) + 2 \times (\%\text{ error in } v) + (\%\text{ error in } r)$$
Substituting the values:
$$\%\text{ error in } F = 1\% + 2 \times (0.1\%) + 2.5\%$$
$$\%\text{ error in } F = 1\% + 0.2\% + 2.5\% = 3.7\%$$
The percentage error in the centripetal force is $3.7\%$.
Q.9. The time period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt{\dfrac{L}{g}}$. $L$ is about $10\text{ cm}$ and is known to $1\text{ mm}$ accuracy. The time period of oscillation is about $0.5\text{ s}$. The time of $100$ oscillations is measured with a wrist watch of $1\text{ s}$ resolution. What is the accuracy in determination of $g$?
Solution.
1. Calculate Percentage Error in Length ($L$):
Measured Length ($L$) = $10\text{ cm}$
Accuracy ($\Delta L$) = $1\text{ mm} = 0.1\text{ cm}$$$\%\text{ error in } L = \frac{\Delta L}{L} \times 100\% = \frac{0.1}{10} \times 100\% = 1\%$$
2. Calculate Percentage Error in Time Period ($T$):
Time for $100$ oscillations ($t$) = $100 \times 0.5\text{ s} = 50\text{ s}$
Resolution of watch ($\Delta t$) = $1\text{ s}$
The percentage error in the time period $T$ is the same as the percentage error in the total measured time $t$.$$\%\text{ error in } T = \frac{\Delta t}{t} \times 100\% = \frac{1}{50} \times 100\% = 2\%$$
3. Calculate Percentage Error in $g$:
Rearranging the formula $T = 2\pi \sqrt{\dfrac{L}{g}}$ gives $g = \dfrac{4\pi^2 L}{T^2}$.
The percentage error in $g$ is the sum of the percentage error in $L$ and twice the percentage error in $T$:
$$\%\text{ error in } g = (\%\text{ error in } L) + 2 \times (\%\text{ error in } T)$$
$$\%\text{ error in } g = 1\% + 2 \times (2\%) = 1\% + 4\% = 5\%$$
The accuracy (percentage error) in the determination of $g$ is $5\%$.
Q.10. A capacitor of capacitance C = (2.0 ± 0.1) µF is charged to a voltage V = (20 ± 0.2) V. What will be the charge Q on the capacitor?
Solution: Charge on capacitor, Q = CV
Q = CV = 2.0×10-6 × 20 C = 4.0×10-5 Coulomb.
Relative error in C = (ΔC/C) = (0.1/2)
Percentage error in C = (0.1/2) ×100 =5 %
Relative error in V = (ΔV/V) = (0.2/20)
Percentage error in V = (0.2/20)×100 =1%
Charge on capacitor, Q = CV
(ΔQ/Q) = (ΔC/C) + (ΔV/V)
Percentage error in Q = 5%+1% = 6%
Charge = 4.0×10-5 ± 6% Coulomb = (4.0±0.24)×10-5 Coulomb
Q.11. Two resistances R1=(100±5) ohm and R2=(200±10) ohm are connected in series. Find the equivalent resistance of the series combination.
Solution: Since, it is known that,
Equivalent resistance = R= R1+R2
Given that, the resistance is:
R1 = (100 ± 5) ohm
R2 = (200 ±10) ohm
Therefore,
R = (100 ± 5) + (200 ± 10)
R = (300 ± 15) ohm
Q.12. The mass and the length of one side of a cube are measured and its density is calculated. If the percentage errors in the measurement of mass and length are 1% and 2% respectively, then what is the percentage error in the density?
Solution: If the mass of the cube is m and the length of its one side is l, then its density,
d = m/l3
So, (Δd/d) = (Δm/m) + 3(Δl/l)
Thus, Percentage error in density = (1+3×2)% = 7%
Q.13. In an experiment, the values of two resistances were measured to be as given below :
R1=5± 0·2 ohm; R2 = 10± 0·1 ohm
Find the values of total resistance in (i) series and (ii) parallel with limits of possible percentage error in each case.
Solution :
Resistance 1 ($R_1$): $5 \pm 0.2\text{ ohm}$
As percentage error: $5\text{ ohm} \pm (\frac{0.2}{5} \times 100\%) = 5\text{ ohm} \pm 4\%$
Resistance 2 ($R_2$): $10 \pm 0.1\text{ ohm}$
As percentage error: $10\text{ ohm} \pm (\frac{0.1}{10} \times 100\%) = 10\text{ ohm} \pm 1\%$
(i) Resistance in Series ($R_s$)
When resistors are connected in series, the total resistance is the sum of individual resistances. In addition, the absolute errors are added together.
$$R_s = R_1 + R_2$$
$$R_s = (5 \pm 0.2) + (10 \pm 0.1)$$
Sum of values: $5 + 10 = 15\text{ ohm}$
Sum of absolute errors: $0.2 + 0.1 = 0.3\text{ ohm}$
Expressing $R_s$ with percentage error:
$$\%\text{ error in } R_s = \frac{0.3}{15} \times 100\% = 2\%$$
Final Series Value: $15\text{ ohm} \pm 2\%$
(ii) Resistance in Parallel ($R_p$)
The formula for two resistors in parallel is $R_p = \dfrac{R_1 R_2}{R_1 + R_2}$. To find the percentage error, we apply the rules for multiplication and division, where percentage errors are added.
1. Calculate the Product ($R_1 R_2$):
Value: $5 \times 10 = 50\text{ ohm}^2$
Percentage Error: $4\% + 1\% = 5\%$
Product $= 50\text{ ohm}^2 \pm 5\%$
2. Calculate the Sum ($R_1 + R_2$):
From the series calculation: $15\text{ ohm} \pm 2\%$
3. Calculate the Quotient ($R_p = \dfrac{\text{Product}}{\text{Sum}}$):
Value: $R_p =\dfrac{50}{15} = 3.333… \approx 3.3\text{ ohm}$
Total Percentage Error: $(\%\text{ error in Product}) + (\%\text{ error in Sum})$
$$\%\text{ error in } R_p = 5\% + 2\% = 7\%$$
Final Parallel Value: $R_p = 3.3\text{ ohm} \pm 7\%$
Q.14. The resistance R = V/I where V = (200 ± 5) V and I = (20 ± 0.2) A. Find the percentage error in R.
Solution:
Relative error in V = (ΔV/V) = (5/200)
Percentage error in V = (5/200)×100% = 2.5%
Relative error in I = (ΔI/I) = (0.2/20)
Percentage error in I = (0.2/20) ×100% = 1%
So, Percentage error in R = 2.5%+1% = 3.5%
Why do we always add absolute errors in sums and differences?
Since errors can be positive or negative, taking their sum ensures that the total uncertainty accounts for the worst possible deviation.
How does power affect error propagation?
When a quantity is raised to a power, the percentage error gets multiplied by that power, making careful measurements crucial.
Why is error in division treated the same way as multiplication?
Fractional errors always add up whether quantities are multiplied or divided.
Explain why fractional errors are added in multiplication and division.
When quantities are multiplied or divided, small changes in one quantity proportionally affect the result, leading to the sum of fractional errors.
Why should we measure the variable with the highest power most accurately?
The percentage error contribution increases with power, so errors in such variables amplify the overall error.
If all measurements in an experiment are performed up to the same number of times, then a maximum error occurs due to which measurement?
The maximum error occurs due to the measurement of the quantity which appears with maximum power in the formula. If all the quantities in the formula have the same powers, then a maximum error occurs due to the measurement of the quantity whose magnitude is least.
If the length of the pencil is given by (4.16 ± 0.01) cm. What does it mean?
It means that the true value of the length of the pencil is unlikely to be less than 4.15 cm or greater than 4.17 cm.
Units and Measurements – Chapter Summary
The chapter Units and Measurements forms the base of physics for JEE exam by explaining how physical quantities are defined, measured, and expressed in a standardized way. It starts with the concept of physical quantities and their classification into fundamental and derived quantities, followed by a detailed study of SI units and other systems like CGS, MKS, and FPS. Proper rules for writing units, prefixes, and unit conversions are also covered to ensure consistency in measurements.
The chapter then focuses on measurement techniques, including accuracy, precision, and different types of errors such as absolute error, relative error, percentage error, and propagation of errors. It also introduces significant figures and rounding-off rules, which are essential for maintaining the correct level of precision in calculations.