Questions & Answers: "error analysis"

To find the maximum percentage error in a product/quotient involving powers, we use the formula: $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}$. Note that errors always add up to find the 'maximum' possible error, and constants or denominators are treated with positive coefficients. Substituting the given percentage errors: Percentage error in $P = 3(1\%) + 2(3\%) + 0.5(4\%) + 1(2\%) = 3\% + 6\% + 2\% + 2\% = 13\%$. Therefore, 13% is the correct answer. The other options are incorrect as they result from arithmetic mistakes or failing to multiply by the correct exponents.

Density $\rho$ is defined as $\rho = \frac{Mass}{Volume}$. For a cube of side $L$, the volume $V = L^3$. Thus, $\rho = \frac{M}{L^3}$. The relative error in density is given by $\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L}$. Converting this to percentage error: Percentage error in $\rho = (\text{Percentage error in } M) + 3 \times (\text{Percentage error in } L)$. Substituting the values: $3\% + 3(2\%) = 3\% + 6\% = 9\%$. Option 5% is incorrect because it simply adds the errors without accounting for the cubic relationship of length. Option 7% is incorrect calculation, and 11% is irrelevant.

The formula for the time period of a simple pendulum is $T = 2\pi\sqrt{L/g}$, which gives $g = 4\pi^2 \frac{L}{T^2}$. The relative error in $g$ is $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}$. Given: $L = 100$ cm, $\Delta L = 1$ mm = $0.1$ cm. Total time $t = 200$ s for $n=100$ oscillations, so $T = t/n$ and $\Delta T/T = \Delta t/t$. Here $\Delta t = 1$ s. Substituting: $\frac{\Delta g}{g} = \frac{0.1}{100} + 2\left(\frac{1}{200}\right) = 0.001 + 0.01 = 0.011$. Expressed as a percentage: $0.011 \times 100 = 1.1\%$. Option 0.1% only considers length, and 2.1% or 0.6% result from incorrect application of the power rule for $T$ or arithmetic errors.

According to the rules for significant figures: (1) In $0.007$, the leading zeros are not significant; only the '7' is significant (1 figure). (2) In scientific notation $2.64 \times 10^{24}$, only the base number $2.64$ counts; the power of 10 is irrelevant for significance (3 figures). (3) In $0.2370$, the leading zero is not significant, but the trailing zero after the decimal point is significant (4 figures). Hence, the counts are 1, 3, and 4.

The relative error in $P$ is given by the sum of the relative errors of its components multiplied by their respective powers: $\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d}$. Substituting the given percentage errors: Percentage error in $P = (3 \times 1\%) + (2 \times 3\%) + (0.5 \times 4\%) + (1 \times 2\%) = 3\% + 6\% + 2\% + 2\% = 13\%$. Note that errors are always added to find the maximum possible error, even if the variable is in the denominator.

According to NCERT/scientific rounding rules: If the digit to be dropped is 5, then the preceding digit is left unchanged if it is even, and increased by 1 if it is odd. For $2.745$, the digit preceding 5 is '4' (even), so it remains $2.74$. For $2.735$, the digit preceding 5 is '3' (odd), so it is rounded up to $2.74$. Option A is correct.

Area $A = l \times b = 16.2 \times 10.1 = 163.62\text{ cm}^2$. Relative error $\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1} \approx 0.00617 + 0.00990 = 0.01607$. Absolute error $\Delta A = A \times 0.01607 = 163.62 \times 0.01607 \approx 2.63\text{ cm}^2$. Since the least significant figure in the measurements is at the tenths place and the product must follow the least significant figures of the factors (3 sig figs here), we round 163.62 to 164. The error is rounded to 3 to match the precision. Thus, $164 \pm 3$ is the most appropriate representation according to NCERT guidelines for propagation and rounding.

1. Find the mean: $g_{mean} = (9.8 + 9.7 + 10.0 + 9.9 + 9.6)/5 = 49.0 / 5 = 9.8\text{ m/s}^2$. 2. Find absolute errors for each: $|9.8-9.8|=0, |9.7-9.8|=0.1, |10.0-9.8|=0.2, |9.9-9.8|=0.1, |9.6-9.8|=0.2$. 3. Mean absolute error = $(0 + 0.1 + 0.2 + 0.1 + 0.2) / 5 = 0.6 / 5 = 0.12\text{ m/s}^2$.

The volume of a sphere is $V = \frac{4}{3}\pi r^3$. The percentage error in $V$ is $\frac{\Delta V}{V} \times 100 = 3 \left( \frac{\Delta r}{r} \times 100 \right)$. Given the percentage error in $r$ is $2\%$, the percentage error in $V = 3 \times 2\% = 6\%$. Constants like $4/3$ and $\pi$ do not contribute to the error.

For division $R = V/I$, the relative error is the sum of the relative errors of the components: $\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$. Percentage error in $R = \left( \frac{5}{100} \times 100 \right) + \left( \frac{0.2}{10} \times 100 \right) = 5\% + 2\% = 7\%$. Even though the quantities are divided, their percentage errors are added.

Density $\rho = \text{mass}/\text{volume}$. The relative error in density is $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}$. Plugging in the values: $\frac{\Delta \rho}{\rho} = \frac{0.05}{5.00} + \frac{0.05}{1.00} = 0.01 + 0.05 = 0.06$. Multiplying by $100$ to get the percentage error gives $6\%$. This represents the maximum possible uncertainty in the density value.

Mean $T = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5 = 13.12 / 5 = 2.624 \approx 2.62\text{ s}$. Absolute errors: $|2.62-2.63|=0.01, |2.62-2.56|=0.06, |2.62-2.42|=0.20, |2.62-2.71|=0.09, |2.62-2.80|=0.18$. Mean absolute error $\Delta T = (0.01+0.06+0.20+0.09+0.18)/5 = 0.54/5 = 0.108 \approx 0.11\text{ s}$. Relative error = $\Delta T / T = 0.11 / 2.62 \approx 0.0419$. Rounding to one or two significant figures common in error analysis, $0.04$ is the most appropriate option.

Systematic errors are those that tend to be in one direction (either positive or negative) and can often be identified and corrected (instrumental errors, personal errors, imperfection in technique). Random errors, however, are those that occur irregularly and are due to random and unpredictable fluctuations in experimental conditions like temperature, voltage, or mechanical vibrations. Therefore, random fluctuations in line voltage are classified as random errors, not systematic errors.