In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a condu

Q: Why do we multiply the error in diameter by 2?

In the formula for area, the diameter is squared ($d^2$). According to the rules of error propagation, the relative error of a quantity raised to a power is multiplied by that power.

Q: How did we know the absolute error for voltage is 0.1 V?

When a measurement is given as $5.0\text{ V}$, the last decimal place indicates the precision of the instrument used. Therefore, the least count, or absolute error, is $0.1\text{ V}$.

Q: Do we need to convert the units of length and diameter to meters?

No, it's not necessary here. We are calculating relative errors (like $\frac{\Delta L}{L}$), which are ratios of the same unit. The units cancel out, making the relative error dimensionless.

Q: Why do we add the relative errors instead of subtracting them?

We want to find the *maximum permissible* error. To ensure we account for the worst-case scenario where all errors compound, we always add the absolute values of the relative errors.

Q: What happens to the constants $\pi$ and 4 in the error equation?

Constants have exact values, meaning their absolute error is zero. Therefore, they do not contribute to the relative error of the final calculated quantity.

Question

In the experiment of Ohm's law, a potential difference of $5.0	ext{ V}$ is applied across the end of a conductor of length $10.0	ext{ cm}$ and diameter of $5.00	ext{ mm}$. The measured current in the conductor is $2.00	ext{ A}$. The maximum permissible percentage error in the resistivity of the conductor is

Options:

(A) 3.9

(B) 8.4

(C) 7.5

(D) 3.0

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a physics lab, conducting the classic Ohm's law experiment. You have a cylindrical wire, and you are measuring its physical dimensions and electrical properties. The problem gives us four measured quantities: the potential difference $V = 5.0	ext{ V}$, the current $I = 2.00	ext{ A}$, the length $L = 10.0	ext{ cm}$, and the diameter $d = 5.00	ext{ mm}$.

Our goal is to find the maximum permissible percentage error in the calculated resistivity ($ho$) of this conductor. To do this, we first need to establish a mathematical relationship between resistivity and the quantities we have measured.

### The Master Equation
We know two fundamental equations for resistance ($R$). From the physical properties of the wire, $R = \frac{ho L}{A}$, where $A$ is the cross-sectional area. Since the wire is cylindrical, $A = \frac{\pi d^2}{4}$.

From Ohm's law, we also know that $R = \frac{V}{I}$.

By equating these two expressions, we get:

$$\frac{ho L}{\left(\frac{\pi d^2}{4}ight)} = \frac{V}{I}$$

Rearranging this to solve for resistivity ($ho$), we obtain our master equation:

$$ho = \frac{V \pi d^2}{4 I L}$$

### The Catch: Deducing Absolute Errors
Here is where many students get stuck. The problem asks for the error, but it doesn't explicitly give us the absolute errors (like $\Delta V$ or $\Delta I$).

**This is a classic JEE trap.** When errors are not given, we must deduce them from the significant figures of the provided measurements. The least count of the measuring instrument is implied by the last decimal place of the recorded value.

- For $V = 5.0	ext{ V}$, the least count is $\Delta V = 0.1	ext{ V}$.
- For $I = 2.00	ext{ A}$, the least count is $\Delta I = 0.01	ext{ A}$.
- For $L = 10.0	ext{ cm}$, the least count is $\Delta L = 0.1	ext{ cm}$.
- For $d = 5.00	ext{ mm}$, the least count is $\Delta d = 0.01	ext{ mm}$.

### Error Analysis
Now, we apply the rules of error propagation to our master equation. For a quantity calculated by multiplication and division, the maximum relative error is the sum of the relative errors of the individual measured quantities. Constants like $\pi$ and $4$ have exact values, so their error is zero.

Crucially, because the diameter $d$ is squared in the formula, its relative error is multiplied by $2$:

$$\frac{\Delta ho}{ho} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}$$

### Final Calculation
Let's carefully substitute our values into the error equation. Notice that we don't need to convert units to SI because the relative errors are dimensionless ratios (e.g., $	ext{cm}/	ext{cm}$ cancels out).

$$\frac{\Delta ho}{ho} = \frac{0.1}{5.0} + \frac{0.01}{2.00} + 2\left(\frac{0.01}{5.00}ight) + \frac{0.1}{10.0}$$

Calculating each term individually:

$$\frac{\Delta ho}{ho} = 0.02 + 0.005 + 0.004 + 0.01$$

Adding them up gives the total relative error:

$$\frac{\Delta ho}{ho} = 0.039$$

To find the **percentage error**, we simply multiply the relative error by $100$:

$$	ext{\% Error} = 0.039 	imes 100 = 3.9\%$$

The maximum permissible percentage error in the resistivity of the conductor is **$3.9\%$**.

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Visualized Solution

$Given Parameters$

$Formulas for Resistance$

$Master Equation for Resistivity$

$Relative Error Equation$

$Deducing Absolute Errors$

$Substitution$

$Calculation$

$Final Percentage Error$

The Sigma Insight: Errors in Measurement

Analyzing the Setup Imagine you are in a physics lab, conducting the classic Ohm's law experiment

The Master Equation We know two fundamental equations for resistance ( $R$ )

The Catch

Error Analysis Now, we apply the rules of error propagation to our master equation

Final Calculation Let's carefully substitute our values into the error equation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formulas for Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Master Equation for Resistivity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Error Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Deducing Absolute Errors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Percentage Error