The density ρ of a uniform cylinder is determined by measuring its mass m, length l and diameter d. The measured values of m, l and d are 97.42 ± 0.02 g, 8.35 ± 0.05 mm and 20.20 ± 0.02 mm, respectively. Calculated percentage fractional error in ρ is _______.

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q26 – JEE-MAIN 2026

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Home Physics Units & MeasurementsQuestion #570

The density

ρ

of a uniform cylinder is determined by measuring its mass

m

, length

l

and diameter

d

. The measured values of

m

l

and

d

are

97.42 \pm 0.02

8.35 \pm 0.05

mm and

20.20 \pm 0.02

mm, respectively. Calculated percentage fractional error in

ρ

is _______.

Clue 1 of 3: Understand the relationship

The density of a cylinder depends on its mass, length, and diameter. The formula for density needs to be established first.

🧠 Full Solution Path

Step 1: Formulate Density and Error Equation✦ Active

The density of a uniform cylinder is given by $ρ = \frac{m}{V}$ . The volume of a cylinder is $V = π r^{2} l$ . Since $r = \frac{d}{2}$ , the volume can be written as $V = π {(\frac{d}{2})}^{2} l = \frac{π d^{2} l}{4}$ . Substituting this into the density formula, we get:

ρ = \frac{m}{\frac{π d^{2} l}{4}} = \frac{4 m}{π d^{2} l}

Using the error propagation formula for a product/quotient, the fractional error in density is:

\frac{Δ ρ}{ρ} = \frac{Δ m}{m} + 2 \frac{Δ d}{d} + \frac{Δ l}{l}

Step 2: Calculate Individual Fractional Errors○ Expand

Calculate the fractional errors for mass, length, and diameter using the given measured values and their absolute errors:

\frac{Δ m}{m} = \frac{0.02}{97.42} \approx 0.0002053

\frac{Δ l}{l} = \frac{0.05}{8.35} \approx 0.0059880

\frac{Δ d}{d} = \frac{0.02}{20.20} \approx 0.0009901

💡 Teacher's Secret Hint

Ensure consistent units for calculations, though for fractional errors, the units cancel out.

Step 3: Calculate Total Percentage Fractional Error○ Expand

Substitute the individual fractional errors into the error propagation formula and multiply by $100 %$ to get the percentage error:

\frac{Δ ρ}{ρ} = 0.0002053 + 2 (0.0009901) + 0.0059880

\frac{Δ ρ}{ρ} = 0.0002053 + 0.0019802 + 0.0059880 = 0.0081735

Percentage fractional error $= \frac{Δ ρ}{ρ} \times 100 % = 0.0081735 \times 100 % \approx 0.81735 % \approx 0.82 %$ . Comparing this with the given options, option 2 is the closest.

💡 Teacher's Secret Hint

Pay close attention to the coefficient '2' for the diameter's fractional error, as diameter is squared in the volume formula.

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Physics Question 26 – JEE-MAIN 2026

Clue 1 of 3: Understand the relationship

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