Physics Q26 – JEE-MAIN 2026

Home Physics Units & MeasurementsQuestion #557

The percentage error in the calculated volume of a sphere, if there is

2 %

error in its diameter measurement, is _______.

Clue 1 of 3: Understand the relationship between volume, radius, and diameter

The volume of a sphere depends on the cube of its radius or diameter.

🧠 Full Solution Path

Step 1: Express Volume in terms of Diameter✦ Active

The volume of a sphere is given by $V = \frac{4}{3} π r^{3}$ . The diameter $D$ is related to the radius $r$ by $D = 2 r$ , so $r = \frac{D}{2}$ .

V = \frac{4}{3} π {(\frac{D}{2})}^{3} = \frac{4}{3} π \frac{D^{3}}{8} = \frac{1}{6} π D^{3}

Step 2: Apply Error Propagation Rule○ Expand

For a quantity $Q$ that depends on another quantity $X$ as $Q = k X^{n}$ (where $k$ is a constant), the relative error is given by $\frac{Δ Q}{Q} = n \frac{Δ X}{X}$ . In this case, $V \propto D^{3}$ , so $n = 3$ .

\frac{Δ V}{V} = 3 \frac{Δ D}{D}

💡 Teacher's Secret Hint

Remember that constants like $\frac{1}{6} π$ do not contribute to the relative error.

Step 3: Calculate Percentage Error in Volume○ Expand

The percentage error in volume is obtained by multiplying the relative error by $100 %$ . We are given that the percentage error in diameter is $2 %$ .

\frac{Δ V}{V} \times 100 % = 3 (\frac{Δ D}{D} \times 100 %) = 3 \times (2 %) = 6 %

💡 Teacher's Secret Hint

Ensure you convert relative errors to percentage errors correctly by multiplying by $100 %$ .

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Physics Question 26 – JEE-MAIN 2026

Clue 1 of 3: Understand the relationship between volume, radius, and diameter

🧠 Full Solution Path