Physics Q27 – JEE-MAIN 2026

Home Physics Units & MeasurementsQuestion #548

The diameter of a wire measured by a screw gauge of least count

0.001 cm

0.08 cm

. The length measured by a scale of least count

0.1 cm

150 cm

. When a weight of

100 N

is applied to the wire, the extension in length is

0.5 cm

, measured by a micrometer of least count

0.001 cm

. The error in the measured Young's modulus is

α \times 10^{9} {N/m}^{2}

. The value of

α

is _______. (Ignore the contribution of the load to Young's modulus error calculation)

Clue 1 of 3: Error Propagation

Understand how errors in individual measurements propagate to the calculated quantity, Young's modulus.

🧠 Full Solution Path

Step 1: Identify the formula for Young's Modulus and its error propagation✦ Active

Young's modulus is given by $Y = \frac{Stress}{Strain} = \frac{F / A}{Δ l / L}$ . Since the area $A = \frac{π D^{2}}{4}$ , the formula becomes $Y = \frac{4 F L}{π D^{2} Δ l}$ . The fractional error in $Y$ is given by the sum of fractional errors of the independent variables. Ignoring the contribution of the load ( $F$ ) to the error, the formula for fractional error is:

\frac{Δ Y}{Y} = \frac{Δ L}{L} + 2 \frac{Δ D}{D} + \frac{Δ (Δ l)}{Δ l}

Step 2: Calculate the fractional errors for each measurement and the value of Young's Modulus○ Expand

Given the measurements and their least counts (absolute errors):

\frac{Δ L}{L} = \frac{0.1 cm}{150 cm} = \frac{1}{1500}

\frac{Δ D}{D} = \frac{0.001 cm}{0.08 cm} = \frac{1}{80}

\frac{Δ (Δ l)}{Δ l} = \frac{0.001 cm}{0.5 cm} = \frac{1}{500}

Substitute these into the fractional error formula for $Y$ :

\frac{Δ Y}{Y} = \frac{1}{1500} + 2 (\frac{1}{80}) + \frac{1}{500} = \frac{1}{1500} + \frac{1}{40} + \frac{1}{500}

Finding a common denominator (LCM = 6000):

\frac{Δ Y}{Y} = \frac{4}{6000} + \frac{150}{6000} + \frac{12}{6000} = \frac{4 + 150 + 12}{6000} = \frac{166}{6000} = \frac{83}{3000}

Now, calculate the value of Young's Modulus $Y$ using the given values in SI units:

F = 100 N

L = 150 cm = 1.5 m

D = 0.08 cm = 8 \times 10^{- 4} m

Δ l = 0.5 cm = 5 \times 10^{- 3} m

Y = \frac{4 \times 100 \times 1.5}{π \times (8 \times 10^{- 4})^{2} \times (5 \times 10^{- 3})} = \frac{600}{π \times 64 \times 10^{- 8} \times 5 \times 10^{- 3}} = \frac{600}{π \times 320 \times 10^{- 11}} = \frac{15}{8 π} \times 10^{11} {N/m}^{2}

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before calculating Y. Pay close attention to powers of 10.

Step 3: Calculate the absolute error

Δ Y

and determine

α

○ Expand

The absolute error $Δ Y$ is given by:

Δ Y = Y \times (\frac{Δ Y}{Y}) = (\frac{15}{8 π} \times 10^{11}) \times (\frac{83}{3000})

Δ Y = \frac{15 \times 83}{8 π \times 3000} \times 10^{11} = \frac{1245}{24000 π} \times 10^{11}

Using $π \approx 3.14159$ :

Δ Y \approx \frac{1245}{24000 \times 3.14159} \times 10^{11} \approx \frac{1245}{75398.16} \times 10^{11} \approx 0.016512 \times 10^{11} {N/m}^{2}

We need to express $Δ Y$ in the form $α \times 10^{9} {N/m}^{2}$ :

Δ Y = 0.016512 \times 10^{11} = 0.016512 \times 10^{2} \times 10^{9} = 1.6512 \times 10^{9} {N/m}^{2}

Comparing this with $α \times 10^{9} {N/m}^{2}$ , we find $α \approx 1.65$ .

💡 Teacher's Secret Hint

Double-check your arithmetic, especially when dealing with fractions and powers of 10. Round the final answer to two decimal places as suggested by the options.

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Physics Question 27 – JEE-MAIN 2026

Clue 1 of 3: Error Propagation

🧠 Full Solution Path