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Units and Measurements question

2022 · 27 Jun · Shift 1 · Q44

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Units and Measurements
2022 · 27 Jun · Shift 1 · Q44

Units and Measurements question

2022 · 27 Jun · Shift 1 · Q44

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

A silver wire has a mass (0.6

\pm

0.006) g, radius (0.5

\pm

0.005) mm and length (4

\pm

0.04) cm. The maximum percentage error in the measurement of its density will be :

A
4%
B
3%
C
6%
D
7%

View written solutionFree

Correct answer: A

Write the formula for density

For a wire of cylindrical shape,

\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}

where:

$m$ = mass
$r$ = radius
$l$ = length

Use error propagation for maximum percentage error

If

\rho = \frac{m}{\pi r^2 l},

then the maximum fractional error is

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}

(Note: $\pi$ is a constant, so it contributes no error.)

Calculate individual percentage errors

(i) Mass

Given:

m = 0.6\,\text{g}, \qquad \Delta m = 0.006\,\text{g}

So,

\frac{\Delta m}{m} = \frac{0.006}{0.6} = 0.01 = 1\%

(ii) Radius

Given:

r = 0.5\,\text{mm}, \qquad \Delta r = 0.005\,\text{mm}

So,

\frac{\Delta r}{r} = \frac{0.005}{0.5} = 0.01 = 1\%

Since $r^2$ appears in the denominator, contribution is:

2\times 1\% = 2\%

(iii) Length

Given:

l = 4\,\text{cm}, \qquad \Delta l = 0.04\,\text{cm}

So,

\frac{\Delta l}{l} = \frac{0.04}{4} = 0.01 = 1\%

Add the percentage errors

\text{Maximum percentage error in density} = 1\% + 2\% + 1\% = 4\%

Evaluate options

A: $4\%$ ✅
B: $3\%$ ❌
C: $6\%$ ❌
D: $7\%$ ❌

Therefore, the correct option is:

\boxed{\text{A: }4\%}

Comparison with stored correct answer

Stored correct answer: A

My derived answer is also A, so they agree.

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