In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

Units and Measurements PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q56 with Solution

View written solutionFree

Correct answer: C

Formula for Young’s modulus

For a वायर under load,

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A\, \Delta L}

Since $F = mg$ and area of cross-section $A = \pi r^2$ , we get

Y = \frac{mgL}{\pi r^2 \, l}

where $l$ is the elongation.

So,

Y \propto \frac{mL}{r^2 l}

Fractional error formula

For multiplication/division powers,

\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}

(Here $g$ and $\pi$ are constants, so no error contribution from them.)

Compute individual fractional errors

(i) Error in mass

Mass measured:

m = 1\,\text{kg} = 1000\,\text{g}

Least count = $1\,\text{g}$ , so

\frac{\Delta m}{m} = \frac{1}{1000} = 0.001 = 0.1\%

(ii) Error in length of wire

Length:

L = 1\,\text{m} = 100\,\text{cm}

Least count = $1\,\text{mm} = 0.1\,\text{cm}$ , so

\frac{\Delta L}{L} = \frac{0.1}{100} = 0.001 = 0.1\%

(iii) Error in radius

Radius:

r = 0.2\,\text{cm}

Least count = $0.001\,\text{cm}$ , so

\frac{\Delta r}{r} = \frac{0.001}{0.2} = 0.005 = 0.5\%

Since $r^2$ appears,

2\frac{\Delta r}{r} = 2(0.005) = 0.01 = 1\%

(iv) Error in elongation

Elongation:

l = 0.5\,\text{cm}

Least count = $0.001\,\text{cm}$ , so

\frac{\Delta l}{l} = \frac{0.001}{0.5} = 0.002 = 0.2\%

Add all fractional errors

\frac{\Delta Y}{Y} = 0.001 + 0.001 + 0.01 + 0.002 = 0.014

In percentage,

0.014 \times 100 = 1.4\%

Compare with options

The fractional error in Young’s modulus is

\boxed{1.4\%}

So the correct option is C.

More from Units and Measurements

View written solutionFree

Correct answer: C

Formula for Young’s modulus

For a वायर under load,

Y = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A\, \Delta L}

Since $F = mg$ and area of cross-section $A = \pi r^2$ , we get

Y = \frac{mgL}{\pi r^2 \, l}

where $l$ is the elongation.

So,

Y \propto \frac{mL}{r^2 l}

Fractional error formula

For multiplication/division powers,

\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}

(Here $g$ and $\pi$ are constants, so no error contribution from them.)

Compute individual fractional errors

(i) Error in mass

Mass measured:

m = 1\,\text{kg} = 1000\,\text{g}

Least count = $1\,\text{g}$ , so

\frac{\Delta m}{m} = \frac{1}{1000} = 0.001 = 0.1\%

(ii) Error in length of wire

Length:

L = 1\,\text{m} = 100\,\text{cm}

Least count = $1\,\text{mm} = 0.1\,\text{cm}$ , so

\frac{\Delta L}{L} = \frac{0.1}{100} = 0.001 = 0.1\%

(iii) Error in radius

Radius:

r = 0.2\,\text{cm}

Least count = $0.001\,\text{cm}$ , so

\frac{\Delta r}{r} = \frac{0.001}{0.2} = 0.005 = 0.5\%

Since $r^2$ appears,

2\frac{\Delta r}{r} = 2(0.005) = 0.01 = 1\%

(iv) Error in elongation

Elongation:

l = 0.5\,\text{cm}

Least count = $0.001\,\text{cm}$ , so

\frac{\Delta l}{l} = \frac{0.001}{0.5} = 0.002 = 0.2\%

Add all fractional errors

\frac{\Delta Y}{Y} = 0.001 + 0.001 + 0.01 + 0.002 = 0.014

In percentage,

0.014 \times 100 = 1.4\%

Compare with options

The fractional error in Young’s modulus is

\boxed{1.4\%}

So the correct option is C.

Units and Measurements question

Units and Measurements question

(i) Error in mass

(ii) Error in length of wire

(iii) Error in radius

(iv) Error in elongation

More from Units and Measurements

Units and Measurements question

Units and Measurements question

(i) Error in mass

(ii) Error in length of wire

(iii) Error in radius

(iv) Error in elongation

More from Units and Measurements