In an experiment to determine the Young's modulus of wire of a length exactly 1 ~m , the extension in the length of the wire is measured as 0.4 ~mm with an uncertainty of \pm\, 0.02 ~mm when a load of 1 ~kg is applied. The diameter of the wire is measured as 0.4 ~mm with an uncertainty of \pm \,0.01 ~mm . The error in the measurement of Young's modulus (\Delta Y) is found to be x \times 10^10\, Nm^-2 . The value of x is \underline\hspace2cm . (. take g=10 ~ms^-2 )

Properties of Matter PYQ — JEE Main 2022 · 26 Jul · Shift 1 · Q66 with Solution

View written solutionFree

Correct answer: 2

Formula for Young’s modulus

For a wire,

Y = \frac{FL}{A\,\Delta l}

where:

$F = mg$
$L = 1\,\text{m}$
$A = \frac{\pi d^2}{4}$
$\Delta l$ is extension.

So,

Y = \frac{4FL}{\pi d^2 \Delta l}

Given data

$m = 1\,\text{kg}$
$g = 10\,\text{m s}^{-2}$
$F = 10\,\text{N}$
$L = 1\,\text{m}$ (exactly, so no error)
$d = 0.4\,\text{mm} = 4\times 10^{-4}\,\text{m}$
$\Delta d = 0.01\,\text{mm}$
$\Delta l = 0.4\,\text{mm} = 4\times 10^{-4}\,\text{m}$
$\Delta(\Delta l) = 0.02\,\text{mm}$

Calculate Young’s modulus

Since $d = \Delta l = 4\times 10^{-4}\,\text{m}$ ,

Y = \frac{4\times 10\times 1}{\pi (4\times 10^{-4})^2(4\times 10^{-4})}

(4\times 10^{-4})^3 = 64\times 10^{-12} = 6.4\times 10^{-11}

Thus,

Y = \frac{40}{\pi \times 6.4\times 10^{-11}} = \frac{6.25}{\pi}\times 10^{11} \approx 1.99\times 10^{11}\,\text{N m}^{-2}

So,

Y \approx 2\times 10^{11}\,\text{N m}^{-2}

Fractional error in $Y$

Since

Y \propto \frac{1}{d^2\,\Delta l}

maximum fractional error is

\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta l)}{\Delta l}

Now,

\frac{\Delta d}{d} = \frac{0.01}{0.4} = 0.025

\frac{\Delta(\Delta l)}{\Delta l} = \frac{0.02}{0.4} = 0.05

Hence,

\frac{\Delta Y}{Y} = 2(0.025) + 0.05 = 0.05 + 0.05 = 0.10

Absolute error in $Y$

\Delta Y = 0.10 \times Y \approx 0.10 \times 2\times 10^{11}

\Delta Y = 2\times 10^{10}\,\text{N m}^{-2}

Thus,

x = 2

Comparison with stored answer

Stored correct answer: $2$

Our derived answer matches it.

More from Properties of Matter

View written solutionFree

Correct answer: 2

Formula for Young’s modulus

For a wire,

Y = \frac{FL}{A\,\Delta l}

where:

$F = mg$
$L = 1\,\text{m}$
$A = \frac{\pi d^2}{4}$
$\Delta l$ is extension.

So,

Y = \frac{4FL}{\pi d^2 \Delta l}

Given data

$m = 1\,\text{kg}$
$g = 10\,\text{m s}^{-2}$
$F = 10\,\text{N}$
$L = 1\,\text{m}$ (exactly, so no error)
$d = 0.4\,\text{mm} = 4\times 10^{-4}\,\text{m}$
$\Delta d = 0.01\,\text{mm}$
$\Delta l = 0.4\,\text{mm} = 4\times 10^{-4}\,\text{m}$
$\Delta(\Delta l) = 0.02\,\text{mm}$

Calculate Young’s modulus

Since $d = \Delta l = 4\times 10^{-4}\,\text{m}$ ,

Y = \frac{4\times 10\times 1}{\pi (4\times 10^{-4})^2(4\times 10^{-4})}

(4\times 10^{-4})^3 = 64\times 10^{-12} = 6.4\times 10^{-11}

Thus,

Y = \frac{40}{\pi \times 6.4\times 10^{-11}} = \frac{6.25}{\pi}\times 10^{11} \approx 1.99\times 10^{11}\,\text{N m}^{-2}

So,

Y \approx 2\times 10^{11}\,\text{N m}^{-2}

Fractional error in $Y$

Since

Y \propto \frac{1}{d^2\,\Delta l}

maximum fractional error is

\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta l)}{\Delta l}

Now,

\frac{\Delta d}{d} = \frac{0.01}{0.4} = 0.025

\frac{\Delta(\Delta l)}{\Delta l} = \frac{0.02}{0.4} = 0.05

Hence,

\frac{\Delta Y}{Y} = 2(0.025) + 0.05 = 0.05 + 0.05 = 0.10

Absolute error in $Y$

\Delta Y = 0.10 \times Y \approx 0.10 \times 2\times 10^{11}

\Delta Y = 2\times 10^{10}\,\text{N m}^{-2}

Thus,

x = 2

Comparison with stored answer

Stored correct answer: $2$

Our derived answer matches it.

Properties of Matter question

Properties of Matter question

More from Properties of Matter

Properties of Matter question

Properties of Matter question

More from Properties of Matter