Question

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

• A. $9\%$
• B. $1.4\%$
• C. $0.9\%$
• D. $0.14\%$

Answer 1

The correct option is B

$1.4\%$

Solution:

Step 1: Given information

Radius $=0.2cm$

Length $=1m$

Mass $=1kg$

Elongation $=0.5cm$

Step 2: Calculate the fractional error in the value of Young’s Modulus determined by this experiment

Knows, $Y=\frac{Stress}{Strain}=\frac{FL}{A\mathcal{l}}=\frac{mg.L}{{\mathrm{\pi r}}^2.\mathcal{l}}$

Where, $Y=$Elongation, $F=$Force, $A=$Area, and $L=$Length

Therefore,

$\Rightarrow Y=\frac{mgL}{A.\mathcal{l}}$

$\Rightarrow \frac{\Delta Y}{Y}=\left(\frac{\Delta m}{m}\right)+\left(\frac{\Delta g}{g}\right)+\left(\frac{\Delta A}{A}\right)+\left(\frac{\Delta l}{l}\right)+\left(\frac{\Delta L}{L}\right)$

$\frac{∆Y}{Y}=$$\left(\frac{1g}{1kg}\right)+0+2+\left(\frac{∆r}{r}\right)+\left(\frac{∆\mathcal{l}}{\mathcal{l}}\right)+\left(\frac{∆L}{L}\right)$

$\frac{∆Y}{Y}=$$\left(\frac{1g}{1kg}\right)+2\left(\frac{0.001cm}{0.2cm}\right)+\left(\frac{0.001cm}{0.5cm}\right)+\left(\frac{0.001m}{1m}\right)$

Simplify further,

$\frac{∆Y}{Y}=$$\left(\frac{1}{1000}\right)+2\left(\frac{1\times 10}{2\times {10}^3}\right)+\left(\frac{1\times {10}^2}{5\times {10}^3}\right)+\left(\frac{1}{{10}^3}\right)$

$\frac{∆Y}{Y}=$$\frac{1}{1000}+\frac{1}{100}+\frac{2}{{10}^3}+\frac{1}{{10}^3}$

$\frac{∆Y}{Y}=$$\frac{1+10+2+1}{1000}$

Simplify further,

$\frac{∆Y}{Y}=$$\frac{14}{1000}\times 100\%$

$\frac{∆Y}{Y}=$$1.4\%$

Hence, the correct option is (B).