Question

Young's modulus is determined by the equation given by:$Y = \frac{49000 \, M}{\ell} \, \text{dyne/cm}^2$ where $M$ is the mass and $\ell$ is the extension of the wire used in the experiment. Now, the error in Young's modulus ($Y$) is estimated by taking data from the $M-\ell$ plot on graph paper. The smallest scale divisions are $5 \, \text{g}$ and $0.02 \, \text{cm}$ along the load axis and extension axis, respectively. If the value of $M$ and $\ell$ are $500 \, \text{g}$ and $2 \, \text{cm}$, respectively, then the percentage error of $Y$ is:

• A. 0.2 %
• B. 0.02 %
• C. 2 %
• D. 0.5 %

Answer 1

Answer: (C)

2 %

Answer 2

The error in Young’s modulus (Y) is given by:
$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta \ell}{\ell}.$

Here:
$\Delta M = 5 \, \text{g}, \, M = 500 \, \text{g}, \, \Delta \ell = 0.02 \, \text{cm}, \, \ell = 2 \, \text{cm}.$

Calculate the fractional errors:
$\frac{\Delta M}{M} = \frac{5}{500} = 0.01 = 1\%.$
$\frac{\Delta \ell}{\ell} = \frac{0.02}{2} = 0.01 = 1\%.$

The total percentage error is:
$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta \ell}{\ell} = 1\% + 1\% = 2\%.$

Final Answer: 2%.