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

The Correct Option is C

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%.