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

To determine the percentage error in Young's modulus \( Y \), we must calculate the fractional errors in mass \( M \) and extension \( \ell \), and subsequently combine these to determine the error in \( Y \).

Young's modulus is defined by the formula:

\(Y = \frac{49000 \, M}{\ell}\)

Assume the errors in \( M \) and \( \ell \) are \(\Delta M\) and \(\Delta \ell\), respectively. The smallest scale divisions are \( 5 \, \text{g} \) for \( M \) and \( 0.02 \, \text{cm} \) for \( \ell \). Consequently, we have:

• \(\Delta M = 5 \, \text{g}\)
• \(\Delta \ell = 0.02 \, \text{cm} \

The measured values for \( M \) and \( \ell \) are \( 500 \, \text{g} \) and \( 2 \, \text{cm} \), respectively.

The percentage error in a product or quotient is the sum of the individual percentage errors of its components. Therefore, the percentage error in \( Y \) is calculated as:

\(\frac{\Delta Y}{Y} \times 100 = \left( \frac{\Delta M}{M} + \frac{\Delta \ell}{\ell} \right) \times 100\)

Upon substituting the values, we obtain:

• \(\frac{\Delta M}{M} = \frac{5}{500} = 0.01\)
• \(\frac{\Delta \ell}{\ell} = \frac{0.02}{2} = 0.01\)

The total percentage error in \( Y \) is thus:

\(\left(0.01 + 0.01 \right) \times 100 = 0.02 \times 100 = 2\%\)

The calculated percentage error for \( Y \) is 2%.