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 need to calculate the fractional errors in the mass \( M \) and the extension \( \ell \) and then combine them to find the error in \( Y \).

The formula for Young's modulus is given by:

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

Let's assume that the errors in \( M \) and \( \ell \) are \(\Delta M\) and \(\Delta \ell\) respectively. The given smallest scale divisions are \( 5 \, \text{g} \) for \( M \) and \( 0.02 \, \text{cm} \) for \( \ell \). Thus, we have:

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

The values of \( M \) and \( \ell \) are given as \( 500 \, \text{g} \) and \( 2 \, \text{cm} \) respectively.

The percentage error in any product or quotient is the sum of the percentage errors in each of the factors. Therefore, the percentage error in \( Y \) is given by:

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

Substituting the values, we get:

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

Therefore, the total percentage error in \( Y \) is:

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

Thus, the percentage error of \( Y \) is 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%.