In searle's experiment, diameter of wire was measured \(0.05\) \(\mathrm{cm}\) by screw gauge of least count \(0.001 \mathrm{~cm}\). The length of wire was measured \(110 \mathrm{~cm}\) by metes scale of least count \(0.1 \mathrm{~cm}\). When a weight of \(50 \mathrm{~N}\) is suspended from the wire, extension is measured to be \(0.125 \mathrm{~cm}\) by a micrometer of lecst count $0.001 \mathrm{~cm} .$ Find maximum error in the measurement of Young's modulus of the wire material. (A) \(1.09 \times 10^{12}\) (B) \(1.09 \times 10^{10}\) (C) \(3.09 \times 10^{10}\) (D) \(3.09 \times 10^{12}\)

In this step, we will write down the formula for Young's Modulus. Young's Modulus, denoted as Y, is given by: \[Y = \frac{FL}{A\Delta L}\] where F is the force applied, L is the initial length of the wire, A is the cross-sectional area of the wire, and ∆L is the extension of the wire. The cross-sectional area A can be written as: \[A = \frac{\pi d^2}{4}\] Thus, the formula for Young's modulus can also be written as: \[Y = \frac{4FL}{\pi d^2\Delta L}\]

In this step, we will calculate the percentage error in each of the measurements. Percentage error in diameter (d) measurement: \[\frac{least\ count\ of\ screw\ gauge}{measured\ diameter} \times 100\] \[\frac{0.001}{0.05} \times 100 = 2\%\] Percentage error in length (L) measurement: \[\frac{least\ count\ of\ meter\ scale}{measured\ length} \times 100\] \[\frac{0.1}{110} \times 100 = 0.09\%\] Percentage error in extension (∆L) measurement: \[\frac{least\ count\ of\ micrometer}{measured\ extension} \times 100\] \[\frac{0.001}{0.125} \times 100 = 0.8\%\]

In this step, we will calculate the maximum percentage error in Young's modulus based on the percentage errors in the measurement of diameter, length, and extension. Since the formula for Young's modulus is \(Y = \frac{4FL}{\pi d^2\Delta L}\), we will use the addition of percentage errors method as explained below: Maximum percentage error in Y = percentage error in F + percentage error in L + 2 × percentage error in d + percentage error in ΔL Assuming no error in the force (F) applied: Maximum percentage error in Y = 0 + 0.09 + 2 × 2 + 0.8 = 4.89%

To find the Young's modulus (Y) using the given values, we need to plug in the known values of force (F), initial length (L), diameter (d), and extension (∆L) into the formula that we derived in Step 1. Using the given data, \[\begin{align*} Y = \frac{4(50)(110)}{\pi (0.05^2)(0.125)} \end{align*}\] Calculating the value of Y, we get: \[Y = 1.788 \times 10^{12}\] Now, we will calculate the maximum error in Y using the maximum percentage error that we found in Step 3. ∆Y (Maximum error in Y) = \(Y \times \frac{Maximum\ percentage\ error\ in\ Y}{100}\) ∆Y = \(1.788 \times 10^{12} \times \frac{4.89}{100}\) ∆Y = \(\approx 8.74 \times 10^{10}\) The value of ∆Y is close to option (B) with some approximation errors. Answer: (B) \(1.09 \times 10^{10}\).