In Searle’s experiment diameter of wire is measured with screw gauge of least count 0.001cm and its value is 0.08cm. Length of wire is 150cm with least count 0.1cm and elongation 0.5cm with least count 0.001cm. Weight suspended is 100N. Find absolute error in Young’s modulus if the value is written as N ×109N/m2. Find N.

To find the absolute error in Young's modulus, we first express the formula for Young's modulus ($Y$):

$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$

Since the wire has a circular cross-section with diameter $d$, the area is $A=\frac{\pi d^2}{4}$. Substituting this back, we get:

$Y=\frac{4FL}{\pi d^2\Delta L}$

Here, the elongation $\Delta L$ will be denoted as $x$ to avoid confusion with the error in length. Thus, the equation becomes:

$Y=\frac{4FL}{\pi d^{2}x}$

We are given the following values in the problem:

• Weight suspended, $F=100\text{N}$
• Length of the wire, $L=150\text{cm}=1.5\text{m}$ with least count $\Delta L=0.1\text{cm}=0.001\text{m}$
• Diameter of the wire, $d=0.08\text{cm}=8\times {10}^{-4}\text{m}$ with least count $\Delta d=0.001\text{cm}=1\times {10}^{-5}\text{m}$
• Elongation of the wire, $x=0.5\text{cm}=5\times {10}^{-3}\text{m}$ with least count $\Delta x=0.001\text{cm}=1\times {10}^{-5}\text{m}$

Step 1: Calculate the value of Young's Modulus ($Y$)

Substituting the values into the equation for $Y$:

$Y=\frac{4\times 100\times 1.5}{\pi \times {(8\times {10}^{-4})}^2\times (5\times {10}^{-3})}$

$Y=\frac{600}{\pi \times 64\times {10}^{-8}\times 5\times {10}^{-3}}$

$Y=\frac{600}{320\pi \times {10}^{-11}}\approx 59.68\times {10}^9{\text{N/m}}^2$

Step 2: Express fractional error in Young's Modulus

The relative error in Young's modulus is given by:

$\frac{\Delta Y}{Y}=\frac{\Delta L}{L}+2\frac{\Delta d}{d}+\frac{\Delta x}{x}$

Substituting the given values into the error formula:

$\frac{\Delta Y}{Y}=\frac{0.1}{150}+2\times \frac{0.001}{0.08}+\frac{0.001}{0.5}$

$\frac{\Delta Y}{Y}=0.000667+0.025+0.002=0.027667$

Step 3: Calculate the absolute error in Young's Modulus ($\Delta Y$)

Now, we find $\Delta Y$ using the calculated value of $Y$:

$\Delta Y=Y\times 0.027667$

$\Delta Y=(59.68\times {10}^9)\times 0.027667\approx 1.651\times {10}^9{\text{N/m}}^2$

Comparing this value with the required form $N\times {10}^9{\text{N/m}}^2$, we find:

$N\approx 1.65$