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Step-by-Step Solution
Step 1: Write the formula for fractional error in Young’s Modulus
Young’s Modulus for a wire under a load can be written as
$Y = \frac{F \times l}{A \times \Delta L}$,
where
$F$ is the force (here $F = mg$),
$l$ is the original length of the wire,
$A$ is the cross-sectional area ($\pi r^2$ for a circular wire),
$\Delta L$ is the extension produced by the force.
When we deal with measured quantities, the fractional (or relative) error in $Y$ can be combined as:
$ \frac{\Delta Y}{Y} \;=\; \frac{\Delta F}{F} \;+\; \frac{\Delta l}{l} \;+\; \frac{\Delta A}{A} \;+\; \frac{\Delta (\Delta L)}{\Delta L}.$
Since $F = mg$, the fractional error in $F$ is
$ \frac{\Delta F}{F} = \frac{\Delta m}{m} + \frac{\Delta g}{g}.$
In most experiments, $g$ is taken as a constant with negligible error, so we consider
$ \frac{\Delta g}{g} \approx 0.$
Hence,
$ \frac{\Delta F}{F} \approx \frac{\Delta m}{m}.$
Step 2: Identify and list all measured quantities with their least counts
From the question, we have:
Mass $m = 1\,\text{kg}$ with least count $= 1\,\text{g} = 0.001\,\text{kg}.$
Radius of wire $r = 0.2\,\text{cm}$ with least count $= 0.001\,\text{cm}.$
Original length of wire $l = 1\,\text{m}$ with least count $= 1\,\text{mm} = 0.001\,\text{m}.$
Extension $\Delta L = 0.5\,\text{cm}$ with least count $= 0.001\,\text{cm}.$
We will convert fractional errors by taking each absolute error over its respective measurement.
Step 3: Compute the individual fractional errors
1. Fractional error in mass ($m$):
$
\frac{\Delta m}{m} = \frac{0.001\,\text{kg}}{1\,\text{kg}} = 0.001 = 0.1\%.
$
2. Fractional error in $g$ (assumed negligible):
$
\frac{\Delta g}{g} \approx 0.
$
3. Fractional error in radius ($r$):
$
\frac{\Delta r}{r} = \frac{0.001\,\text{cm}}{0.2\,\text{cm}} = 0.005 = 0.5\%.
$
Since $A = \pi r^2$, the fractional error in area $A$ is
$
\frac{\Delta A}{A} = 2 \times \frac{\Delta r}{r} = 2 \times 0.005 = 0.01 = 1\%.
$
4. Fractional error in original length ($l$):
$
\frac{\Delta l}{l} = \frac{0.001\,\text{m}}{1\,\text{m}} = 0.001 = 0.1\%.
$
5. Fractional error in extension ($\Delta L$):
Convert the extension to centimeters: $\Delta L = 0.5\,\text{cm}.$
The least count is $0.001\,\text{cm},$ so
$
\frac{\Delta (\Delta L)}{\Delta L} = \frac{0.001\,\text{cm}}{0.5\,\text{cm}} = 0.002 = 0.2\%.
$
Step 4: Add up the fractional errors
Now, summing all relevant terms for the fractional error in $Y$:
$
\frac{\Delta Y}{Y} = \underbrace{\frac{\Delta m}{m}}_{0.001} \;+\; 0 \;+\; \underbrace{2 \frac{\Delta r}{r}}_{0.01} \;+\; \underbrace{\frac{\Delta l}{l}}_{0.001} \;+\; \underbrace{\frac{\Delta (\Delta L)}{\Delta L}}_{0.002}.
$
Let’s add them carefully:
$
0.001 + 0.01 + 0.001 + 0.002 = 0.014.
$
Hence, in percentage form:
$
0.014 \times 100\% = 1.4\%.
$
Thus, the fractional error in the value of Young's Modulus determined by the experiment is $\boxed{1.4\%}$.
