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Step-by-Step Solution
Step 1: Recall the relationship for extension under a load
When a wire of length $L$, cross-sectional area $A$, and Young's modulus $Y$ is stretched by a load $F$, the extension $l$ is given by:
$ l \;=\; \frac{F \, L}{A \, Y} $
Step 2: Apply the condition for identical extensions
Given that wires A and B stretch by the same amount for the same force $F$, we use the equivalence of the quantity $\frac{A \, Y}{L}$ for both wires (because $F$ and $l$ are the same):
$ \frac{A_A \, Y_A}{L_A} \;=\; \frac{A_B \, Y_B}{L_B} $
Step 3: Substitute the known parameters
Ratio of Young's moduli: $ \frac{Y_A}{Y_B} = \frac{7}{4} $
Wire A: $L_A = 2 \text{ m}$, cross-sectional area $A_A = \pi R^2$
Wire B: $L_B = 1.5 \text{ m}$, cross-sectional area $A_B = \pi \,(2\,\text{mm})^2$
Putting these into the equality:
$ \frac{\pi R^2 \, Y_A}{2}
\;=\;
\frac{\pi \,(2\,\text{mm})^2 \, Y_B}{1.5}
$
But $ Y_A = \frac{7}{4} Y_B $. Substituting this in:
$ \frac{\pi R^2 \, \left(\frac{7}{4}Y_B\right)}{2}
\;=\;
\frac{\pi \,(2\,\text{mm})^2 \, Y_B}{1.5}
$
Step 4: Simplify and solve for $R$
Cancel out the common factor $\pi \, Y_B$ from both sides:
$ \frac{R^2 \,\frac{7}{4}}{2}
\;=\;
\frac{(2\,\text{mm})^2}{1.5}
$
$ \frac{7 \, R^2}{4 \times 2}
\;=\;
\frac{4 \,\text{mm}^2}{1.5}
$
$ \frac{7 \, R^2}{8}
\;=\;
\frac{4}{1.5}
=
\frac{8}{3}
$
$ 7 \, R^2
=
8 \times \frac{8}{3}
=
\frac{64}{3}
$
$ R^2
=
\frac{64}{3 \times 7}
=
\frac{64}{21}
$
$ R
=
\sqrt{\frac{64}{21}}
\approx 1.74\,\text{mm}
$
Step 5: Conclude the radius value
The radius $R$ of wire A is approximately $1.7\,\text{mm}$.
