\[\begin {array} {1 1} (a)\;4\;F \\ (b)\;6\;F \\ (c)\;9\;F \\ (d)\;F \end {array}\]

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Since the two wires have same volume.

Length of wire 1 $=l$

Length of wire 2 $=\large\frac{l}{3}$

Area of wire 1 $=A$

Area of wire 2 $=3A$

wire 1:

$Y=\large\frac{F/A}{\Delta x /l}$---------(1)

wire 2:

$Y=\large\frac{F'/A}{\Delta x /l/3}$-------------(2)

from (1) and (2)

$\large\frac{F}{A} \times \frac{l}{\Delta x}=\frac{F^1}{3A} \times \frac{l}{3 \Delta x}$

$F^1=9F$

Hence c is the correct answer.

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