Copper wire of certain length and resistance is drawn out to three times its length without change in volume, the new resistance of wire becomes

$$ rac{1}{9}$$ times
3 times
9 times
Unchanged

The correct answer is: A. $\frac{1}{9}$ times.

The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area. When the wire is drawn out to three times its length, its cross-sectional area is reduced to one-third of its original value. This is because the volume of the wire is unchanged, so the length and cross-sectional area must be in the ratio of 3:1.

The resistance of the wire is therefore reduced to $\frac{1}{3}$ times its original value, or $\frac{1}{9}$ times the resistance of the wire if it were not drawn out.

Here is a more detailed explanation of each option:

  • Option A: $\frac{1}{9}$ times. This is the correct answer. As explained above, the resistance of the wire is reduced to $\frac{1}{3}$ times its original value, or $\frac{1}{9}$ times the resistance of the wire if it were not drawn out.
  • Option B: 3 times. This is incorrect. The resistance of the wire is not increased by drawing it out. In fact, it is reduced.
  • Option C: 9 times. This is also incorrect. The resistance of the wire is not increased by drawing it out. In fact, it is reduced.
  • Option D: Unchanged. This is incorrect. The resistance of the wire is not unchanged by drawing it out. It is reduced.