A wire of length 6 m is stretched such that its radius is reduced by 20%. Which one of the following is the value of increase in its length?

Correct — B, 56.25%. Stretching a wire does not change the amount of metal in it, so the volume is constant. Volume is π r² L, so r² L stays fixed. The radius falls by twenty per cent, leaving 0.8r, and the cross-sectional area therefore falls to 0.8² = 0.64 of what it was. For the product to be unchanged, the length must rise by the reciprocal: L' = L ÷ 0.64 = 1.5625L. That is an increase of 0.5625, which is 56.25 per cent. The six metres in the stem is not needed at all — the answer is a percentage, so the original length cancels; putting it in gives 6 ÷ 0.64 = 9.375 metres, an increase of 3.375 metres, and 3.375 ÷ 6 is again 56.25 per cent.

• (a)50% — A fifty per cent increase would need the area to fall to 1 ÷ 1.5 = 0.667 of its value, which corresponds to a radius reduction of about eighteen per cent, not twenty.
• (c)62.25% — A near miss on the digits. It resembles 1 ÷ 0.64 misread, but the correct reciprocal is 1.5625 and the increase is 56.25 per cent, not 62.25.
• (d)75% — This is what you get by applying the twenty per cent reduction to the radius but then treating the area as falling to 0.8 × 0.8 × 0.8 = 0.512 as though volume scaled with the cube. Only the two radial dimensions shrink; the length is what absorbs the change.

Wire-drawing problems are conservation-of-volume problems. A cylinder has volume π r² L, and any operation that reshapes the metal without adding or removing it must leave that product alone. Because the area depends on the square of the radius, a modest change in radius produces a large change in length — cutting the radius to eighty per cent cuts the area to sixty-four per cent and so raises the length by more than half.

The reciprocal is the step to be careful with. Falling to a fraction f of the area means rising by a factor 1/f in length, and 1/f is not the same as adding back the percentage that was lost: losing thirty-six per cent of the area gains fifty-six and a quarter per cent of length, not thirty-six. The same asymmetry appears throughout percentage work, and it is why an increase and a decrease of the same percentage never cancel. The identical physics governs the resistance of a stretched wire, which is why this item and its electrical cousin are solved the same way.

• Stretching conserves volume, so π r² L is unchanged and r² L is constant.
• A radius reduced by 20% leaves 0.8r, so the cross-sectional area falls to 0.64 of its original value.
• Length must rise by the reciprocal: 1 ÷ 0.64 = 1.5625, an increase of 56.25 per cent.
• The original length of six metres is irrelevant to a percentage answer; it becomes 9.375 m if computed.
• Falling to a fraction f of the area raises the length by 1/f, which is never the same as adding back the lost percentage.

• Adding back the lost thirty-six per cent instead of taking the reciprocal.
• Treating the volume as scaling with the cube of the radius when only two dimensions shrink.
• Trying to use the six metres, which cancels out of a percentage answer.

Asked as a mensuration item with a redundant number in the stem, so a candidate who reaches for the figure rather than the ratio wastes time.

• practice — not a real PYQ

  A wire is stretched so that its radius is halved. By what factor does its length increase?

  • (a)2
  • (b)4
  • (c)8
  • (d)16

  Answer(b) 4 — the area falls to one-fourth, so the length must rise fourfold to keep the volume constant.

• practice — not a real PYQ

  If a wire is stretched to twice its original length, its new cross-sectional area is what fraction of the original?

  • (a)One-half
  • (b)One-fourth
  • (c)Two-thirds
  • (d)Unchanged

  Answer(a) One-half — volume is constant, so doubling the length halves the area.