Ratio of lengths of two brass wires is 3 : 4; their areas of cross section are in the ratio 2:3. When same force is applied on them, the elongations produced will be in the ratio:

9 : 8

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8 : 9

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\sqrt{2}

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1 : 1

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Solution

The correct option is A 9 : 8
$\frac{L_{1}}{L_{2}} = \frac{3}{4}, \frac{A_{1}}{A_{2}} = \frac{2}{3}$
$\frac{Δ L_{1}}{Δ L_{2}} = \frac{3}{4} \times \frac{3}{2}$ $= \frac{9}{8}$

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Ratio of lengths of two brass wires is 3 : 4; their areas of cross section are in the ratio 2:3. When same force is applied on them, the elongations produced will be in the ratio:

The correct option is A 9 : 8L1L2=34,A1A2=23ΔL1ΔL2=34×32=98

The correct option is A 9 : 8
$\frac{L_{1}}{L_{2}} = \frac{3}{4}, \frac{A_{1}}{A_{2}} = \frac{2}{3}$
$\frac{Δ L_{1}}{Δ L_{2}} = \frac{3}{4} \times \frac{3}{2}$ $= \frac{9}{8}$