Two steel wires have equal volumes. Their diameters are in the ratio $2 : 1$ . When same force is applied on them, the elongation produced will be in the ratio of

1 : 8

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8 : 1

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1 : 16

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16 : 1

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Solution

The correct option is D $1 : 16$
Given,
Force is same.
diameter ratio, $\frac{D_{1}}{D_{2}} = \frac{2}{1}$
Volume ratio $\frac{V_{1}}{V_{2}} = 1$ . . . . .(1)
Young modulus ratio, $\frac{Y_{1}}{Y_{2}} = 1$ . . . . .(2)
Area ratio, $\frac{A_{1}}{A_{2}} = (\frac{D_{1}}{D_{2}})^{2} = \frac{4}{1}$
From equation (1),
$\frac{V_{1}}{V_{2}} = \frac{A_{1} l_{1}}{A_{2} l_{2}}$
$\frac{l_{1}}{l_{2}} = \frac{1}{4}$
From equation (2), we get
$\frac{Y_{1}}{Y_{2}} = \frac{\frac{F}{A_{1}} \times \frac{l_{1}}{Δ l_{1}}}{\frac{F}{A_{2}} \times \frac{l_{2}}{Δ l_{2}}}$
$\frac{Δ l_{1}}{Δ l_{2}} = \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}}$
$\frac{Δ l_{1}}{Δ_{2}} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$
$Δ l_{1} : Δ l_{2} = 1 : 16$
The correct option is C.

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Two steel wires have equal volumes. Their diameters are in the ratio 2:1. When same force is applied on them, the elongation produced will be in the ratio of

Two steel wires have equal volumes. Their diameters are in the ratio $2 : 1$ . When same force is applied on them, the elongation produced will be in the ratio of