There are two wires of same material and same length while diameter of second wire is 2 times diamet

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8.Mechanical Properties of Solids

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There are two wires of same material and same length while the diameter of second wire is $2$ times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be

A

$1:1$

B

$2:1$

C

$1:2$

D

$4:1$

Solution

(d) $l = \frac{{FL}}{{AY}}$ $\Rightarrow$ $l \propto \frac{1}{{{r^2}}}$ $(F, L$ and $Y$ are constant$)$

$\frac{{{l_1}}}{{{l_2}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = {(2)^2} = 4$

Standard 11

Physics

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