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There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
$1 : 1$
$2 : 1$
$1 : 2$
$4 : 1$
Solution
Both wires are same materials so both will have same $Young's$ modulus, and let it
be. $Y.Y = \frac{{stress}}{{strain}} = \frac{F}{{A.\left( {\Delta L/L} \right)}},$
$F = applied\,force$
$A = \,area\,of\,cross – \,section\,of\,wire$
$Now,$
${Y_1} = {Y_2} \Rightarrow \frac{{FL}}{{\left( {{A_1}} \right)\left( {\Delta {L_1}} \right)}} = \frac{{FL}}{{\left( {{A_2}} \right)\left( {\Delta {L_2}} \right)}}$
Since load and length are same for both
$ \Rightarrow r_1^2\Delta {L_1} = r_2^2\Delta {L_2},$
$\left( {\frac{{\Delta {L_1}}}{{\Delta {L_2}}}} \right) = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = 4\,\Delta {L_1}\,:\,\Delta {L_2} = 4:1$
