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},$
$\left( {\frac{{\Delta {L_1}}}{{\Delta {L_2}}}} \right) = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} = 4\,\Delta {L_1}\,:\,\Delta {L_2} = 4:1$
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Assertion $(A)$ : $A$ cyclist always bends inwards while negotiating a curve
Reason $(R)$ : By bending he lowers his centre of gravity Of these statements,