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8.Mechanical Properties of Solids
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Two wires $A$ and $B$ are of same materials. Their lengths are in the ratio $1 : 2$ and diameters are in the ratio $2 : 1$ when stretched by force ${F_A}$ and ${F_B}$ respectively they get equal increase in their lengths. Then the ratio ${F_A}/{F_B}$ should be
A
$1:2$
B
$1:1$
C
$2:1$
D
$8:1$
Solution
(d) $F = Y \times A \times \frac{l}{L}$$⇒$ $F \propto \frac{{{r^2}}}{L}$ $(Y$ and $l$ are constant$)$
$\frac{{{F_A}}}{{{F_B}}} = {\left( {\frac{{{r_A}}}{{{r_B}}}} \right)^2} \times \left( {\frac{{{L_B}}}{{{L_A}}}} \right) = {\left( {\frac{2}{1}} \right)^2} \times \left( {\frac{2}{1}} \right) = \frac{8}{1}$
Standard 11
Physics
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