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8.Mechanical Properties of Solids
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Two wires $‘A’$ and $‘B’$ of the same material have radii in the ratio $2 : 1$ and lengths in the ratio $4 : 1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is
A
$1:1$
B
$2:1$
C
$1:4$
D
$1:2$
Solution
(a) $F = Y \times A \times \frac{l}{L}$ $⇒$ $F \propto \frac{{{r^2}}}{L}$ $(Y$ and $l$ are constant$)$
$\frac{{{F_1}}}{{{F_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2}\left( {\frac{{{L_2}}}{{{L_1}}}} \right) = {\left( {\frac{2}{1}} \right)^2}\left( {\frac{1}{4}} \right) = 1$ $⇒$ $\frac{{{F_1}}}{{{F_2}}} = 1:1$
Standard 11
Physics
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