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8.Mechanical Properties of Solids
medium
A wire of length $L$ and radius $r$ is rigidly fixed at one end. On stretching the other end of the wire with a force $F$, the increase in its length is $l$. If another wire of same material but of length $2L$ and radius $2r$ is stretched with a force of $2F$, the increase in its length will be
A
$l$
B
$2l$
C
$\frac{l}{2}$
D
$\frac{l}{4}$
(AIIMS-1980)
Solution
(a) $l = \frac{{FL}}{{AY}} = \frac{{FL}}{{\pi {r^2}Y}}\therefore l \propto \frac{{FL}}{{{r^2}}}$ ($Y =$ constant)
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{F_2}}}{{{F_1}}} \times \frac{{{L_2}}}{{{L_1}}}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times 2 \times {\left( {\frac{1}{2}} \right)^2} = 1$
${l_2} = {l_1}$ i.e. increment in its length will be $l.$
Standard 11
Physics
