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8.Mechanical Properties of Solids
hard
A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
A
$1.75$
B
$1.5$
C
$2.5$
D
$5$
Solution
(c) $l = \frac{{FL}}{{\pi {r^2}Y}} \Rightarrow {r^2} \propto \frac{1}{Y}$ $(F,L$ and $l$ are constant$)$
$\frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{{{Y_1}}}{{{Y_2}}}} \right)^{1/2}} = {\left( {\frac{{7 \times {{10}^{10}}}}{{12 \times {{10}^{10}}}}} \right)^{1/2}}$
$⇒$ ${r_2} = 1.5 \times {\left( {\frac{7}{{12}}} \right)^{1/2}}$$= 1.145 mm$
dia $ = 2.29 mm$
$⇒$
Standard 11
Physics
