A copper wire and an aluminium wire have lengths in the ratio 5 : 2, diameters in the ratio 4 : 3 and forces applied in the ratio 4 : 5. Find the ratio of increase in length of the copper wire to that of aluminium wire.(Let YCu=1.1×1011 Nm-2, YAl=0.7×1011 Nm-2 ) - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
$\frac{176}{63}$
B
$\frac{63}{88}$
C
$\frac{189}{110}$
D
$\frac{33}{89}$

Solution:

Given the ratio of copper to aluminum wire length is $l_{1} : l_{2} = 5 : 2$

and the wire's diameter ratio is $d_{1} : d_{2} = 4 : 3$

The increase in length of a wire under tension can be calculated using Hooke's Law:

copper wire,

$∆ l_{1} = \frac{F_{1} l_{1}}{A_{1} Y_{1}}________(1)$

aluminum wire,

$∆ l_{2} = \frac{F_{2} l_{2}}{A_{2} Y_{2}}________(2)$

from equations 1 and 2,

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{l_{1}}{l_{2}} . \frac{A_{2}}{A_{1}} . \frac{Y_{2}}{Y_{1}} . \frac{F_{1}}{F_{2}} [\frac{Y_{2}}{Y_{1}} = \frac{Y_{Al}}{Y_{Cu}} = \frac{0.7}{1.1}]$

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{5}{2} . \frac{{πr}_{2}^{2}}{{πr}_{1}^{2}} . \frac{0.7 \times 10^{11}}{1.1 \times 10^{11}} . \frac{4}{5}$

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{5}{2} \times \frac{{(\frac{3}{2})}^{2}}{{(2)}^{2}} \times \frac{7}{11} \times \frac{4}{5}$

$\frac{∆ l_{1}}{∆ l_{2}} = \frac{63}{88}$

Hence the correct answer is $\frac{63}{88} .$

Solution:

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