$(a)\;1.5\;\alpha\qquad(b)\;(\large\frac{5}{3})\;\alpha\qquad(c)\;1.25\;\alpha\qquad(d)\;(\large\frac{7}{5})\;\alpha$

Answer : $\;(\large\frac{5}{3})\;\alpha$

Explanation :

$L_{1}=L (1+\alpha \bigtriangleup T)$

$L_{2}=2L (1+2\alpha \bigtriangleup T)$

$L_{3}=L_{1}+L_{2}$

$=3L + 5 \alpha \bigtriangleup T$

$=3L\; (1+ \large\frac{5}{3} \alpha \bigtriangleup T)$

Therefore , Coefficient of linear expansion of rod =$\;\large\frac{5}{3} \alpha$

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