A) 20 cm
B) 10 cm
C) 15 cm
D) 18 cm
Correct Answer: B
Solution :
\[\Delta L={{L}_{0}}\,\alpha \Delta \theta \] Rod A : 0.075 = 20 ´ aA ´ 100 Þ \[{{\alpha }_{A}}=\frac{75}{2}\times {{10}^{-6}}/{}^\circ C\] rod B : 0.045 = 20 ´ aB ´ 100 Þ \[{{\alpha }_{B}}=\frac{45}{2}\times {{10}^{-6}}/{}^\circ C\] For composite rod : x cm of A and (20 ? x) cm of B we have 0.060 = x aA ´ 100 + (20 ? x) aB ´ 100 \[=x\,\left[ \frac{75}{2}\times {{10}^{-6}}\times 100+(20-x)\times \frac{45}{2}\times {{10}^{-6}}\times 100 \right]\] On solving we get x = 10 cm.You need to login to perform this action.
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