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# The coefficient of linear expansion of brass and steel are ${\mathrm{\alpha }}_{1}$ and ${\alpha }_{2}$. If we take a brass rod of length ${l}_{1}$ and steel rod of length ${l}_{\mathit{2}}$ at 0oC, their difference in length $\left({l}_{2}-{l}_{1}\right)$will remain the same at all temperatures if

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a
α1l2=α2l1
b
α1l22=α2l12
c
α12l1=α22l2
d
α1l1=α2l2
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detailed solution

Correct option is D

L2=l21+α2Δθ and L1=l11+α1Δθ⇒L2−L1=l2−l1+Δθl2α2−l1α1 Now L2−L1=l2−l1 so, l2α2−l1α1=0α1l1=l2α2

Similar Questions

A steel wire of cross sectional area $2m{m}^{2}$  is just stretched horizontally between two fixed points at a temperature of ${35}^{0}C$ . The temperature of the wire changed to ${25}^{0}C$ . Then (Coefficient of linear expansion of steel$=11×{10}^{-6}{/}^{0}C;\text{\hspace{0.17em}}\text{\hspace{0.17em}}{Y}_{steel}=2×{10}^{11}N{m}^{-2}$ )

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