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The coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ . If we take a brass rod of length $l_{1}$ and steel rod of length $l_{2}$ at 0^oC, their difference in length $(l_{2} - l_{1})$ will remain the same at all temperatures if

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By Expert Faculty of Sri Chaitanya

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 $2 m m^{2}$ 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 $= 11 \times 10^{- 6} /^{0} C; Y_{s t e e l} = 2 \times 10^{11} N m^{- 2}$ )

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