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# The diameter of a metal ring is $\text{'}D\text{'}$  and coefficient of linear expansion is  $\alpha$, if the temperature of the ring is increases by ${1}^{0}C$ , The circumference of the ring will increase by  $x\pi D\alpha$ where  $x=$

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1
b
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detailed solution

Correct option is A

circumference of ring is S=2πR=2πD2 S=πD differentiate ΔS=πΔD             coefficient of linear expansion α=ΔDDΔt⇒ΔD=DαΔt;ΔS=xπDα ; substitute in above equation  xπDα=πDαΔt x=Δt    given Δt   =1 ⇒x=1

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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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