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# A steel scale is correct at 0oC, the length of a brass tube measured by it at 40oC is 5m. The correct length of the tube at 0oC is (Coefficients of linear expansion of steel and brass are 11×10−6/oC and 19×10−6/oC respectively).4.001m4.999m4.501m5.001m

A
4.001m
B
5.001m
C
4.999m
D
4.501m
Solution
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#### l′s=l′b=5 mTf=400cActual length of brass tube, lb=?Ti=0For Length of scale at 00=ls, l′s=ls(1+αsΔT) 5=ls(1+40αs) ls=51+40αsActual length of tube at 400l′b=l′s+Δls =l′s+l′s−ls =2l′s−lsl′b=(2×5−51+40αs)Actual length of tube at 00c=lbl′b=lb(1+α04T)lb=l′b1+αb4T=(10−51+40αs)×11+40αb=(10−51+40αs)×11+40αb=(10−51+40×11×10−6)×(11+40×19×10−6)=(10−51+440×10−6)×(11+760×10−6)=(10−51.000440)×(11+00076)=4.998≅4.999m orLet the length of tube and scale at 0oC be l. Then,Change in measured tube lengthΔl=Thermal expansion of brass− Thermal expansion of steelΔl=lαBΔT−lΔαBΔTΔl=lΔT(αB−αS)Δl=l(40−0)(19×10−6−11×10−6)So, new length measured by scale=l+Δl⇒5=l+l(40)(7×10−6)⇒51+40×7×10−6=l⇒l=4.9986≈4.999mOption C is correct.

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