At temperature To, two metal strips of length /o and thickness d is bolted so that their ends coincide. The upper strip is made up of metal A and have coefficient of expansion ${\mathrm{\alpha }}_{\mathrm{A}}$and lower strip is made up of metal B with coefficient of expansion ${\mathrm{\alpha }}_{\mathrm{B}}\left({\mathrm{\alpha }}_{\mathrm{A}}>{\mathrm{\alpha }}_{\mathrm{B}}\right)$. When temperature of their blastic strip is increased from To to $\left({\mathrm{T}}_0+\mathrm{\Delta T}\right)$, one stop becomes longer than the other and the blastic strip is bend in the form of a circle as shown in .
 

The radius of curvature fi of the strip is

detailed solution

Correct option is A

At temperature T0+ΔT, the increase in length                          LA=l01+αAΔT              ....(1)and                    LB=l01+αBΔT              ....(2)Let RA and RB the radius of the strips, then                         LA=θRA                   ....(3)  and                        LB=θRB          .....(4)where θ = common angle as shown in the fig. (2).From eqs. (1) and (2), we get      LA−LB=l0ΔTαA−αB           ....(5)From eqs. (3) and (4), we get   LA−LB=θRA−RB=θd          ....(6)From eqs. (5) and (6), we get                    θ=l0ΔTαA−αBd       .....(7)From eqs. (1), (2), (3) and (4), we get RA+RB=2l0+l0αA+αBΔTθ      ....(8)∴Mean radius R is given by R=RA+RB2=2l0+l0αA+αBΔT2θ ....(9)Substituting the value of 0 from eq. (7) in eq. (9), we get             R=2l0+l0αA+αBΔT2l0ΔTαA−αB/dor        R=2+αA+αBΔTd2αA−αBΔT