A copper and a tungsten plate having a thickness `sigma=2`mm each are riveted together so that at `0^@`C they form a flat bimetallic plate. Find the average radius of curvature of this plate at `T=200^@C`. The coefficients of linear expansion for copper and tungsten are
`alpha_(cu)=1.7xx 10^-5//K` and `alpha_(W)=0.4xx10^(-5)//K`, respectively.

Options

A. `0.67 m`
B. `0.45 m`
C. `0.79 m`
D. `0.77 m`

Answer

D

Explanation

The average length of copper plate at a temperature `T=200^@C` is `l_C=l_0(1+alpha_cT)`,
where `l_0` is the length of copper plate at `0^@`C. The length of the tungsten plate is `l_t=l_0(1+alpha_tT)`
We shall assume that the edges of plates are not displaced during deformation and that an increase in the plate thickness duw to heating can be neglected.
From Fig we have
`l_C=phi(R+ delta//2)implies l_t=phi(R- delta//2)`
Consequently,
`phi(R+ delta//2)=l_0(1+alpha_CT)` ..(i)
`phi(R-delta//2)=l_0(1+alpha_tT)` ..(ii)
To eliminate the unknown quantities `phi` and `l_0`, we divide Eq. (i) by Eq. (ii) term wise.
`rArr ((R+ delta//2))/((R-delta//2))=((1+ alpha_(c)T))/((1+alpha_(t)T))`
`rArr R=delta ([2+(alpha_(c)+alpha_(t))T])/((alpha_(c)-alpha_(t))T)`
`impliesR=(delta)/((alpha_c-alpha_t)T)`
neglecting `(alpha_C + alpha_t)` in numerator. Sustituting the values in above relation we get `R=0.769m`.