An isosceles triangles is formed with a thin rod of length `l_(1)` and coefficient of linear expansion `alpha_(1)`, as the base and two thin rods each of length `l_(2)` and coefficient of linear expansion `alpha_(2)` as the two sides. The distance between the apex and the midpoint of the base remain unchanged as the temperature is varied. the ratio of lengths `(l_(1))/(l_(2))` is

Options

A. `(L_1)/(L_2)=(alpha_2)/(alpha_1)`
B. `(L_1)/(L_2)=sqrt((alpha_2)/(alpha_1))`
C. `(L_1)/(L_2)=2(alpha_2)/(alpha_1)`
D. `(L_1)/(L_2)=2sqrt((alpha_2)/(alpha_1))`

Answer

D

Explanation

The apex of the isosceles Delta to remain at a constant distance form the knife edge. Thus, DC should remain constant before and after heating
Before expansion: In Delta ADC
`(DC)^2=L_2^2-((L_1)/(2))^2` .(i)
After expansion:
`(DC)^2=[L_2(1+alpha_2t]^2-[(L_1)/(2)(1+alpha_1t)]^2` ..(ii)
Equating Eqs. (i) and (ii) we get
`L_2^2-((L_1)/(2))^2=[L_2(1+alpha_2t)]^2-[(L_1)/(2)(1+alpha_1t)]^2`
`L_2^2-(L_1^2)/(4)=L_2^2+L_2^2xx2alpha_2xxt-(L_1^2)/(4)-(L_1^2)/(4)xx2alpha_1xxt`
(neglecting higher terms)
`implies(L_1^2)/(4)(2alpha_1t)=L_2^2(2alpha_2t)implies(L_1)/(L_2)=2sqrt((alpha_2)/(alpha_1))`