Properties Of Solids And Liquids Question 116
Question: Three rods of equal length l are joined to form an equilateral triangle PQR. O is the mid point of PQ. Distance OR remains same for small change in temperature. Coefficient of linear expansion for PR and RQ is same i.e. $ {\alpha _{2}} $ but that for PQ is $ {\alpha _{1}} $ . Then
Options:
A) $ {\alpha _{2}}=3{\alpha _{1}} $
B) $ {\alpha _{2}}=4{\alpha _{1}} $
C) $ {\alpha _{1}}=3{\alpha _{2}} $
D) $ {\alpha _{1}}=4{\alpha _{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{(OR)}^{2}}={{(PR)}^{2}}-{{(PO)}^{2}}=l^{2}-{{( \frac{l}{2} )}^{2}} $
$ ={{[l(1+{\alpha _{2}}t)]}^{2}}-{{[ \frac{l}{2}(1+{\alpha _{1}}t) ]}^{2}} $
$ l^{2}-\frac{l^{2}}{4}=l^{2}(1+\alpha _{2}^{2}t^{2}+2{\alpha _{2}}t)-\frac{l^{2}}{4}(1+\alpha _{1}^{2}t^{2}+2{\alpha _{1}}t) $
Neglecting $ \alpha _{2}^{2}t^{2} $
and $ \alpha _{1}^{2}t^{2} $
$ 0=l^{2}(2{\alpha _{2}}t)-\frac{l^{2}}{4}(2{\alpha _{1}}t)\Rightarrow 2{\alpha _{2}}=\frac{2{\alpha _{1}}}{4}\Rightarrow ;{\alpha _{1}}=4{\alpha _{2}} $
