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Three rods P, Q and R having identical shape and size and hinged togetherat ends to form an equilateral triangle. Rods P and Q are made of same material having coefficient of linear thermal expansion alpha_(1) while that of material of rod R is alpha_(2). By how any kelvin should the system of rods be increase the angle opposite to rod R by Delta theta. |
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Answer» <P> Solution :When the system is heated,the rods EXPAND and the triangle does not REMAIN equilateral. Let lengths of rods P, Q and R be `l_(1)l_(2) and l_(3)` respectively. Then from cosine rule,`cos theta = (l_(1)^(2) +l_(2)^(2) -l_(3)^(2))/(2l_(1)l_(2)) (or) 2l_(1)^(2) +l_(2)^(2)-l_(3)^(2)""....(a)`  Differentiating equation(a), we get `2l_(1) cos theta dl_(2) +2l_(2) cos theta dl_(1)-2l_(1)l_(2) sin theta d theta` `=2l_(1)dl_(1) +2l_(2) dl_(2) -2l_(3) dl_(3) "" ....(b)` Let the temperature of the system be increased by `DELTAT`. Then `dl_(1)=l_(1)alpha_(1) DeltaT` `dl_(2)=l_(2)alpha_(1) DeltaT and dl_(3)=l_(3) alpha_(2) DeltaT` In case of equilateral triangle `l_(1)=l_(2)=l_(3)=L` (say) and `theta=60^(@)` and for small change in angle `d theta =Delta theta` From equation (b) and (c ), we get `2l^(2)cos (60^(@)) alpha_(1) DeltaT +2l^(2) cos(60^(@)) alpha_(1)DeltaT-2l^(2) sin(60^(@))Delta theta` `=2l^(2)alpha_(1)DeltaT +2l^(2)alpha_(1)DeltaT-2l^(2)alpha_(2)DeltaT` `(alpha_(1))/(2)DeltaT +(alpha_(1))/(2)DeltaT-(sqrt(3))/(2) Delta theta=alpha_(1)DeltaT+alpha_(1) DeltaT-alpha_(2)DeltaT` `alpha_(1) DeltaT -(sqrt(3))/(2) Delta theta =2alpha_(1)DeltaT-alpha_(2) DeltaT` `(alpha_(1)-alpha_(2))DeltaT=(sqrt(3))/(2)Delta theta rArrDeltaT=(sqrt(3)Delta theta)/(2(alpha_(2)-alpha_(1)))` |
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