Three rods A, B and C, having identical shape and size, are hinged together at ends to form an equilateral triangle. Rods A and B are made of same material having coefficient of linear thermal expansion a while that of material of rod C is $α_{2}$ . By how many kelvin must the system of rods be heated to increase the angle opposite to rod C by $Δ θ$ .

Δ T = \frac{\sqrt{3} d θ}{2 (α_{2} - α_{1})}

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Δ T = \frac{\sqrt{3} d θ}{(α_{2} - α_{1})}

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Δ T = \frac{\sqrt{3} d θ}{2 (α_{2})}

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Δ T = \frac{\sqrt{53} d θ}{2 (α_{2} - α_{1})}

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Solution

The correct option is A $Δ T = \frac{\sqrt{3} d θ}{2 (α_{2} - α_{1})}$
When the system is heated, the rods expand and the triangle does not remain equilateral. Let lengths of rods A, B and C be $l_{1}$ , $l_{2}$ and $l_{3}$ respectively. Then

$c o s θ = \frac{l_{1}^{2} + l_{2}^{2} - l_{3}^{2}}{2 l_{1} l_{2}} o r 2 l_{1} l_{2} c o s θ = l_{1}^{2} + l_{2}^{2} - l_{3}^{2} \dots \dots (1) D i f f e r e n t i a t i n g e q . (1), w e g e t 2 l_{1} c o s θ d l_{2} + 2 l_{2} c o s θ d l_{1} - 2 l_{1} l_{2} s i n θ d θ = 2 l_{1} d l_{1} + 2 l_{2} d l_{2} - 2 l_{3} d l_{3} \dots \dots (2)$
Let the temperature of the system be increased by $Δ T$ . Then
$d l_{1} = l_{1} α_{1} Δ T d l_{2} = l_{2} α_{1} Δ T \dots \dots (3) a n d d l_{3} = l_{3} α_{2} Δ T I n c a s e o f e q u i l a t e r a l t r i a n g l e l_{1} = l_{2} = l_{3} = l (s a y) a n d θ = 60^{\circ} f r o m e q n . (2) a n d (3), w e g e t 2 l^{2} c o s 60 α_{1} Δ T + 2 l^{2} c o s 60 α_{1} Δ T \frac{α_{1}}{2} Δ T + \frac{α_{1}}{2} Δ T - \frac{\sqrt{3}}{2} d θ = α_{1} Δ T + α_{1} Δ T - α_{2} Δ T α_{1} Δ T - \frac{\sqrt{3}}{2} d θ = 2 α_{1} Δ T - α_{2} Δ T (α_{2} - α_{1}) Δ T = \frac{\sqrt{3}}{2} d θ Δ T = \frac{\sqrt{3} d θ}{2 (α_{2} - α_{1})}$

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