An isosceles triangle is formed with rods of length L1 L2 as shown in the figure-The coefficient of linear expansion -1 -2- 10-5 -0 C and -2 -8- 10-5 -0 C for the two pieces- If the distance between apex and the midpoint of the base remains unchanged as the temperature is varied- then L1-L2 is equal to-

H^2=L_2^2 L_1/4 Differentiating both LHS and RHS, 2H Δ H=2L_2Δ L_2 2L_1Δ L_1/4 ⇒ 0=L_1L_1α _1Δ T/4 L_2L_2α _2Δ T since Δ H=0 due to H being ...

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Byju's Answer

Standard XII

Physics

Area and Volume Expansion

An isosceles ...

Question

An isosceles triangle is formed with rods of length $L_{1}$ & $L_{2}$ as shown in the figure.The coefficient of linear expansion $α_{1} = 2 \times 10^{- 5} /^{0} C$ and $α_{2} = 8 \times 10^{- 5} /^{0} C$ for the two pieces. If the distance between apex and the midpoint of the base remains unchanged as the temperature is varied, then $\frac{L_{1}}{L_{2}}$ is equal to:

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Solution

$H^{2} = L_{2}^{2} - \frac{L_{1}}{4}$
Differentiating both LHS and RHS,
$2 H Δ H = 2 L_{2} Δ L_{2} - \frac{2 L_{1} Δ L_{1}}{4}$
$\Rightarrow 0 = \frac{L_{1} L_{1} α_{1} Δ T}{4} - L_{2} L_{2} α_{2} Δ T$ since $Δ H = 0$ due to H being constant
$\Rightarrow \frac{L_{1}}{L_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}} = 2 \times 2 = 4$

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An isosceles triangle is formed with a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ for the base and two thin rods each of length $l_{2}$ and coefficient of linear expansion $α_{2}$ for the two pieces, if the distance between the apex and the midpoint of the base remain unchanged as the temperatures varied, we get $\frac{l_{1}}{l_{2}} = x \sqrt{\frac{α_{2}}{α_{1}}}$ . Find ' $x$ '.