A uniform heavy plank of length L is resting on two rods of identical length and area of cross-section, made up of different materials. It is observed that, to keep the planks horizontal, rod A has to be at the end of the plank and rod B has to be at a distance $\frac{L}{4}$ from centre of the plank as shown in figure.

In another scenario, the rods are kept between two rigid walls such that the separation between the walls is equal to the natural length of the rod. Now the temperature of both the rods is gradually increased in the same manner, and it is observed that thermal stress developed in rod A is always thrice the thermal stress developed in rod B. Ratio of coefficient of linear expansion of rod A to coefficient of linear expansion of rod B is. (Answer upto two digits after the decimal point)

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Solution

Let $T_{A} and T_{B}$ be tension developed in both the rods respectively.

By balancing vertical forces we get
$T_{A} + T_{B} = m g$

By balancing torque we get
$T_{A} \frac{l}{2} = T_{B} \frac{l}{4}$

So $T_{A} = m g / 3$ and $T_{B} = 2 m g / 3$

We know $Y_{A} (\frac{Δ l}{l}) = \frac{T_{A}}{A}$ and

$Y_{B} (\frac{Δ l}{l}) = \frac{T_{B}}{A}$

$⟹ \frac{Y_{B}}{Y_{A}} = \frac{T_{B}}{T_{A}} = 2$

Thermal stress in A is $Y_{A} (\frac{Δ l}{l}) = Y_{A} (\frac{Δ l}{l}) α_{A} Δ t$

Thermal stress in B is $Y_{B} (\frac{Δ l}{l}) = Y_{B} (\frac{Δ l}{l}) α_{B} Δ t$

Given thermal stress developed in A is thrice thermal stress developed in B. So

$⟹ Y_{A} α_{A} Δ t = 3 Y_{B} α_{B} Δ t$

$⟹ \frac{α_{A}}{α_{B}} = 3 \frac{Y_{B}}{Y_{A}} = 6$

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