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Three rods of equal of length are joined...

Three rods of equal of length are joined to from an equilateral triangle ABC. `D` is the midpoint of AB. The coefficient of linear expansion is `alpha_(1)` for AB and `alpha_(2)` for `AC` and `BC` . If the distance `DC` remains constant for small changes in temperature,

A

`(alpha_(1) + alpha_(2))Ldeltat`

B

`(2alpha_(1) + alpha_(2))/(2)LDeltat`

C

`((alpha_(1) + 2alpha_(2))Ldeltat)/(2)`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
D
`Dc^(2) = l^(2) = l^(2)(1 + alpha_(2)t)^(2) - [(l)/(2)(1 + alpha_(1)t)]^(2)`
neglecting `alpha_(2)^(2)` & `alpha_(1)^(2)` & solving above we get `alpha_(1) = 4alpha_(2)`
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Knowledge Check

  • An equilateral triangle ABC is formed by joining three rods of equal length and D is the mid-point of AB . The coefficient of linear expansion for AB is alpha_(1) and for AC and BC is alpha_(2) . The relation between alpha_(1) and alpha_(2) , if distance DC remains constant for small changes in temperture is

    A
    `alpha_(1) = alpha_(2)`
    B
    `alpha_(1) = 4alpha_(2)`
    C
    `alpha_(2) = 4alpha_(1)`
    D
    `alpha_(1) = (1)/(2)alpha_(2)`

  • Two rods , one of aluminium and the other made of steel, having initial length l_(1) and l_(2) are connected together to from a sinlge rod of length l_(1)+l_(2) . The coefficient of linear expansion for aluminium and steel are alpha_(a) and alpha_(s) for AC and BC . If the distance DC remains constant for small changes in temperature,

    A
    `(alpha_(s))/(alpha_(a))`
    B
    `(alpha_(a))/(alpha_(s))`
    C
    `(alpha_(s))/((alpha_(a)+alpha_(s)))`
    D
    `(alpha_(a))/((alpha_(a)+alpha_(s)))`

  • Three rods of equal l are joined to from an equilateral trangle 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 for PQ is alpha_(1^.) Then

    A
    `alpha_(2) = 3 alpha_(1)`
    B
    `alpha_(2) = 4 alpha_(1)`
    C
    `alpha_(1) = 3 alpha_(2)`
    D
    `alpha_(1) = 4 alpha_(2)`