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Two rods AB and BC of equal cross-sectio...

Two rods `AB` and `BC` of equal cross-sectional area are joined together and clamped between two fixed supports as shown in the figure. For the rod `AB` and road `BC` lengths are `l_(1)` and `l_(2)` coefficient of linear expansion are `alpha_(1)` and `alpha_(2)`, young's modulus are `Y_(1)` and `Y_(2)`, densities are `rho_(1)` and `rho_(2)` respectively. Now the temperature of the compound rod is increased by `theta`. Assume of that there is no significant change in the lengths of rod due to heating. then the time taken by transverse wave pulse to travel from end `A` to other end `C` of the compound rod is directly proportional to

A

`sqrt(l_(2)Y_(1) + l_(1)Y_(2))`

B

`2sqrt(l_(2)Y_(1) + l_(1)Y_(2))`

C

`sqrt(l_(1)Y_(2) + l_(2)Y_(1))`

D

`sqrt(l_(2)Y_(1) - l_(1)Y_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
Final length of the rod `AB` will be
`l_(1)^(')(1 + alpha_(1)theta)-(Fl_(1))/(Y_(1)A)`
Final length of the rod `BC` will be
`l_(2)^(') = l_(2)(1 + alpha_(2)theta) - (Fl_(2))/(Y_(2)A)`
Now, `l_(1) + l_(2) = l_(1)^(') + l_(2)^(')`
`rArr (F)/(A) = ((l_(1)alpha_(1) + l_(2)alpha_(2))Y_(1)Y_(2)theta)/(lY_(2) + l_(2)Y_(1))`
`t=(i_(1))/(sqrt(F)/(Arho_(1)))+(i_(2))/(sqrt(F)/(Arho_(2)))`
It is given that `l_(1)^(') = l_(1)` and `l_(2) = l_(2)`,
then `t prop sqrt(l_(1)Y_(2) + l_(2)Y_(1))`
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Knowledge Check

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