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A uniform rod of length L, has a mass ...

A uniform rod of length `L`, has a mass per unit length `lambda` and area of cross section `A`. The elongation in the rod is `l` due to its own weight, if it suspended form the celing of a room. The Young's modulus of the rod is

A

`(2 lamda gL)/(Al)`

B

`(lamda g l^2)/(AL)`

C

`(lamda g L^2)/(2Al)`

D

`(2 lamda g L^2)/(Al)`

Text Solution

Verified by Experts

The correct Answer is:
C
Increase in length of the rod due to its own weight (mg) acting at the centre of the rod will be
` l = (mg (L/2) )/(AY) = (mgL)/(2AY) " or " Y = (mgL)/(2Al) = ((lamda L) gL)/(2Al) = (lamda gL^2)/(2Al)`
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