Young's modulus of a rod is `(AgL^2)/(2l)` for which elongation is `lamda` due to its own weight when suspended from the ceiling. L is the length of the rod and A is constant, which is:

Options

A. area
B. mass per unit length
C. mass per unit length per unit area
D. area per unit mass

Answer

C

Explanation

Elongation in (dx) at distance x from lower end
`=((W)/(AY))dx=((mgx)/(L))((dx)/(AY))`
Increase in length due to own weight
`I=int_0^L(mgxdx)/(LAY)=(mgL)/(2A_1Y)`
`=((lamda)/(A_1))(gL^2)/(2l)=(AgL^2)/(2l)`
Where `A=(lamda)/(A_1)=`mass per unit length per unit area.