A uniform rod of mass M and length L, area of cross section A is placed on a smooth horizontal surface. Forces acting on the rod are shown in the digram

Ratio of elastic potential energy stored in section PQ and section QR of the rod is

Options

A. `19:37`
B. `21:39`
C. `23:41`
D. `17:35`

Explanation

`F=((F_(1)-F_(2))/(x_(0)))x+F_(2)=alphax+beta`
Energy density at any x
`(dU)/(dV)=1/2((alphax+beta)/A)((alphax+beta)/(AY))=1/(2A^(2)Y)(alphax+beta)^(2)`
Energy stored on small segment dx
`dU=1/(2A^(2)Y)(alpha^(2)x^(2) +beta^(2)+2alphabetax)Adx`
`U=intdU=1/(2AY) int_(0)^(x_(0))(alpha^(2)x^(2)+beta^(2)+2alphabetax)dx=1/(2AY)((alpha^(2)x_(0)^(3))/3+beta^(2)x_(0)+betax_(0)^(2))`
consider section PQ
`alpha=F//L, beta=F, x_(0) =L//2`
`U_(1)=(19 F^(2)L)/(48AY)` consider section QR
`alpha=F//L , beta=3F//2, x_(0)=L//2`
`U_(2)=(37F^(2)L)/(48AY)`