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A rod of length l, mass M, cross section...

A rod of length `l`, mass `M`, cross section area `A` is placed on a rough horizonatal surface. A horizonatal force `F` is applied to rod as shwon in figure. The coefficient of fricition between rod and surface is `mu`, the Young, modulus of material of rod is `Y`. [Assume that fricition force is distributed uniformly on rod]

Teh elongation in rod if `F gt mu Mg` is

A

`((F- mu Mg)l)/(2AY)`

B

`[(F- (mu Mg)/(2))/(2AY)]l`

C

`(Fl)/(2AY)`

D

None

Text Solution

Verified by Experts

The correct Answer is:
C
In the case the rod is moving with acecleration
`a = (F-mu Mg)/(M)`
As a function of distance `x` form right end
`F = T - f_(1) = (M)/(L) x xx a`
`rArr T = F - (mu Mg)/(L) x - (M)/(L) xa`
`= F - (mu Mg)/(L) x - (M)/(L)x [(F - mu Mg)/(M)] = F [1 - (x)/(l)]`
`Delta l = int Delta (dx) int_(0)^(1) (Fdx)/(AY) = (Fl)/(2AY)`
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