JEE Advanced Model Paper 3 with solutions for online practice

As F < µMg , i.e. , rod is stationary so F = f where f is the friction force. Let Tis the tension in rod at distance x from the end where force has been applied.

Then, F - T - f1 = 0 & T = f2
Where f1+f2 = f = F
As friction force is distributed uniformly,
f1 =

fx

l

⇒ T = F - f1 = F -

f×x

l

= F [1−

x

l

]
Let Δ(dx) be the elongation in dx length of rod, Δ(dx) =

Tdx

AY

=

F(1− x l )dx

AY

Total elongation , Δl = ∫Δ(dx) =

1

∫

0

F

AY

(1−

x

l

) dx =

Fl

2YA