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Correct option is C)

$Hint$: Draw the FBD

$Step 1$:

Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is $μ$.

Let mass of the ladder is $m$.

From the figure: for the vertical equilibrium $N_{1}+μN_{2}=mg$ $(1)$

For horizontal equilibrium $N_{2}=μN_{1}$ $(2)$

The torque about point A is given by:

$mg(2.5)cosθ=N_{2}(4)+μN_{2}(3)$

$mg(2.5×53 )=(4+3μ)N_{2}$, since $cosθ=53 $

$mg(23 )=(4+3μ)N_{2}$ $(3)$

\textbf{Step 2}$$:

From equation $(1)$ and $(2)$

$N_{2}(μ1+μ_{2} )=mg⟹N_{2}=1+μ_{2}μmg $

So, $mg23 =(4+3μ)1+μ_{2}μmg $

$(3+3μ_{2})=8μ+6μ_{2}$

$3μ_{2}+8μ−3=0$

$μ=6−8±sqrt64+36 =6−8±10 $

$μ=31 $

Thus option C is correct.

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