A $60 cm$ metal rod $M_{1}$ is joined to another $100 cm$ metal rod $M_{2}$ to form an $L$ shaped single piece. This piece is hung on a ply at the joint $A$ . The two rods are observed to be equally inclined to the vertical as shown. The two rods are equally thick. If the ratio of density $M_{1}$ to that of $M_{2}$ is $η$ , then value of $\frac{18 η}{25}$ is

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Solution

Torque about joint A is zero, as setup is in equilibrium
$τ_{A} = 0$
$\frac{m_{1} g l_{1} sin 45^{\circ}}{2} = \frac{m_{2} g l_{2} sin 45^{\circ}}{2}$ ........(1)
$m_{1} = ρ_{1} l_{1} A$
$m_{2} = ρ_{2} l_{2} A$ , where $A$ is cross sectional area of rods
From (1)
$ρ_{1} l_{1}^{2} = ρ_{2} l_{2}^{2}$
$η = \frac{ρ_{1}}{ρ_{2}} = {(\frac{5}{3})}^{2} = \frac{25}{9}$
$x = \frac{18}{25} η = 2$

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Torque about joint A is zero, as setup is in equilibrium τA=0 m1gl1sin45∘2=m2gl2sin45∘2 ........(1) m1=ρ1l1A m2=ρ2l2A, where A is cross sectional area of rods From (1) ρ1l21=ρ2l22 η=ρ1ρ2=(53)2=259 x=1825η=2