The KE and moment of inertia about the given end point of a rod of mass m and length l and cross sectional area A which is rotating with ω=gl as shown in the Fig. will be [ density of the rod varies as ρ=ρs1+xl , x is the distance measured from O) ]

=∫0l13dmx2=∫0l13ρ01+xl(Adx)x2=13ρ0A∫0lx2dx+x3ldx =13ρ0Al33+1l×l44=7ρ0Al336

The KE and moment of inertia about the given end point of a rod of mass m and length l and cross sectional area A which is rotating with ω=gl as shown in the Fig. will be [ density of the rod varies as ρ=ρs1+xl , x is the distance measured from O) ]

=∫0l13dmx2=∫0l13ρ01+xl(Adx)x2=13ρ0A∫0lx2dx+x3ldx =13ρ0Al33+1l×l44=7ρ0Al336

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The $K E$ and moment of inertia about the given end point of a rod of mass m and length $l$ and cross sectional area $A$ which is rotating with $ω = \sqrt{\frac{g}{l}}$ as shown in the Fig. will be [ density of the rod varies as $ρ = ρ_{s} (1 + \frac{x}{l})$ , x is the distance measured from O) ]

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$K E = \frac{7}{36} m g l$

$I = \frac{7 ρ_{0} A l^{3}}{36}$

$K E = \frac{7}{39} m g l$

$I = \frac{11 ρ_{0} A l^{3}}{36}$

answer is A, B.

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$= \int_{0}^{l} \frac{1}{3} d m x^{2}$

= $\int_{0}^{l} \frac{1}{3} ρ_{0} (1 + \frac{x}{l}) (A d x) x^{2} = \frac{1}{3} ρ_{0} A \int_{0}^{l} [x^{2} d x + \frac{x^{3}}{l} d x]$

= $\frac{1}{3} ρ_{0} A [\frac{l^{3}}{3} + \frac{1}{l} \times \frac{l^{4}}{4}] = \frac{7 ρ_{0} A l^{3}}{36}$

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The KE and moment of inertia about the given end point of a rod of mass m and length l and cross sectional area A which is rotating with ω=gl as shown in the Fig. will be [ density of the rod varies as ρ=ρs1+xl , x is the distance measured from O) ]

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