JEE Main 2017 · 2 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

The moment of inertia of a uniform cylinder of length $l$ and radius R about its perpendicular bisector is I. What is the ratio $l/R$ such that the moment of inertia is minimum ?

\sqrt{\dfrac32}

\dfrac{\sqrt3}2

\dfrac3{\sqrt2}

Answer and worked solution

Official answer: A

$I=M\left(\dfrac{R^2}4+\dfrac{l^2}{12}\right)$ with the volume $V=\pi R^2l$ fixed, so $R^2=\dfrac V{\pi l}$ :
$I=M\left(\frac V{4\pi l}+\frac{l^2}{12}\right),\qquad\frac{dI}{dl}=-\frac V{4\pi l^2}+\frac l6=0.$
So $l^3=\dfrac{3V}{2\pi}=\dfrac32R^2l$ , giving $\dfrac lR=\sqrt{\dfrac32}$ .

Topics:Moment of inertia

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 5