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JEE Main 20203 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
3 September
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2020 · 3 September · Morning Shift · Physics · Rotational Motion

Moment of inertia of a cylinder of mass m, length L and radius R...

Moment of inertia of a cylinder of mass m,m, length LL and radius RR about an axis passing through its centre and perpendicular to the axis of the cylinder is I=M(R24+L212).I = M\left(\dfrac{{R}^{2}}{4} + \dfrac{{L}^{2}}{12}\right) . If such a cylinder is to be made for a given mass of a material, the ratio LR\dfrac{L}{R} for it to have minimum possible II is:Moment of inertia of a cylinder of mass m, length L and radius R about an axis passing through its centre and

Options

  1. A23\dfrac{2}{3}
  2. B32\dfrac{3}{2}
  3. C32\sqrt{\dfrac{3}{2}}
  4. D23\sqrt{\dfrac{2}{3}}

Correct answer

C. 32\sqrt{\dfrac{3}{2}}

Step-by-step solution

Let a cylinder of mass m1{m}_{1}, length LL and radius RR then take elementary disc of radius RR and trickiness dxdx at distance of xx from axis OO′OO ' then moment of inertia about OO′OO ' as this element

Moment of inertia of a cylinder of mass m, length L and radius R about an axis passing through its centre and

dl=dmR24+dmx2dl = \dfrac{{dmR}^{2}}{4} + {dmx}^{2}

I=∫dl=∫dmR24+∫n=L/2n=−L/2MLdx×x2I = ∫ dl = ∫ \dfrac{{dmR}^{2}}{4} + {∫}_{n = L / 2}^{n = - L / 2}\dfrac{M}{L}dx \times {x}^{2}

⇒I=MR24+ML212⇒ I = \dfrac{{MR}^{2}}{4} + \dfrac{{ML}^{2}}{12}

⇒I=M4×VπL+ML212⇒ I = \dfrac{M}{4} \times \dfrac{V}{\pi L} + \dfrac{{ML}^{2}}{12}

dIdL=−mV4πL2+M×2L12=0\dfrac{dI}{dL} = - \dfrac{mV}{4{\pi L}^{2}} + \dfrac{M \times 2L}{12} = 0

⇒V=23πL3⇒πR2L=23πL3⇒LR=32⇒ V = \dfrac{2}{3}{\pi L}^{3} ⇒ {\pi R}^{2}L = \dfrac{2}{3}{\pi L}^{3} ⇒ \dfrac{L}{R} = \sqrt{\dfrac{3}{2}}(I=MV4πL+ML212)\left(I = \dfrac{MV}{4\pi L} + \dfrac{{ML}^{2}}{12}\right)

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