Question

JEE MAINS PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Let the mass of the cylinder be $\sf{M}$ and $\sf{L}$ and $\sf{R}$ are its length and base radius respectively.

Consider an elemental disc of mass $\sf{dm}$ and width $\sf{dx}$ from the solid cylinder which is at a distance $\sf{x}$ from the axis OO'.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(0,-10)(0,20){2}{\line(1,0){50}}\multiput(0,0)(50,0){2}{\qbezier(-1,-9)(-3.5,0)(-1,9)\qbezier(-1,-9)(0,-11)(1,-9)\qbezier(1,-9)(3.5,0)(1,9)\qbezier(-1,9)(0,11)(1,9)}\multiput(0,0)(0.3,0){2}{\qbezier(34,9)(35,11)(36,9)\qbezier(36.5,7)(37.15,6)(37,5)\qbezier(37.6,2.4)(37.8,0.7)(37.7,0)\qbezier(37.5,-3)(38,-2)(37.1,-5.5)\qbezier(36.3,-8)(35.2,-10.6)(34.5,-9.7)\qbezier(33.67,-8)(33.5,-7)(33.1,-5.5)\qbezier(32.9,-3)(32.9,-2)(32.6,0)\qbezier(32.9,2.4)(33,5)(33.2,5.8)}\put(25,0){\vector(0,-1){20}}\put(25,0){\vector(0,1){20}}\put(25,0){\vector(1,0){10}}\put(28.5,1.5){\sf{x}}\put(50,0){\vector(0,1){10}}\put(52.5,3){\sf{R}}\put(30,10){\vector(1,0){4.7}}\put(40,10){\vector(-1,0){4.7}}\put(33,12){\sf{dx}}\put(23.5,22.5){\sf{O}}\put(23.5,-24.5){\sf{O'}}\put(24,-30){\sf{L}}\put(22.5,-28.5){\vector(-1,0){22.5}}\put(27.5,-28.5){\vector(1,0){22.5}}\end{picture}$

Assume the cylinder to be uniform so it has uniform linear density (area of cross section is uniform).

$\sf{\longrightarrow \dfrac{M}{L}=\dfrac{dm}{dx}}$

$\sf{\longrightarrow dm=\dfrac{M}{L}\ dx}$

The cylinder is of given volume $\sf{V}$ (say).

$\sf{\longrightarrow V=\pi R^2L}$

$\sf{\longrightarrow R^2=\dfrac{V}{\pi L}}$

The moment of inertia of the disc about an axis passing along one of its diameter is,

$\sf{\longrightarrow dI_d=\dfrac{1}{4}\,dm\cdot R^2}$

By parallel axes theorem, the moment of inertia of the disc about OO' is given by,

$\sf{\longrightarrow dI=\dfrac{1}{4}\,dm\cdot R^2+dm\cdot x^2}$

$\sf{\longrightarrow dI=dm\left(\dfrac{R^2}{4}+x^2\right)}$

$\sf{\longrightarrow dI=\dfrac{M}{L}\left(\dfrac{V}{4\pi L}+x^2\right)\ dx}$

Hence the moment of inertia of the whole solid cylinder will be,

$\displaystyle\sf{\longrightarrow I=\dfrac{M}{L}\int\limits_{-\frac{L}{2}}^{\frac{L}{2}}\left(\dfrac{V}{4\pi L}+x^2\right)\ dx}$

$\displaystyle\sf{\longrightarrow I=\dfrac{MV}{4\pi L^2}\big[x\big]_{-\frac{L}{2}}^{\frac{L}{2}}+\dfrac{M}{3L}\left[x^3\right]_{-\frac{L}{2}}^{\frac{L}{2}}}$

$\displaystyle\sf{\longrightarrow I=\dfrac{MV}{4\pi L}+\dfrac{ML^2}{12}}$

The condition for maximum moment of inertia:-

$\sf{\longrightarrow\dfrac{dI}{dL}=0}$

$\displaystyle\sf{\longrightarrow\dfrac{d}{dL}\left[\dfrac{MV}{4\pi L}+\dfrac{ML^2}{12}\right]=0}$

Since the volume is constant,

$\displaystyle\sf{\longrightarrow-\dfrac{MV}{4\pi L^2}+\dfrac{ML}{6}=0}$

$\displaystyle\sf{\longrightarrow\dfrac{L}{6}=\dfrac{V}{4\pi L^2}}$

$\displaystyle\sf{\longrightarrow\dfrac{L}{6}=\dfrac{R^2}{4L}\quad\quad\!\!\left[\because\,V=\pi R^2L\right]}$

$\displaystyle\sf{\longrightarrow\dfrac{R^2}{L^2}=\dfrac{2}{3}}$

$\displaystyle\sf{\longrightarrow\underline{\underline{\dfrac{L}{R}=\sqrt{\dfrac{3}{2}}}}}$

Hence (1) is the answer.