The linear mass density of a thin rod AB of length L varies from A to B as \lambda ( x ) = \lambda _0( 1 + x \over L ) , where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

Rotational Motion PYQ — JEE Main 2020 · 6 Sep · Shift 2 · Q37 with Solution

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Correct answer: C

Given linear mass density

The rod has length $L$ and variable linear mass density
$\lambda(x)=\lambda_0\left(1+\frac{x}{L}\right), \qquad 0\le x\le L$
where $x$ is measured from end $A$ .
Find total mass $M$ of the rod

A small element of length $dx$ at distance $x$ has mass
$dm = \lambda(x)\,dx = \lambda_0\left(1+\frac{x}{L}\right)dx$
So total mass is
$M=\int_0^L dm = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)dx$ $M=\lambda_0\left[\int_0^L 1\,dx + \frac{1}{L}\int_0^L x\,dx\right]$ $M=\lambda_0\left[L + \frac{1}{L}\cdot \frac{L^2}{2}\right] =\lambda_0\left(L+\frac{L}{2}\right) =\frac{3}{2}\lambda_0 L$
Hence,
$\lambda_0 = \frac{2M}{3L}$
Moment of inertia about axis through $A$ and perpendicular to rod

For the small element $dm$ at distance $x$ from $A$ ,
$dI = x^2\,dm$
Therefore,
$I_A = \int_0^L x^2\,dm = \int_0^L x^2\lambda_0\left(1+\frac{x}{L}\right)dx$ $I_A = \lambda_0\left[\int_0^L x^2\,dx + \frac{1}{L}\int_0^L x^3\,dx\right]$
Using
$\int_0^L x^2\,dx = \frac{L^3}{3}, \qquad \int_0^L x^3\,dx = \frac{L^4}{4}$
we get
$I_A = \lambda_0\left(\frac{L^3}{3}+\frac{1}{L}\cdot\frac{L^4}{4}\right) =\lambda_0 L^3\left(\frac{1}{3}+\frac{1}{4}\right)$ $I_A = \lambda_0 L^3\cdot \frac{7}{12}$
Substitute $\lambda_0$ in terms of $M$
$I_A = \frac{7}{12} \lambda_0 L^3 = \frac{7}{12}\cdot \frac{2M}{3L}\cdot L^3$ $I_A = \frac{14}{36}ML^2 = \frac{7}{18}ML^2$
Match with options
$I_A = \frac{7}{18}ML^2$
So the correct option is C.

View written solutionFree

Correct answer: C

Given linear mass density

The rod has length $L$ and variable linear mass density
$\lambda(x)=\lambda_0\left(1+\frac{x}{L}\right), \qquad 0\le x\le L$
where $x$ is measured from end $A$ .
Find total mass $M$ of the rod

A small element of length $dx$ at distance $x$ has mass
$dm = \lambda(x)\,dx = \lambda_0\left(1+\frac{x}{L}\right)dx$
So total mass is
$M=\int_0^L dm = \int_0^L \lambda_0\left(1+\frac{x}{L}\right)dx$ $M=\lambda_0\left[\int_0^L 1\,dx + \frac{1}{L}\int_0^L x\,dx\right]$ $M=\lambda_0\left[L + \frac{1}{L}\cdot \frac{L^2}{2}\right] =\lambda_0\left(L+\frac{L}{2}\right) =\frac{3}{2}\lambda_0 L$
Hence,
$\lambda_0 = \frac{2M}{3L}$
Moment of inertia about axis through $A$ and perpendicular to rod

For the small element $dm$ at distance $x$ from $A$ ,
$dI = x^2\,dm$
Therefore,
$I_A = \int_0^L x^2\,dm = \int_0^L x^2\lambda_0\left(1+\frac{x}{L}\right)dx$ $I_A = \lambda_0\left[\int_0^L x^2\,dx + \frac{1}{L}\int_0^L x^3\,dx\right]$
Using
$\int_0^L x^2\,dx = \frac{L^3}{3}, \qquad \int_0^L x^3\,dx = \frac{L^4}{4}$
we get
$I_A = \lambda_0\left(\frac{L^3}{3}+\frac{1}{L}\cdot\frac{L^4}{4}\right) =\lambda_0 L^3\left(\frac{1}{3}+\frac{1}{4}\right)$ $I_A = \lambda_0 L^3\cdot \frac{7}{12}$
Substitute $\lambda_0$ in terms of $M$
$I_A = \frac{7}{12} \lambda_0 L^3 = \frac{7}{12}\cdot \frac{2M}{3L}\cdot L^3$ $I_A = \frac{14}{36}ML^2 = \frac{7}{18}ML^2$
Match with options
$I_A = \frac{7}{18}ML^2$
So the correct option is C.