The moment of inertia of the uniform rod about an axis through one end and perpendicular to length is
$$I = \frac{{m{l^2}}}{3}$$
where, $$m$$ is mass of rod and $$l$$ its length.
Torque $$\left( {\tau = I\alpha } \right)$$   acting on centre of gravity of rod is given by
$$\tau = mg\frac{l}{2}$$
As we know that $${\tau = I\alpha }$$
$$\eqalign{ & {\text{so}}\,I\alpha = mg\frac{l}{2}\,\,{\text{or}}\,\,\frac{{m{l^2}}}{3}\alpha = mg\frac{l}{2} \cr & \therefore \alpha = \frac{{3g}}{{2l}} \cr} $$

Consider a rigid body rotating about a given axis with a uniform angular velocity $$\omega .$$ Let the body consists of $$n$$ particles of masses $${m_1},{m_2},{m_3},....{m_n}$$    at perpendicular distances $${r_1},{r_2},{r_3},....{r_n}$$    respectively from the axis of rotation.

As the body is rigid, angular velocity $$\omega $$ of all the particles is the same. However, as the distances of the particles from the axis of rotation are different, their linear velocities are different. If $${v_1},{v_2},{v_3},....{v_n}$$    are the linear velocities of the particles respectively, then
$$\eqalign{ & {v_1} = {r_1}\omega \cr & {v_2} = {r_2}\omega \cr & {v_3} = {r_3}\omega .... \cr} $$
The linear momentum of this particle of mass $${m_1}$$ is $${p_1} = {m_1}{v_1} = {m_1}\left( {{r_1}\omega } \right)$$
The angular momentum of this particle about the given axis
$$\eqalign{ & = {p_1} \times {r_1} = \left( {{m_1}{r_1}\omega } \right) \times {r_1} \cr & = {m_1}r_1^2\omega \cr} $$
Similarly, angular momenta of other particles of the body about the given axis are $${m_2}r_2^2\omega ,{m_3}r_3^2\omega ,....{m_n}r_n^2\omega $$
∴ Angular momentum of the body about the given axis $$L = {m_1}r_1^2\omega + {m_2}r_2^2\omega + {m_2}r_3^2\omega + ... + {m_n}r_n^2\omega $$
$$\eqalign{ & = \left( {{m_1}r_1^2 + {m_2}r_2^2 + {m_3}r_3^2 + ... + {m_n}r_n^2} \right)\omega \cr & {\text{or}}\,\,L = \left( {\sum\limits_{i = 1}^n {{m_i}r_i^2} } \right)\omega \cr & {\text{or}}\,\,L = I\omega \cr} $$
where, $$I = \sum\limits_{i = 1}^n {{m_i}r_i^2} $$   is moment of inertia of the body about the given axis.

The equation of the line is $$Y = X + 4$$
$${\text{or}}\,\,X - Y + 4 = 0$$
Length of perpendicular from origin on this line is
$$R = \frac{{0 - 0 + 4}}{{\sqrt {{1^2} + {1^2}} }} = \frac{4}{{\sqrt 2 }}$$
$$\therefore $$ Angular momentum
$$L = mvR = 5 \times 3\sqrt 2 \times \frac{4}{{\sqrt 2 }} = 60\,unit$$
Alternative

$$Y = X + 4$$   line is shown in the figure.
$$\eqalign{ & {\text{When}}\,X = 0,\,\,Y = 4 \cr & {\text{so}}\,\,OS = 4 \cr} $$
To find slope of this line comparing this with equation of line
$$\eqalign{ & y = m'x + c \cr & \therefore {\text{Slope,}}\,\,m' = \tan \theta = 1 \cr & \Rightarrow \theta = {45^ \circ } \cr} $$
Length of perpendicular $$= OP$$
$$\eqalign{ & {\text{In}}\,\,\Delta PSO,\,\,\frac{{OP}}{{OS}} = \sin {45^ \circ } \cr & \therefore OP = OS\sin {45^ \circ } \cr & = 4 \times \frac{1}{{\sqrt 2 }} = \frac{4}{{\sqrt 2 }} \cr} $$
∴ Angular momentum of particle going along this line
$$\eqalign{ & = mvR \cr & = 5 \times 3\sqrt 2 \times \frac{4}{{\sqrt 2 }} \cr & = 60{\text{ }}unit \cr} $$

Use theorem of parallel axes.

$$ \bullet $$  When $$n=0,\,x=k$$   where $$k$$ is a constant. This means that the linear mass density is constant. In this case the centre of mass will be at the middle of the rod ie at $$\frac{L}{2}.$$
Therefore (C) is ruled out.
$$ \bullet $$  $$n$$ is positive and as its value increases, the rate of increase of linear mass density with increase in $$x$$ increases. This shows that the centre of mass will shift towards that end of the rod where $$n = L$$  as the value of $$n$$ increases.
Therefore graph (B) is ruled out.
$$ \bullet $$  The linear mass density $$\lambda = k{\left( {\frac{x}{L}} \right)^n}{\text{ Here }}\frac{x}{L} \leqslant 1$$
With increase in the value of $$n,$$ the centre of mass shift towards the end $$x = L$$  such that first the shifting is at a higher rate with increase in the value of $$n$$ and then the rate decreases with the value of $$n.$$
These characteristics are represented by graph (A).
$$\eqalign{ & {x_{CM}} = \frac{{\int\limits_0^L {x\,dm} }}{{\int\limits_0^L {dm} }} \cr & = \frac{{\int\limits_0^L {x\,\left( {\lambda \,dx} \right)} }}{{\int\limits_0^L {\lambda \,dx} }} \cr & = \frac{{\int\limits_0^L {k\,{{\left( {\frac{x}{L}} \right)}^n}\,.\,xdx} }}{{\int\limits_0^L {k\,{{\left( {\frac{x}{L}} \right)}^n}\,.\,dx} }} \cr & = \frac{{k\left[ {\frac{{{x^{n + 2}}}}{{\left( {n + 2} \right){L^n}}}} \right]_0^L}}{{\left[ {\frac{{k\,{x^{n + 1}}}}{{\left( {n + 1} \right){L^n}}}} \right]_0^L}} \cr & = \frac{{L\left( {n + 1} \right)}}{{n + 2}} \cr & {\text{For }}n = 0,\,\,\,\,{x_{CM}} = \frac{L}{2};\,\,\,\,n = 1, \cr & {x_{CM}} = \frac{{2L}}{3};\,\,\,\,n = 2,\,\,\,\,{x_{CM}} = \frac{{3L}}{4};\,..... \cr} $$

Angular momentum $$\left( L \right) = $$   (linear momentum) x (perpendicular distance of the line of action of momentum from the axis of rotation)
$$ = mv \times r$$   [ Here $$r= 0$$  because the line of action of momentum passes through the axis of rotation]
$$\eqalign{ & = mv \times 0 \cr & = 0 \cr} $$

By the concept of energy conservation
$$\eqalign{ & \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = mg\left( {\frac{{3{v^2}}}{{4g}}} \right) \cr & \therefore \frac{1}{2}m{v^2} + \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2}\,\,\,\,\,\left[ {\because v = R\omega } \right] \cr & \therefore \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2} - \frac{1}{2}m{v^2} = \frac{1}{4}m{v^2} \cr & \Rightarrow I = \frac{1}{2}m{R^2} \cr} $$
This is the formula of the moment of inertia of the disc.

As two point masses $${m_1}$$ and $${m_2}$$ are placed at opposite ends of a rigid rod of length $$L$$ and negligible mass as shown in figure. Total moment of inertia of the rod
$$\eqalign{ & I = {m_1}{x^2} + {m_2}{\left( {L - x} \right)^2} \cr & I = {m_1}{x^2} + {m_2}{L^2} + {m_2}{x^2} - 2{m_2}Lx \cr} $$
As, $$I$$ is minimum i.e.
$$\eqalign{ & \frac{{dI}}{{dx}} = 2{m_1}x + 0 + 2x{m_2} - 2{m_2}L = 0 \cr & \Rightarrow x\left( {2{m_1} + 2{m_2}} \right) = 2{m_2}L \cr & \Rightarrow x = \frac{{{m_2}L}}{{{m_1} + {m_2}}} \cr} $$
When $$I$$ is minimum, then work done on rotating a rod $$\frac{1}{2}I{\omega ^2}$$  with angular velocity $${\omega _0}$$ will be minimum.
Shortcut Way
The position of point $$P$$ on rod through which the axis should pass, so that the work required to set the rod rotating with minimum angular velocity $${\omega _0}$$ is their centre of mass, we have
$${m_1}x = {m_2}\left( {L - x} \right) \Rightarrow x = \frac{{{m_2}L}}{{{m_1} + {m_2}}}$$

$$\eqalign{ & {m_2}g - T = {m_2}{a_2}\,......\left( {\text{i}} \right) \cr & TR = \frac{{{m_2}{R^2}}}{2}{\alpha _2}\,......\left( {{\text{ii}}} \right) \cr & {\alpha _1}R = {a_2} - {\alpha _2}R\,......\left( {{\text{iii}}} \right) \cr & TR = \left( {\frac{{{m_1}{R^2}}}{2}} \right){\alpha _1}\,......\left( {{\text{iv}}} \right) \cr & {\alpha _2} = \frac{{2T}}{{{m_2}R}} = \frac{2}{{{m_2}R}}.\left( {{m_2}{a_2} + {m_2}g} \right) \cr & = \frac{{2\left( {{a_2} + g} \right)}}{R} \cr} $$

$${I_{nn'}} = M.I$$   due to the point mass at $$B + M.I$$  due to the point mass at $$D + M.I$$  due to the point mass at $$C.$$
$$ = 2 \times m{\left( {\frac{\ell }{{\sqrt 2 }}} \right)^2} + m{\left( {\sqrt 2 \ell } \right)^2} = m{\ell ^2} + 2m{\ell ^2} = 3m{\ell ^2}$$