The concepts to form the integrand to find the moment of inertia are to be followed. Now, the origin is fixed to the left 'end of the rod and the limits are to be taken from 0 to l.

I =  \(\frac { M }{ l } \int _{ 0 }^{ l }{ { x }^{ 2 }dx } =\frac { M }{ l } \left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ l }=\frac { M }{ l } \left[ \frac { { l }^{ 3 } }{ 3 } \right] \)
I = \(\frac { 1 }{ 3 } \)Ml²

(i) Consider a rigid body rotating about a fixed axis. A point the body rotating about an axis perpendicular to the plane of the page. A tangential force F is applied on the body.
(ii) It produces a small displacement ds on the body. The work done (dw) by the force is,
dw=Fds
(iii) As the distance ds, the angle of rotation d\(\theta\) and radius r are related by the expression
ds=r d\(\theta\)
The expression for work done now becomes,
dw=F ds; dw=F r d\(\theta\)
(iv) The term (Fr) is the torque ፒ produced by the force on the body.
dw=\(\tau\)d\(\theta\)
This expression gives the work done by the external torque ፒ, which acts on the body rotating about a fixed axis through an angle d\(\theta\).

(i) A rotating rigid body with respect to any axis, is considered to be made up of point masses m₁, m₂, m₃, ... m_n at perpendicular distances (or positions) r₁, r₂, r₃....r_n respectively.
(ii) The moment of inertia of that object can be written as,
I=\(\sum { { m }_{ i }{ r }_{ i }^{ 2 } } ={ m }_{ 1 }{ r }_{ 1 }^{ 2 }+{ m }_{ 2 }{ r }_{ 2 }^{ 2 }+{ m }_{ 3 }{ r }_{ 3 }^{ 2 }+....+{ m }_{ n }{ r }_{ n }^{ 2 }\)
If all the n number of individual masses to be equal,
m=m₁=m₂=m₃=....=m_n
then,
I=\({ mr }_{ 1 }^{ 2 }+{ mr }_{ 2 }^{ 2 }+{ mr }_{ 3 }^{ 2 }+...+{ mr }_{ n }^{ 2 }\)
=\(m({ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+{ r }_{ n }^{ 2 })\)
=\(nm\left( \frac { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+r_{ n }^{ 2 } }{ n } \right) \)
I=MK²
where, nm is the total mass M of the body and K is the radius of gyration.
K=\(\sqrt { \frac { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 3 }^{ 2 }+...+{ r }_{ n }^{ 2 } }{ n } } \)
(iii) The expression for radius of gyration indicates that it is the root mean square (rms) distance of the particles of the body from the axis of rotation.

(i) The principle for beam balance used for weighing goods with the condition d₁ = d₂ ; F₁ =F₂·
\(\therefore \frac { { F }_{ 1 } }{ { F }_{ 2 } } =\frac { { d }_{ 2 } }{ { d }_{ 1 } } \)
(ii) If F₁ is the load and F₂ is our effort, we get advantage when, d₁ < d₂. This implies that F₁>F₂. Hence, we could lift a large load with small effort. The ratio \(\left( \frac { { d }_{ 2 } }{ d_{ 1 } } \right) \) is called mechanical advantage of the simple lever The pivoted point is called fulcrum. Mechanical Advantage (MA)=\(\frac { { d }_{ 2 } }{ { d }_{ 1 } } \) .