Question

On the marks of 20 cm and 70 cm of a light meter scale, weights of 1kg and 4 kg respectively are placed. The M.I in Kg $m^2$ about the vertical axis through the 100 cm. mark will be :-

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$Given:$ $M_1=1kg$ ; $M_2=4kg$

$Solution:$ We know that,

Center of mass, $CM=\frac{{\sum }_{i=1}^n{m}_i{r}_i}{M}$

Where,

$m$ is the mass of individual points,

$r$ is the distance from the origin.

$M$ is the total mass of the system.

We can find the moment of inertia of a system as -

i.e.

Moment of inertia.

$I={\sum }_{i=1}^n{m}_i{r}_i^2$

Now, we are given two points at two marks on a

scale. They are -

$m_1=1kg,r_1=0.8m$

$m_2=4kg,r_2=0.3m$

Now, let us substitute this in the formula to find the

moment of inertia -

$I=m_1{r}_1^2+m_2{r}_2^2$

$\Rightarrow I=1\times (0.8{)}^2+4\times (0.3{)}^2$

$\Rightarrow I=0.64+0.36$

$\therefore I=1{kgm}^{-2}$

The moment of inertia is $1{kgm}^{-2}$

$So,the$ $correct$ $option:A$