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Last updated date: 03rd Dec 2023
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MVSAT Dec 2023

Two small spheres of mass $10\,kg$ and $30\,kg$ are joined by a rod of length $0.5\,m$ and of the system negligible mass. The M.I. of the system about an axis passing through centre of rod and normal to it is
A. $2.5\,kg{m^2}$
B. $2.5\,kg{m^2}$
C. $2.5\,kg{m^2}$
D. $2.5\,kg{m^2}$

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Hint:Before going through the question let us have some idea about centre of mass. A point at which the entire mass of a body or all the masses of a system of particles appears to be concentrated is known as the centre of mass of a body or system of particles.The centre of mass of the system of particles is this location. S.I. unit of centre of mass is the meter.

Complete step by step answer:
In physics, a moment of inertia is a measurable indicator of a body's rotational inertia—that is, the body's resistance to having its speed of rotation about an axis changed by the application of a torque (turning force). The axis can be internal or external, and it can be set or not.

The moment of inertia (I) is calculated by multiplying the mass of each particle of matter in a given body by the square of its distance from the axis with respect to that axis. The moment of the inertia unit is a composite measurement unit. In the International System (SI), m is measured in kilogrammes and r is measured in metres, with I (moment of inertia) having the kilogram-meter square dimension. The system's M.I. is measured around a standard axis via the system's centre.

Now, according to the given question,
$I = {m_1}{r_1}^2 + {m_2}{r_2}^2$
\[\Rightarrow I=10 \times {\left( {0.25} \right)^2} + 30 \times {\left( {0.25} \right)^2}\]
\[\Rightarrow I = 40 \times {\left( {0.25} \right)^2}\]
\[\therefore I = 2.5\,kg\,{m^2}\]

Hence the correct option is B.

Note:The mass, as well as the distribution of mass along the axis along which it is measured, determine the moment of inertia. Different moments of inertia will exist for an object along different axes.