Match List-I with List-II.
List - I | List - II | ||
A. | Moment of inertia of solid sphere of Radius R about any tangent | i. | \(\frac{5}{3}MR^2\) |
B. | Moment of inertia of hollow sphere of radius (R) about any tangent | ii. | \(\frac{7}{5}MR^2\) |
C. | Moment of inertia of circular ring of radius (R) about its diameter | iii. | \(\frac{1}{4}MR^2\) |
D. | Moment of inertia of circular disc of radius (R) about any diameter | iv. | \(\frac{1}{2}MR^2\) |
Choose the correct answer from the options given below.
The correct answer is (A) : A-II, B-I, C-IV, D-III
(A) Moment of inertia of solid sphere of radius R about a tangent
\(= \frac{2}{5} MR² + MR² = \frac{7}{5} MR²\)
⇒ A – (II)
(B) Moment of inertia of hollow sphere of radius R about a tangent
\(= \frac{2}{3} MR² + MR² = \frac{5}{3} MR²\)
⇒ B – (I)
(C) Moment of inertia of circular ring of radius (R) about its diameter
= \(\frac{(MR²) }{ 2}\)
⇒ C – (IV)
(D) Moment of inertia of circular ring of radius (R) about any diameter
= \(\frac{\frac{MR²}{2} }{2} = \frac{MR²}{4}\)
⇒ D – (III)
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
In general form, the moment of inertia can be expressed as,
I = m × r²
Where,
I = Moment of inertia.
m = sum of the product of the mass.
r = distance from the axis of the rotation.
M¹ L² T° is the dimensional formula of the moment of inertia.
The equation for moment of inertia is given by,
I = I = ∑mi ri²
To calculate the moment of inertia, we use two important theorems-