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 circular disc has radius \( R_1 \) and thickness \( T_1 \). Another circular disc made of the same material has radius \( R_2 \) and thickness \( T_2 \). If the moments of inertia of both the discs are same and \[ \frac{R_1}{R_2} = 2, \quad \text{then} \quad \frac{T_1}{T_2} = \frac{1}{\alpha}. \] The value of \( \alpha \) is __________.
A solid cylinder of radius $\dfrac{R}{3}$ and length $\dfrac{L}{2}$ is removed along the central axis. Find ratio of initial moment of inertia and moment of inertia of removed cylinder. 
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop 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-