The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If
\(\frac{I_1}{I_2}\) is \(\frac{xπ²}{3}\)
then the value of x will be ______.
The correct answer is 8
\(I_1 = \frac{ML^2}{3} ...... (i)\)
For ring :
\(I_2 = \frac{MR^2}{2}\)and 2πR = L
\(⇒ I_2 = \frac{M}{2} (\frac{L²}{4π²}) ....... (ii)\)
From equation (i) and (ii) , we get
\(⇒ \frac{I_1}{I_2} = \frac{8π²}{3}\)
⇒ x = 8
The remainder when \( 64^{64} \) is divided by 7 is equal to:
x mg of Mg(OH)$_2$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer) (Given: Mg(OH)$_2$ is assumed to dissociate completely in H$_2$O)
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-