A solid sphere $A$ of radius $R$ and mass $M$ is attached at a point to a smaller solid sphere $B$ of radius $r < R$ and mass $m < M$. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of $A$ is $I_A$ and that calculated about a vertical axis passing through the centre of $B$ is $I_B$. The difference $I_A - I_B$ is :
To find the difference between the moments of inertia \( I_A - I_B \), we need to calculate the moment of inertia of the system about the respective axes through the centers of spheres \( A \) and \( B \).
The correct answer is:
$0$

A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
Bob B of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10\text{ m}$, as shown in the figure. Point mass A of mass $m$ travelling horizontally with speed $10\text{ ms}^{-1}$ hits bob B elastically. The bob B rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\text{ ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is :
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
