Animated Solution for Physics - System of Particles: As shown in figure, when a spherical cavity (centred at $O$) of radius 1 is cut out of a uniform sphere of radius $R$ (centred at $C$), the centre of mass of remaining (shaded) part of sphere is at $G$, i.e. on the surface of the cavity. $R$ can be determined by the equation

Visualizing the Geometry

Imagine a perfectly uniform, solid sphere of radius $R$. Now, suppose we carve out a smaller spherical cavity of radius $1$ from it. The problem states that this cavity touches the outer edge of the large sphere.

To make our mathematical lives easier, let's place the center of the original large sphere, $C$, exactly at the origin $(0,0)$. Because the cavity touches the outer boundary of the large sphere, its center $O$ must be located at a distance of $R-1$ from the origin.

We are given a fascinating constraint: the new center of mass of the remaining shape, $G$, lies exactly on the surface of the cavity. Geometrically, the leftmost edge of the cavity is at a distance of $1$ from its center $O$. Therefore, the coordinate of $G$ is $x_G=(R-1)-1=R-2$.

Negative Mass

How do we find the center of mass of a body with a hole in it? We use the brilliant concept of negative mass. Instead of dealing with complex integrals over an awkward shape, we treat the system as a superposition of two complete objects: 1. The original solid sphere (with positive mass $M$). 2. The removed cavity (treated as having a negative mass $-m$).

The center of mass of this composite system is given by the weighted average:

Let's define the masses in terms of the uniform density $\rho$: - Mass of the original sphere: $M=\frac{4}{3}\pi R^3\rho$ - Mass of the removed cavity: $m=\frac{4}{3}\pi (1{)}^3\rho$

We know that the new center of mass $x_{CM}$ is at $G$, so $x_{CM}=R-2$. Since we placed the original center $C$ at the origin, $x_C=0$. Substituting these into our center of mass formula gives:

Cross-multiplying to clear the denominator, we get:

Now, let's substitute our expressions for $M$ and $m$:

Notice how the constant term $\frac{4}{3}\pi \rho$ appears in every single term? We can safely divide it out from both sides of the equation. This leaves us with a beautifully simple algebraic relation:

Here, we encounter a classic algebraic identity: the difference of cubes. We can expand $R^3-1$ as:

Substituting this expansion back into our equation yields:

For a cavity of radius $1$ to exist inside a sphere of radius $R$, it is physically required that $R>1$. Therefore, $(R-1)$ is strictly non-zero, and we can confidently divide both sides by $(R-1)$:

To match the format of the given multiple-choice options, we can absorb the negative sign on the right side into the $(R-2)$ term on the left side. Multiplying both sides by $-1$ flips $(R-2)$ to $(2-R)$:

This is the exact equation that determines the radius $R$, perfectly matching option (a). The negative mass trick transforms a seemingly complex calculus problem into an elegant algebraic puzzle!