As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered 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 detemined by the equation :

Center of Mass PYQ — JEE Main 2020 · 8 Jan · Shift 2 · Q55 with Solution

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Correct answer: D

Set up the geometry

Let the big sphere have:

radius $R$
center $C$

A smaller spherical cavity of radius $1$ is removed, with center $O$ .

From the figure/context, the cavity is internally tangent to the outer sphere, so the distance between centers is $CO = R-1.$

The center of mass $G$ of the remaining body lies on the line $CO$ and is given to be on the surface of the cavity.

Since the cavity surface is at distance $1$ from $O$ , and from the usual figure $G$ lies on the far side of $O$ from $C$ , we have $OG=1.$ Hence $CG = CO + OG = (R-1)+1 = R.$ So the COM of the remaining part is at distance $R$ from $C$ .

Use negative mass method

Treat the removed cavity as a sphere of negative mass.

Let density be $\rho$ .

Then

mass of full sphere: $M = \frac{4}{3}\pi \rho R^3$
mass of removed cavity: $m = \frac{4}{3}\pi \rho$

So effectively, a mass $m$ is removed from point $O$ , whose distance from $C$ is $R-1$ .

Take $C$ as origin and positive direction toward $O$ . Then COM of remaining body is

= -\frac{m(R-1)}{M-m}.$$ Thus its distance from $C$ is $$CG = \frac{m(R-1)}{M-m}.$$ Substitute $M = mR^3$: $$CG = \frac{m(R-1)}{m(R^3-1)} = \frac{R-1}{R^3-1}.$$ But this is only the dimensionless fraction if cavity radius were scaled out incorrectly. Let us write it carefully including actual masses: Since $$M = kR^3,\qquad m=k$$ with $k=\frac{4}{3}\pi\rho$, we get $$CG = \frac{k(R-1)}{k(R^3-1)} = \frac{R-1}{R^3-1}.$$ This simplifies to $$CG = \frac{1}{R^2+R+1}.$$ This is the shift from $C$ toward the side opposite the cavity, so the point $G$ lies on the side away from $O$. Hence distance from $O$ to $G$ is $$OG = CO + CG = (R-1) + \frac{1}{R^2+R+1}.$$ Given that $G$ is on the cavity surface, $$OG = 1.$$ Therefore, $$(R-1) + \frac{1}{R^2+R+1} = 1.$$ So $$\frac{1}{R^2+R+1} = 2-R.$$ Rearranging, $$(R^2+R+1)(2-R)=1.$$ --- 3. **Match with options** This is exactly: $$\boxed{(R^2+R+1)(2-R)=1}$$ So the correct option is **D**. --- 4. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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View written solutionFree

Correct answer: D

Set up the geometry

Let the big sphere have:

radius $R$
center $C$

A smaller spherical cavity of radius $1$ is removed, with center $O$ .

From the figure/context, the cavity is internally tangent to the outer sphere, so the distance between centers is $CO = R-1.$

The center of mass $G$ of the remaining body lies on the line $CO$ and is given to be on the surface of the cavity.

Use negative mass method

Treat the removed cavity as a sphere of negative mass.

Let density be $\rho$ .

Then

mass of full sphere: $M = \frac{4}{3}\pi \rho R^3$
mass of removed cavity: $m = \frac{4}{3}\pi \rho$

So effectively, a mass $m$ is removed from point $O$ , whose distance from $C$ is $R-1$ .

Take $C$ as origin and positive direction toward $O$ . Then COM of remaining body is

= -\frac{m(R-1)}{M-m}.$$ Thus its distance from $C$ is $$CG = \frac{m(R-1)}{M-m}.$$ Substitute $M = mR^3$: $$CG = \frac{m(R-1)}{m(R^3-1)} = \frac{R-1}{R^3-1}.$$ But this is only the dimensionless fraction if cavity radius were scaled out incorrectly. Let us write it carefully including actual masses: Since $$M = kR^3,\qquad m=k$$ with $k=\frac{4}{3}\pi\rho$, we get $$CG = \frac{k(R-1)}{k(R^3-1)} = \frac{R-1}{R^3-1}.$$ This simplifies to $$CG = \frac{1}{R^2+R+1}.$$ This is the shift from $C$ toward the side opposite the cavity, so the point $G$ lies on the side away from $O$. Hence distance from $O$ to $G$ is $$OG = CO + CG = (R-1) + \frac{1}{R^2+R+1}.$$ Given that $G$ is on the cavity surface, $$OG = 1.$$ Therefore, $$(R-1) + \frac{1}{R^2+R+1} = 1.$$ So $$\frac{1}{R^2+R+1} = 2-R.$$ Rearranging, $$(R^2+R+1)(2-R)=1.$$ --- 3. **Match with options** This is exactly: $$\boxed{(R^2+R+1)(2-R)=1}$$ So the correct option is **D**. --- 4. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** They agree.

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