Question

A sphere of radius R is removed from a bigger sphere of radius 2R such that the circumferences of the spheres coincide. The centre of mass of the new sphere is $\alpha$ R from the centre of the bigger disc. The value of$\alpha$ is
A. 1/3
B. 1/7
C. 1/6
D. 1/4

Answer 1

Here, as the smaller sphere is getting detached from the larger sphere, the total mass of the system will decrease and hence there will be negative sign in the formula of the centre of mass. In addition to this, we are taking the centre of the larger sphere as origin so the position of its centre of mass would be zero.

Complete step by step answer:
We have a bigger sphere which has a radius of 2R. Now, there is also a smaller sphere which has a radius of R inside the bigger sphere. Now, both the spheres’ circumference coincides and the smaller sphere has been detached from the larger sphere, and we need to find the centre of mass after the sphere has been detached. Here, we take$\rho$as the surface charge density of the spheres which is constant. Centre of mass for two bodies if one body is detached can be described from the following formula;
$X = \dfrac{{{M_l}{x_l} - {M_s}{x_s}}}{{{M_l} - {M_s}}}$
Here, X is the centre of mass from the centre of the larger sphere after smaller sphere is detached, ${M_l}$is the mass of the larger sphere, ${x_l}$is the distance of centre of mass of larger sphere from the large sphere, ${M_s}$is the mass of the smaller sphere and${x_s}$is the centre of mass of smaller sphere from the larger sphere. Now, the masses of the spheres can be obtained by multiplying the surface charge density with respective areas, hence the centre of mass will be as follows:

$$X=\dfrac{(\rho \centerdot \pi {{(2R)}^{2}})\times 0-(\rho \centerdot \pi {{(R)}^{2}})\times R}{(\rho \centerdot \pi {{(2R)}^{2}})\times -(\rho \centerdot \pi {{(R)}^{2}})} \\\ \therefore X=\dfrac{R}{3}$$

Now, from the question, we have $X = \alpha R$. Thus, $\alpha = \dfrac{1}{3}$.

Hence option A is the correct answer.

Note: We have assumed that the surface charge density for both the spheres is the same, that is it is uniform. This is because the sphere is made up of the same material and then the other sphere got detached from it. Also, it is important to remember to use the surface charge density and not volume charge density.