The correct option is D Vc=q4πϵ0(r+Rr2+R2)
Let charge on inner sphere be q1 and, on outer shell be q2.
According to the given data in the problem,
q=q1+q2 ……(1)
Since the surface charge density (σ) on both the spheres are equal, so
σ=q14πr2=q24πR2
⇒q1q2=r2R2 ........(2)
Adding +1 on both sides we get,
⇒q1+q2q2=r2+R2R2
From equation (1),
qq2=r2+R2R2
∴q2=qR2r2+R2...(3)
Substituting this in (2),
q1=qr2r2+R2...(4)
The potential at the common centre is
Vc=V1+V2
⇒Vc=14πϵ0[q1r+q2R]
Substituting the values from (3) and (4)
⇒Vc=14πϵ0[(qr2+R2)[r2r+R2R]]
∴Vc=q4πϵ0(r+Rr2+R2)
Hence, option (d) is the correct answer.