Thank you for reporting, we will resolve it shortly
Q.
If a charged spherical conductor of radius 10 cm has potential V at a point distant 5 cm from its centre, then the potential at a point distant 15 cm from the centre will be:
We know that the potential inside and on the surface of a spherical conductor is same so $ V={{V}_{surface}}=\frac{q}{10}\text{stat}\,\text{volt} $ $ {{V}_{out}}=\frac{q}{15}\text{stat}\,\text{volt} $ $ \frac{{{V}_{out}}}{V}=\frac{2}{3} $ so $ {{V}_{out}}=\frac{2}{3}V $