A conducting sphere of radius 10cm is given a charge of 10 μF. Another uncharged sphere of radius 20cm is allowed to touch it for some time. After that if the spheres are separated, then surface density of charges, on the spheres will be in the ratio of

When two conducting spheres of different radii come into contact, charge redistributes between them until they reach the same electric potential. Given a charged sphere of radius 10 cm with a charge of 10 μC and an uncharged sphere of radius 20 cm, upon touching and separating, the surface charge densities (charge per unit area) on the spheres will be in the ratio 2:1.

Explanation:

1. Equalizing Potential: When the spheres touch, charges flow between them until both spheres have the same electric potential.
2. Charge Distribution: The charge Q on a sphere is related to its radius r and surface charge density σ by the formula: 
   Q = σ × 4πr².
3. Surface Charge Density: Since the potential V is equalized and V ∝ Q/r, the charge Q is proportional to the radius r. Therefore, the surface charge density σ = Q / 4πr² is inversely proportional to the radius.
4. Ratio Calculation: For spheres with radii in the ratio 1:2, the surface charge densities will be in the inverse ratio, i.e., 2:1.

Thus, after separation, the surface charge density on the smaller sphere will be twice that on the larger sphere, resulting in a ratio of 2:1.

Final Answer: (c) 2:1