Question

Two metallic spheres of radii r1 and r2 are connected by a thin wire if + 21 + 32 are the charges on the two spheres then

Answer 1

You haven't said what to find but I assume it is R1 and R2 so

For a Sphere V=kq/R

When connected by a thin wire the two spheres attain a common or same potential so equating the potentials

q1/R1=q2/R2

that is 21/32=R1/R2

-> Match with options .

Answer 2

• Potential at the surface of spherical conductor of radius r carrying charge q , V = q / 4π∈r
• The charges on sphere with radii r1 and r2 are +21 and +32 respectively.
• When these two charged spherical conductors are connected by  a wire , the potential at their surfaces becomes equal ∵ V1 = V2

Therefore comparing equations of both the spheres ,

                         $q_{1}$ /4π∈$r_{1}$  = $q_{2}$ / 4π∈$r_{2}$

                                               ⇒ $\frac{q_{1} }{q_{2} }$ =$\frac{r_{1} }{r_{2} }$

                                                ⇒ $\frac{r_{1} }{r_{2} }$ = $\frac{21}{32}$

                                                        =0.65

Answer : Two metallic spheres of radii r1 and r2 are connected by a wire . If q1 is +21and q2 is +32 . Then the ratio of their radii would be 0.65