If the frequency of light in a photoelectric experiment is doubled the stopping potential will be:

becomes more than the double

The correct answer is Option (3) → becomes more than the double

According to Einstein's Photoelectric equation -

$K.E._{max}=hv-\phi$   ...(1)

and, 

$K.E._{max}=eV_s$ [$V_s$ = Stopping potential]   ...(2)

from (1) and (2) we get,

$V_0=\frac{hv-\phi}{e}$

if the frequency of the light is doubled,

$V_0'=\frac{h(2v)-\phi}{e}=2hv-\phi$

$V_0'=2V_0+\frac{\phi}{e}$

Since, $\frac{\phi}{e}$ is constant for a given material, the stopping potential becomes slightly more than double.