$n = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}\rho }}} = \frac{1}{l}\sqrt {\frac{T}{{\pi {d^2}\rho }}} $              ($d =$ Diameter of string)

==> $\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}}\sqrt {\frac{{{T_1}}}{{{T_2}}} \times {{\left( {\frac{{{d_2}}}{{{d_1}}}} \right)}^2} \times \left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right)} $

$ = \frac{{35}}{{36}}\sqrt {\frac{8}{1} \times {{\left( {\frac{1}{4}} \right)}^2} \times \frac{2}{1}} = \frac{{35}}{{36}}$$ \Rightarrow {n_2} = \frac{{36}}{{35}} \times 360 = 370$

Hence beat frequency = ${n_2} - {n_1} = 10$