If the tension and diameter of a sonometer wi | vedclass

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(c) $n = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}\rho }}} \propto \sqrt {\frac{T}{{{r^2}\rho }}} $ 

==>$\frac{{{n_1}}}{{{n_2}}} = \sqrt {\l

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$n$

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(c) $n = \frac{1}{{2l}}\sqrt {\frac{T}{{\pi {r^2}ho }}} \propto \sqrt {\frac{T}{{{r^2}ho }}} $ 

==>$\frac{{{n_1}}}{{{n_2}}} = \sqrt {\left( {\frac{{{T_1}}}{{{T_2}}}} ight)\,{{\left( {\frac{{{r_2}}}{{{r_1}}}} ight)}^2}\left( {\frac{{{ho _2}}}{{{ho _1}}}} ight)} = \sqrt {\left( {\frac{1}{2}} ight)\,{{\left( {\frac{2}{1}} ight)}^2}\left( {\frac{1}{2}} ight)} = 1$  

$ 	herefore$ ${n_1} = {n_2}$

If the tension and diameter of a sonometer wire of fundamental frequency $n$ are doubled and density is halved then its fundamental frequency will become

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